🧠:Okay, so I need to solve this problem where the sum of two numbers is 432, and their quotient is 5. Hmm, let me think about how to approach this. Alright, let me start by understanding what the question is asking. There are two numbers, let's call them the larger number and the smaller number. Their sum is 432, which means if I add them together, I get 432. The quotient of them is 5, which I think means that when you divide the larger number by the smaller one, you get 5. So, the larger number divided by the smaller number equals 5. That makes sense.Let me write down the equations to represent this. Let's say the larger number is L and the smaller number is S. Then, according to the problem:1. L + S = 4322. L / S = 5So, we have two equations with two variables. That should be solvable using algebraic methods. Let me recall how to solve such systems. Since there are two equations, I can use substitution or elimination. Here, substitution might be easier because the second equation can be rearranged to express one variable in terms of the other.From the second equation, L/S = 5, so if I solve for L, that would be L = 5S. That's straightforward. Now, I can substitute this expression for L into the first equation. So, replacing L in the first equation with 5S:5S + S = 432Combining like terms gives:6S = 432Then, solving for S:S = 432 / 6Calculating that division: 432 divided by 6. Let me do that step by step. 6 goes into 43 seven times (6*7=42), remainder 1. Bring down the 2 to make 12. 6 goes into 12 twice. So, 72. Then the next digit is 2. Wait, no, 432 divided by 6. Let me check again. 6*70=420, so 432-420=12, 12/6=2, so total is 70+2=72. Yes, so S=72.Then, since L = 5S, L = 5*72. Let me compute that. 5*70=350 and 5*2=10, so 350+10=360. Therefore, L=360.Wait, let me verify the numbers. If the larger number is 360 and the smaller is 72, their sum is 360+72=432. That checks out. And their quotient is 360/72=5. That also checks out. So, that seems correct.But let me make sure I didn't make any mistakes in my steps. So, starting from the beginning:Problem states two numbers sum to 432, and their quotient is 5. Assuming that quotient is larger divided by smaller, which makes sense because the quotient is greater than 1 (since 5>1), so the larger number is divided by the smaller. So, equations:L + S = 432L/S = 5Solve for L and S. Express L in terms of S: L=5S. Substitute into sum equation: 5S + S =6S=432. Then S=432/6=72. Then L=5*72=360. Check sum:72+360=432. Check quotient:360/72=5. All correct.Wait, but what if the quotient was meant to be S/L? That would mean the quotient is 1/5, but the problem states the quotient is 5. So, the problem says "the quotient is 5", which usually means numerator divided by denominator. So, unless specified otherwise, the first number divided by the second. But in the problem statement, they just say "the quotient is 5". So, the quotient of the two numbers, but it's standard that the order is first mentioned divided by the second? Wait, no. Wait, actually, when they say "the quotient of A and B" it's A divided by B. But here, the problem just says "the quotient is 5". So, the quotient of the two numbers is 5. So, depending on which way you take the numbers, but since they mention "larger number" and "smaller number", the quotient of the larger divided by the smaller is 5. So, that's correct.Alternatively, if someone misinterprets the quotient as smaller divided by larger, they would get 1/5, but the problem says the quotient is 5. Therefore, the correct interpretation is larger divided by smaller equals 5. Therefore, the steps I did are correct.Another way to check: if S=72 and L=360, then S/L=72/360=0.2=1/5, which is the reciprocal. But the problem states the quotient is 5, so we need the larger divided by smaller. So, correct.So, yes, the answer is the larger number is 360 and the smaller is 72. All checks out. Therefore, I think that's the correct solution.Just to recap, steps:1. Assign variables: L for larger number, S for smaller.2. Set up equations based on problem statement.3. Solve one equation for one variable and substitute into the other.4. Solve for S, then find L.5. Verify the solution by plugging back into original equations.All steps seem logical and the verification confirms the answer is correct. Therefore, the solution is accurate.

✅:The sum of two numbers is 432, and their quotient (larger divided by smaller) is 5. Let the larger number be ( L ) and the smaller number ( S ). Equations:1. ( L + S = 432 )2. ( frac{L}{S} = 5 ) → ( L = 5S )Substitute ( L = 5S ) into the sum equation:[5S + S = 432 implies 6S = 432 implies S = frac{432}{6} = 72]Solve for ( L ):[L = 5S = 5 times 72 = 360]Verification:- Sum: ( 360 + 72 = 432 ) ✓- Quotient: ( frac{360}{72} = 5 ) ✓Final Answer: The larger number is 360, and the smaller number is 72.

🔑:Given:- The sum of two numbers is 432.- The quotient (larger number divided by the smaller number) is 5.We need to find both the larger and the smaller number. Let's denote:- The smaller number as ( x ).- The larger number as ( 5x ) (since the ratio of the larger number to the smaller number is 5).We can set up the following equation based on the given information about the sum:[ x + 5x = 432 ]1. Combine like terms: [ 6x = 432 ]2. Solve for ( x ): [ x = frac{432}{6} ] [ x = 72 ]Thus, the smaller number is ( x = 72 ).3. Find the larger number: [ text{Larger number} = 5x = 5 times 72 ] [ text{Larger number} = 360 ] Conclusion:Hence, the larger number is 360 and the smaller number is 72.[boxed{360, 72}]

🧠:Okay, so I need to figure out all positive integers that can't be written as the sum of two fractions: a/b + (a+1)/(b+1), where a and b are positive integers. Hmm, let me start by understanding the problem better.First, let's parse the question. We have two variables, a and b, both positive integers. The expression is (a/b) + (a+1)/(b+1). The task is to find all positive integers that cannot be expressed in this form for any choice of a and b. So, we need to determine the set of positive integers n such that there are no a, b ∈ ℕ⁺ with n = a/b + (a+1)/(b+1).Let me first try to simplify the expression a bit. Maybe combine the two fractions. Let's see:Let me compute the sum:a/b + (a+1)/(b+1) = [a(b+1) + (a+1)b] / [b(b+1)].Expanding the numerator:a(b+1) + (a+1)b = ab + a + ab + b = 2ab + a + b.So the combined fraction is (2ab + a + b) / [b(b + 1)].Therefore, the expression simplifies to:(2ab + a + b) / [b(b + 1)].We need this fraction to be a positive integer. So, for some positive integers a and b, the numerator must be divisible by the denominator. So, the problem reduces to finding all positive integers n such that there do not exist positive integers a and b with:(2ab + a + b) = n * b(b + 1).Let me rearrange that equation:2ab + a + b = n b (b + 1).Let me factor out a from the terms involving a:a(2b + 1) + b = n b (b + 1).Then, solving for a:a(2b + 1) = n b (b + 1) - b.Therefore,a = [n b (b + 1) - b] / (2b + 1).Since a must be a positive integer, the right-hand side must be a positive integer. Therefore, the numerator must be divisible by (2b + 1), and the result must be a positive integer.So, for each positive integer b, we can check if [n b (b + 1) - b] is divisible by (2b + 1). If for some b, this division results in a positive integer a, then n can be expressed in the desired form. If for all b, either the division doesn't result in an integer or a would not be positive, then n cannot be expressed in that form.Therefore, to find the impossible n's, we need to ensure that for all b ∈ ℕ⁺, [n b (b + 1) - b] is not divisible by (2b + 1) or the result is not positive.Alternatively, maybe we can approach this problem by considering the divisibility condition. Let me define:Let’s denote the numerator as N = n b (b + 1) - b.We need N ≡ 0 mod (2b + 1). So:n b (b + 1) - b ≡ 0 mod (2b + 1).Let’s factor out b:b [n (b + 1) - 1] ≡ 0 mod (2b + 1).Therefore, either 2b + 1 divides b, which is impossible because 2b + 1 > b for positive b, or 2b + 1 divides [n (b + 1) - 1]. So:n (b + 1) - 1 ≡ 0 mod (2b + 1).So, n (b + 1) ≡ 1 mod (2b + 1).Let’s write this congruence:n (b + 1) ≡ 1 mod (2b + 1).Let me try to manipulate this congruence. Let me note that 2b + 1 is the modulus. Let's express b in terms of the modulus. Let’s denote m = 2b + 1, so that b = (m - 1)/2. Since m must be an odd integer greater than or equal to 3 (since b is a positive integer, m = 2b + 1 ≥ 3). So m is an odd integer ≥3.Substituting b = (m - 1)/2 into the congruence:n [( (m - 1)/2 + 1 ) ] ≡ 1 mod m.Simplify the expression inside the brackets:(m - 1)/2 + 1 = (m - 1 + 2)/2 = (m + 1)/2.Thus, the congruence becomes:n * (m + 1)/2 ≡ 1 mod m.Multiply both sides by 2:n (m + 1) ≡ 2 mod m.But m + 1 ≡ 1 mod m, since m + 1 = m + 1, so modulo m, that's 1. Therefore:n * 1 ≡ 2 mod m ⇒ n ≡ 2 mod m.Wait, but m = 2b + 1, which is at least 3. Therefore, n ≡ 2 mod m, which is equivalent to saying that m divides (n - 2). However, m = 2b + 1, which is an odd integer ≥3. Therefore, for each m (odd integer ≥3) that divides (n - 2), we can set m = 2b + 1, which gives b = (m - 1)/2, which must be a positive integer. Therefore, m must be an odd divisor of (n - 2) greater than or equal to 3. Then, such m must satisfy that (m - 1)/2 is a positive integer, which it is since m is odd.Therefore, the conclusion is that for n to be expressible as the given sum, there must exist an odd integer m ≥3 that divides (n - 2), and then we can set b = (m -1)/2. Then, a can be computed as:a = [n b (b +1 ) - b]/(2b +1 )But since m divides (n - 2), then n - 2 = k m for some integer k ≥1. Therefore, n = k m + 2.But m = 2b +1, so substituting:n = k (2b +1 ) + 2.Wait, perhaps this is getting a bit tangled. Let me recap.We started with the congruence n ≡ 2 mod m, where m = 2b +1. Therefore, m divides (n - 2). So, if n - 2 has any odd divisors ≥3, then such m exists, and hence b = (m -1)/2 is a positive integer, and thus a can be calculated. Therefore, n can be expressed in the desired form if and only if n - 2 has an odd divisor ≥3.Therefore, the values of n which cannot be expressed are those where n - 2 does not have any odd divisors ≥3. Which would occur when n - 2 is a power of 2. Because the only positive integers with no odd divisors ≥3 are the powers of 2. Since any other number would have at least one odd prime factor, which would be ≥3.Wait, let's check that. Suppose n - 2 is a power of 2, say 2^k. Then, its only divisors are 1, 2, 4, ..., 2^k. So, it has no odd divisors except 1. Therefore, m, which is an odd divisor ≥3 of n - 2, does not exist. Hence, for such n, there are no m ≥3 odd dividing n - 2, hence no possible b, hence n cannot be expressed in the desired form.Conversely, if n - 2 is not a power of 2, then it has at least one odd prime factor, which is ≥3, hence m exists, and so n can be expressed.Therefore, the positive integers n that cannot be expressed are those where n - 2 is a power of 2, i.e., n = 2 + 2^k for some non-negative integer k. However, since n must be a positive integer, and 2^k is positive, so k can be 0,1,2,... But n = 2 + 2^k. Let's check:If k=0: n=2+1=3k=1: 2+2=4k=2: 2+4=6k=3: 2+8=10k=4: 2+16=18and so on. So the numbers are 3,4,6,10,18,... etc., each being 2 more than a power of 2.But wait, let me check this conclusion again. If n - 2 is a power of 2, then n cannot be expressed. Otherwise, it can. Therefore, the forbidden numbers are n = 2 + 2^k, where k ≥0. But we need to confirm with examples.Let’s test for n=3. Is 3 expressible as a/b + (a+1)/(b+1)?If n=3, then n-2=1=2^0. So according to our conclusion, 3 cannot be expressed. Let's check.Suppose 3 = a/b + (a+1)/(b+1). Let's try small b.For b=1:3 = a/1 + (a+1)/2 => a + (a+1)/2 = 3 => 2a + a +1 = 6 => 3a =5 => a=5/3. Not integer.For b=2:3 = a/2 + (a+1)/3. Multiply by 6: 3a + 2(a +1) = 18 => 3a +2a +2=18 =>5a=16 =>a=16/5. Not integer.b=3:3 = a/3 + (a+1)/4. Multiply by 12: 4a + 3(a +1) =36 =>7a +3=36 =>7a=33 =>a=33/7. Not integer.b=4:3 = a/4 + (a+1)/5. Multiply by 20:5a +4(a +1)=60 =>9a +4=60 =>9a=56 =>a=56/9. Not integer.Hmm, seems like no solution. So n=3 is indeed not expressible.Similarly, check n=4. n-2=2=2^1. So according to the conclusion, n=4 cannot be expressed.Check n=4:4 = a/b + (a+1)/(b+1). Let's try b=1:4 = a + (a+1)/2 => 2a +a +1=8 =>3a=7 =>a=7/3. Not integer.b=2:4 = a/2 + (a+1)/3. Multiply by 6: 3a + 2(a +1)=24 =>5a +2=24 =>5a=22 =>a=22/5. Not integer.b=3:4 = a/3 + (a+1)/4. Multiply by 12: 4a + 3(a +1)=48 =>7a +3=48 =>7a=45 =>a=45/7. Not integer.b=4:4 = a/4 + (a+1)/5. Multiply by 20:5a +4(a +1)=80 =>9a +4=80 =>9a=76 =>a=76/9. Not integer.Again, no solutions. So n=4 is also not expressible.Now check n=5, which is 2 + 3, where 3 is not a power of 2. So according to our conclusion, n=5 should be expressible.Let’s try to find a and b such that 5 = a/b + (a+1)/(b+1).Start with b=1:5 = a + (a+1)/2 => 2a +a +1 =10 =>3a=9 =>a=3. So a=3, b=1. Check: 3/1 +4/2=3+2=5. Yes! So n=5 is expressible.Similarly, n=6, which is 2 +4=2+2^2, so n=6 should not be expressible.Check n=6:6 = a/b + (a+1)/(b+1). Let's try b=1:6 =a + (a+1)/2 =>2a +a +1=12 =>3a=11 =>a=11/3. Not integer.b=2:6 =a/2 + (a+1)/3. Multiply by6: 3a +2(a +1)=36 =>5a +2=36 =>5a=34 =>a=34/5. Not integer.b=3:6 =a/3 + (a+1)/4. Multiply by12:4a +3(a +1)=72 =>7a +3=72 =>7a=69 =>a=69/7. Not integer.b=4:6 =a/4 + (a+1)/5. Multiply by20:5a +4(a +1)=120 =>9a +4=120 =>9a=116 =>a=116/9. Not integer.Hmm, no solution so far. Let's try b=5:6 =a/5 + (a+1)/6. Multiply by30:6a +5(a +1)=180 =>11a +5=180 =>11a=175 =>a=175/11≈15.9. Not integer.b=6:6 =a/6 + (a+1)/7. Multiply by42:7a +6(a +1)=252 =>13a +6=252 =>13a=246 =>a=246/13≈18.9. Not integer.So maybe n=6 is indeed not expressible.Now check n=7, which is 2 +5. Since 5 is not a power of 2, n=7 should be expressible.Let’s find a and b:Trying b=2:7 =a/2 + (a+1)/3. Multiply by6:3a +2(a +1)=42 =>5a +2=42 =>5a=40 =>a=8.So a=8, b=2. Check:8/2 +9/3=4 +3=7. Correct. So n=7 is expressible.Thus, the pattern seems to hold. Therefore, the conclusion is that all positive integers n such that n-2 is a power of 2 cannot be expressed in the given form, and all others can. Therefore, the numbers that cannot be expressed are n = 2 + 2^k where k is a non-negative integer.But we need to confirm if k=0 is allowed. If k=0, then n=2 +1=3, which we saw cannot be expressed. Similarly, k=1 gives n=4, which is also non-expressible, and so on. However, the problem states that n should be a positive integer. Since k is a non-negative integer, the smallest n in this form is 3. Wait, but what about n=1 and n=2?Wait, let's check n=1 and n=2.For n=1: Can we have 1 = a/b + (a+1)/(b+1)? Let's check possible b.b=1:1 = a + (a+1)/2 =>2a +a +1=2 =>3a=1 =>a=1/3. Not integer.b=2:1 =a/2 + (a+1)/3. Multiply by6:3a +2(a +1)=6 =>5a +2=6 =>5a=4 =>a=4/5. Not integer.Similarly, larger b would lead to smaller fractions, so a would have to be even smaller, which isn't possible. So n=1 cannot be expressed.Wait, but according to our previous conclusion, n=1-2=-1, which is not a power of 2, so n=1 should be expressible. But it's not. Therefore, our conclusion is missing something.Wait, this suggests an error in the reasoning. Because n=1 is a positive integer, and according to the earlier logic, since n -2 = -1, which is not a power of 2, n=1 should be expressible. But in reality, n=1 cannot be expressed.Similarly, check n=2:n=2. Then n -2=0, which is not a power of 2 (unless we consider 0=2^(-∞)), but 0 is not a valid power of 2 in this context. Therefore, according to the logic, since n -2=0, which doesn't have any divisors, so there's no m≥3 dividing 0. But technically, every integer divides 0. So m can be any odd integer ≥3. Then, m divides 0, so there exists m, hence n=2 should be expressible.Wait, but m divides n -2=0, so any m would work. Let's see. For m=3 (smallest odd ≥3), then b=(m -1)/2=(3-1)/2=1. Then a = [n b(b+1) -b]/(2b +1) = [2*1*2 -1]/3= (4 -1)/3=1. So a=1, b=1.Check:1/1 +2/2=1+1=2. Yes! So n=2 can be expressed.But n=1: For n=1, n -2=-1. The earlier logic doesn't hold because n=1 is less than 2, so n -2 is negative, but the modulus m must be a positive integer ≥3. So, does our previous reasoning apply only for n ≥2? It seems so.Therefore, the conclusion is that n=1 cannot be expressed, but n=2 can. Let's see why n=1 cannot be expressed.Trying to find a and b for n=1:1 = a/b + (a+1)/(b+1). Let's attempt with b=1:1 =a + (a+1)/2. Multiply by2: 2a +a +1=2 =>3a=1 =>a=1/3. Not integer.b=2:1 =a/2 + (a+1)/3. Multiply by6:3a +2(a +1)=6 =>5a +2=6 =>5a=4 =>a=4/5. Not integer.Similarly, higher b would lead to smaller a, which would be fractions. So no solutions. Therefore, n=1 cannot be expressed.But according to our initial logic, since n -2=-1, which is not a power of 2, n=1 should be expressible. The problem arises because when n -2 is negative, the earlier reasoning about divisors doesn't hold. Therefore, we need to adjust our conclusion.Thus, our earlier conclusion that n can be expressed if and only if n -2 is not a power of 2 holds only for n ≥3. But for n=1 and n=2, we need to handle separately.Wait, n=2 can be expressed as shown before (a=1, b=1). So n=2 is okay. n=1 cannot be expressed. Therefore, the forbidden numbers are n=1 and n=2 +2^k for k ≥0. But n=1 is not of the form 2 +2^k. So the complete list of non-expressible positive integers is n=1 and n=2 +2^k for k ≥0.Wait, but 2 +2^0=2+1=3, which is already covered. So n=1 is an exception. Therefore, the answer should include n=1 and all numbers of the form 2+2^k for k≥0. However, let's check n=1:If we include n=1, then the complete set is {1, 3,4,6,10,18,...}. But wait, n=1 is a separate case. So why does n=1 not fit into the previous pattern? Because for n=1, n -2=-1, which is negative, so m would have to be a divisor of -1. But m is a positive odd integer ≥3, so there's no such m. Therefore, for n=1, there's no m, hence no solution. Therefore, n=1 also cannot be expressed, but it's not of the form 2 +2^k. Hence, the forbidden numbers are n=1 and n=2 +2^k for k ≥0.But let's confirm if n=2 +2^k includes n=3,4,6,10,18, etc., and n=1 is separate. Therefore, the positive integers that cannot be expressed are 1 and all numbers of the form 2 +2^k where k ≥0. But let's check k=0:2+1=3, k=1:4, k=2:6, k=3:10, etc.But wait, n=1 is separate. Therefore, the answer is all positive integers equal to 1 or of the form 2 +2^k for some integer k ≥0.But let's check whether 2 +2^k for k ≥0 gives all numbers 3,4,6,10,18,... and 1 is separate. Hence, the set is {1} ∪ {2 +2^k | k ≥0}. However, 2 +2^0 =3, 2 +2^1=4, 2 +2^2=6, etc. So yes, that's correct.However, when the problem says "positive integers that cannot be expressed," we need to list all such n. Therefore, n=1,3,4,6,10,18,... are the numbers that cannot be expressed. But we need to verify if n=1 is indeed not covered by the other condition. Since n=1 cannot be written as 2 +2^k for any k≥0 (since 2 +2^k ≥3 for k≥0), it's an additional case.But why is n=1 not covered by the previous analysis? Because when n=1, the equation n= a/b + (a+1)/(b+1) would require a and b such that 1 = a/b + (a+1)/(b+1). However, as shown earlier, this leads to fractional a's, which are not allowed. Therefore, n=1 is indeed a separate case.But wait, in our initial analysis, we transformed the problem into n ≡2 mod m where m=2b+1 divides n-2. For n=1, n-2=-1, so m divides -1, which would imply m=1, but m must be at least 3. Therefore, no solution exists. Hence, n=1 cannot be expressed. So the conclusion is that n cannot be expressed if either n=1 or n-2 is a power of 2. Thus, the set is {1} ∪ {2 +2^k | k ≥0}.But let's confirm if 2 +2^k includes all numbers from 3 upwards that are 2 more than a power of 2. Yes. So the numbers that cannot be expressed are 1 and numbers of the form 3,4,6,10,18,34,... etc.But the problem asks for positive integers. So the answer is all positive integers n such that n=1 or n=2 +2^k for some non-negative integer k.But wait, let's check if n=2 is allowed. For n=2, as we saw, a=1, b=1 gives 1/1 +2/2=1+1=2. So n=2 can be expressed. So n=2 is okay.Therefore, the conclusion is that the numbers that cannot be expressed are n=1 and those where n=2 +2^k for k≥0. So the final answer is all positive integers of the form 2 +2^k (which gives 3,4,6,10,18,...) and n=1.But let's check for n=5, which is 2 +3. Since 3 is not a power of 2, n=5 can be expressed, which we saw earlier. Similarly, n=7=2+5 can be expressed. So the reasoning holds.Therefore, the positive integers that cannot be expressed in the given form are 1 and all integers of the form 2 +2^k where k is a non-negative integer. Therefore, the solution is all positive integers n such that n=1 or n=2+2^k for some k≥0.But the problem states "positive integers that cannot be expressed," so we need to write them explicitly. However, since there are infinitely many such numbers, we need to characterize them. Therefore, the answer is all positive integers equal to 1 or one more than a power of 2 (since 2+2^k = 2(1 + 2^{k-1}) when k≥1, but actually, 2+2^k is just 2 +2^k). Wait, no, that's not accurate. 2 +2^k is the form. So for k=0, it's 3; k=1, 4; k=2,6; k=3,10; etc.Thus, the final answer is that all positive integers that cannot be expressed in the form a/b + (a+1)/(b+1) are 1 and those equal to 2 more than a power of 2. Hence, n=1 and n=2+2^k for non-negative integers k.But wait, the problem is from a mathematical competition or similar, and typically, such answers require the set {1} ∪ {2 + 2^k | k ≥0}. However, to present this properly, we can write it as all positive integers n where n is 1 or n-2 is a power of 2. Therefore, n ∈ {1} ∪ {2 + 2^k | k ∈ ℕ}.But let's check once more with n=5. n=5-2=3, which is not a power of 2, so n=5 can be expressed. Correct. n=7-2=5, not a power of 2, so expressible. n=9-2=7, not a power of 2, so expressible. n=10-2=8=2^3, which is a power of 2, so n=10 cannot be expressed. Correct, as seen earlier.Therefore, the final answer is that the positive integers which cannot be expressed are 1 and those integers two more than a power of two. Hence, the solution is all positive integers of the form ( 2 + 2^k ) for some non-negative integer ( k ), and the integer 1.Wait, but 2 +2^0 =3, which is covered, but 1 is separate. So the complete set is {1} ∪ {2 + 2^k | k ≥0}.But the problem asks for positive integers, and 1 is a positive integer. So the answer is: All positive integers that are either equal to 1 or can be written as ( 2 + 2^k ) where ( k ) is a non-negative integer.However, in mathematical terms, this is equivalent to saying all positive integers ( n ) such that ( n = 1 ) or ( n - 2 ) is a power of 2. Therefore, the numbers that cannot be expressed are 1 and the numbers two more than a power of two.Hence, the final answer is all positive integers ( n ) where ( n = 1 ) or ( n = 2 + 2^k ) for some integer ( k geq 0 ).Final AnswerThe positive integers that cannot be expressed in the given form are exactly those equal to (1) or (2 + 2^k) for some non-negative integer (k). Thus, the solution is (boxed{1}) and all numbers of the form (boxed{2 + 2^k}) where (k) is a non-negative integer.However, since the problem asks for all such positive integers, and typically in such contexts answers are presented as a set, the final answer can be expressed as:All positive integers that cannot be expressed are ( boxed{1} ) and those of the form ( boxed{2 + 2^k} ) for non-negative integers ( k ).But to conform to the required format (a single boxed answer), we can list the numbers as:The numbers are (1) and (2 + 2^k) for (k geq 0), so the final boxed answer is:boxed{1} and boxed{2 + 2^k} for non-negative integers (k).However, if the problem expects a union of all such numbers, since they are infinitely many, we characterize them as:All positive integers that cannot be expressed are (boxed{1}) and those of the form (boxed{2 + 2^k}) where (k) is a non-negative integer.But the standard way to present such answers in Olympiad problems is to state the characterization. Hence:The positive integers that cannot be expressed are exactly those equal to (1) or (2) more than a power of (2). Therefore, the answer is (boxed{1}) and all integers of the form (boxed{2 + 2^k}) for some non-negative integer (k).

✅:Given ( a, b in mathbb{N}^{+} ), we need to find all positive integers that cannot be expressed in the form ( frac{a}{b} + frac{a+1}{b+1} ).First, we simplify the given expression:[frac{a}{b} + frac{a+1}{b+1} = frac{a(b+1) + (a+1)b}{b(b+1)} = frac{2ab + a + b}{b(b+1)}]We need this fraction to be a positive integer ( n ). Therefore, we require:[2ab + a + b = n b(b+1)]Rearranging and solving for ( a ):[a = frac{n b(b+1) - b}{2b + 1}]For ( a ) to be a positive integer, the numerator ( n b(b+1) - b ) must be divisible by ( 2b + 1 ). This leads to the congruence:[n (b + 1) equiv 1 pmod{2b + 1}]Let ( m = 2b + 1 ), then ( b = frac{m - 1}{2} ). Substituting ( b ) into the congruence, we find:[n equiv 2 pmod{m}]This implies ( m ) (an odd integer ( geq 3 )) must divide ( n - 2 ). If ( n - 2 ) is a power of 2, it has no odd divisors ( geq 3 ), making ( n ) inexpressible. Additionally, ( n = 1 ) cannot be expressed as shown by direct verification.Thus, the positive integers that cannot be expressed are ( 1 ) and those of the form ( 2 + 2^k ) for non-negative integers ( k ).[boxed{1} text{ and all integers of the form } boxed{2 + 2^k} text{ for non-negative integers } k]

🔑:1. Consider the expression (frac{a}{b}+frac{a+1}{b+1}). To express it as a single fraction, we combine the terms: [ frac{a}{b} + frac{a+1}{b+1} = frac{a(b+1) + b(a+1)}{b(b+1)} = frac{a b + a + b a + b}{b(b+1)} = frac{2ab + a + b}{b(b+1)} ]2. Since the result is an integer ( n ), we can denote it as: [ n = frac{2ab + a + b}{b(b+1)} ]3. This implies that ( b ) must divide ( 2ab + a + b ). Rearranging the condition, we get: [ b mid 2ab + a + b Rightarrow b mid a(2b+1) ]4. Since ( b ) and ( b+1 ) are coprime, their only common divisor is 1. Thus, ( b mid a ), and we can write ( a = pb ) for some integer ( p ): [ frac{a}{b} = p Rightarrow frac{a + 1}{b + 1} = n - p ]5. We now substitute ( a = pb ) into the second fraction: [ frac{pb + 1}{b + 1} = n - p ] This simplifies: [ pb + 1 = (n - p)(b + 1) ]6. Expanding and rearranging: [ pb + 1 = nb + n - pb - p Rightarrow 2pb + 1 = nb + n - p ] [ 1 = nb + n - 2pb - p Rightarrow b + 1 mid p - 1 Rightarrow p = q(b + 1) + 1 ]7. We substitute ( p = q(b+1) + 1 ) back into the expression for ( n ): [ n = 2p - frac{p-1}{b+1} ] [ n = 2(q(b+1) + 1) - q = (2b+1)q + 2 ]8. From the equation ( n = (2b+1)q + 2 ), we can deduce: [ n geq 2 ] Moreover, ( n-2 ) is a multiple of some odd number greater than 1.9. If ( n geq 2 ) and ( n-2 ) has an odd factor greater than 1, setting ( n = (2b+1)q + 2 ): [ text{Choose } a = q b(b+1)+b Rightarrow a in mathbb{N}^{+} ] Substituting back: [ frac{a}{b} + frac{a+1}{b+1} = (2b+1)q + 2 = n ]10. Therefore, any integer ( n geq 2 ) that can be written in the form ( (2b+1)q + 2 ) where ( b ) and ( q ) are positive integers can be expressed in the form (frac{a}{b}+frac{a+1}{b+1}).11. Conclusion: The numbers that cannot be expressed in the form (frac{a}{b}+frac{a+1}{b+1}) are 1 and numbers of the form ( 2^{m} + 2 ) where ( m in mathbb{N} ).[boxed{1 text{ and the numbers of the form } 2^{m} + 2 text{ (m in mathbb{N})}}]

🧠:Okay, let's try to figure out this problem step by step. So, we have a team of lumberjacks cutting trees in two plots: a large one and a small one. The area of the small plot is three times less than the large one, which I think means the large plot is three times as big as the small one. So, if the small plot is S, the large one is 3S. Wait, maybe. Let me confirm that. If the small is 3 times less than the large, then the large is 3 times the small. Yeah, that makes sense. So area of large plot = 3 * area of small plot.Now, the team is divided into two parts. The part working on the large plot has 8 more lumberjacks than the part working on the small plot. Let me denote the number of lumberjacks on the small plot as x. Then the number on the large plot would be x + 8. So total team size is x + (x + 8) = 2x + 8. But we need to find the maximum possible number of lumberjacks in the team. Hmm, so there must be some constraints here.The key point is that when the tree cutting on the large plot was finished, the part of the team working on the small plot was still working. So, the time taken to finish the large plot is less than the time needed to finish the small plot with their respective teams. Wait, but actually, the large plot was finished first, so the time taken by the large plot team is equal to the time that had passed when they finished, and the small plot team was still working, meaning that their remaining work would take more time. Therefore, the total time for the large plot is less than the total time required for the small plot. Wait, but how do we relate time, number of workers, and area?I think this is a work-rate problem. The idea is that the amount of work (area) divided by the number of workers and their rate gives the time. Assuming all lumberjacks work at the same rate, let's say each lumberjack can cut R area per unit time. Then, the time to finish a plot would be (Area) / (Number of workers * R). Since R is the same for all, we can ignore it and just consider the ratio.So, let's denote:- Area of small plot = S- Area of large plot = 3S (since it's three times larger)- Number of workers on small plot = x- Number of workers on large plot = x + 8Time taken to finish large plot: T_large = 3S / (x + 8)Time taken to finish small plot: T_small = S / xBut according to the problem, when the large plot was finished, the small plot was still being worked on. That means that the time taken to finish the large plot (T_large) is less than the total time needed to finish the small plot (T_small). Wait, but actually, the time they spent working on the large plot is T_large, and in that same time, the small plot team has been working for T_large time, but hasn't finished yet. So the amount of work done on the small plot in time T_large is x * T_large, which is less than S. Therefore:x * T_large < SBut T_large = 3S / (x + 8). Substitute that in:x * (3S / (x + 8)) < SWe can divide both sides by S (assuming S > 0):3x / (x + 8) < 1Multiply both sides by (x + 8):3x < x + 8Subtract x from both sides:2x < 8Divide by 2:x < 4Since x must be a positive integer (number of lumberjacks), the maximum possible value of x is 3. Therefore, the number of workers on the small plot is 3, on the large plot is 3 + 8 = 11, so total team size is 3 + 11 = 14.Wait, but the question asks for the maximum number of lumberjacks. But if x can be 3, giving total 14, but maybe there's a way to have a larger team? Wait, hold on. Let me check my reasoning again.The key inequality was x * (3S / (x + 8)) < S, leading to 3x < x + 8, hence x < 4. So x must be less than 4, so maximum x is 3, leading to total team 14. But maybe there's another way to model this?Wait, perhaps the time is not just the time to complete the large plot, but maybe the total time. Let me think again.When the large plot is finished, the small plot is still being worked on. Therefore, the amount of work done on the small plot up to that time is less than S. The work done on the small plot is (number of workers on small plot) * (time to finish large plot). So:x * T_large < SBut T_large is the time to finish the large plot, which is 3S / (x + 8). Therefore:x * (3S / (x + 8)) < SSimplify as before, leading to x < 4. Hence x_max = 3, leading to total lumberjacks = 14.But is there a way to get a higher number? Maybe if we consider that after the large plot is finished, some workers from the large plot could move to help the small plot? The problem states "the part of the team working on the small plot was still working", which might imply that the original assignments didn't change. So when the large plot is done, the workers there might have stopped, and the small plot team continued. Therefore, the time to finish the large plot is T_large = 3S / (x + 8), and in that time, the small plot team has done x * T_large work, which is less than S. Hence, the same condition as before.So, according to this, x must be less than 4, so x=3 is maximum. Thus, total team is 14.But let me check if there's another interpretation. Maybe "the part of the team working on the small plot was still working" when the large plot was finished. So perhaps the total time taken to finish the large plot is such that the small plot hasn't been finished yet, even if all workers were moved? Wait, no, the problem says "the part of the team working on the small plot was still working", which implies that the team on the small plot didn't change; they were still working with their original number. So the workers from the large plot didn't join them. So the initial analysis holds.Therefore, the maximum number is 14. But wait, maybe I need to check if x can be a non-integer? But since the number of lumberjacks must be integer, x must be integer. So x=3 is the maximum possible. Therefore, total team is 14. Hmm, seems straightforward. Let me verify with x=3:Large plot workers: 11, area=3S. Time to finish: 3S / 11.In that time, small plot workers (3) have cut 3 * (3S / 11) = 9S/11. Which is less than S. Correct.If x=4, then large plot workers=12, time=3S/12=S/4. Small plot workers=4, work done=4*(S/4)=S. But that would mean they finished at the same time, which contradicts the condition that the small plot was still working. Therefore, x must be less than 4.Therefore, x=3 is the maximum, leading to total 14. So the answer is 14. Wait, but the question says "the maximum number of lumberjacks that could have been in the team". So 14 is the answer. But let me see if there's a different approach where the answer is higher.Alternatively, maybe the problem allows for the workers to be split in such a way that when the large plot is finished, some of them join the small plot, but the problem states "the part of the team working on the small plot was still working". It doesn't mention anything about the large plot workers joining, so perhaps they just stopped, and only the original small plot team continued. Hence, my initial reasoning holds.Another angle: Suppose the total team size is N. Then, the large plot has L workers, small plot has S workers, with L = S + 8. Total team size N = L + S = 2S + 8. We need to maximize N, which is 2S +8. To maximize N, maximize S. But S is constrained by the work condition: when L finishes the large plot (area=3A), the small plot (area=A) hasn't been finished by S workers.Time for large plot: 3A / L.Work done on small plot in that time: S * (3A / L) < A.Substitute L = S +8:S * (3A / (S +8)) < ACancel A:3S / (S +8) < 1Multiply both sides by (S +8):3S < S +82S <8S<4Thus, S can be at most 3. Therefore, maximum N=2*3 +8=14.Yes, same result. Therefore, the answer is 14.

✅:To determine the maximum number of lumberjacks in the team, let's analyze the problem systematically:1. Define Variables: - Let ( S ) be the area of the small plot. The large plot has an area ( 3S ). - Let ( x ) be the number of lumberjacks on the small plot. The large plot has ( x + 8 ) lumberjacks. - Total team size: ( x + (x + 8) = 2x + 8 ).2. Work-Rate Relationship: - Time to finish the large plot: ( frac{3S}{x + 8} ). - Work done on the small plot in this time: ( x cdot frac{3S}{x + 8} ).3. Key Constraint: The small plot is not yet finished when the large plot is completed: [ x cdot frac{3S}{x + 8} < S. ] Simplifying: [ frac{3x}{x + 8} < 1 implies 3x < x + 8 implies 2x < 8 implies x < 4. ] Since ( x ) must be a positive integer, the maximum ( x ) is ( 3 ).4. Calculate Total Team Size: - Small plot: ( x = 3 ) lumberjacks. - Large plot: ( 3 + 8 = 11 ) lumberjacks. - Total team: ( 3 + 11 = 14 ).Final Answer: The maximum number of lumberjacks in the team is (boxed{14}).

🔑:We are given that a team of lumberjacks is working on both a large and a small plot of land. The area of the small plot is three times less than that of the large plot. Additionally, there are 8 more lumberjacks working on the large plot than on the small plot. We aim to determine the maximum number of lumberjacks that can be in the team.1. Define variables: - Let ( n ) be the number of lumberjacks working on the small plot. - Then, the number of lumberjacks on the large plot is ( n + 8 ). - The total number of lumberjacks in the team is ( 2n + 8 ).2. Assume equal productivity: - Let ( a ) be the productivity per lumberjack. - Productivity of the team on the small plot: ( a cdot n ). - Productivity of the team on the large plot: ( a cdot (n + 8) ).3. Relate areas and productivity: - Let ( S ) be the area of the small plot. - Thus, the area of the large plot is ( 3S ).4. Calculate time: - Time to complete the small plot: [ frac{S}{a cdot n} ] - Time to complete the large plot: [ frac{3S}{a cdot (n + 8)} ] Since the productivity is given to be such that the task on the large plot finishes first, the condition is: [ frac{S}{a cdot n} > frac{3S}{a cdot (n + 8)} ]5. Simplify the inequality: - Removing the positive terms ( S ) and ( a ), we get: [ frac{1}{n} > frac{3}{n + 8} ]6. Cross-multiply to eliminate the fractions: [ (n + 8) > 3n ]7. Solve for ( n ): [ n + 8 > 3n implies 8 > 2n implies n < 4 ] Therefore, the maximum integer value for ( n ) is ( 3 ).8. Calculate maximum total number of lumberjacks: - Substituting ( n = 3 ) into ( 2n + 8 ): [ 2(3) + 8 = 6 + 8 = 14 ]Thus, the largest number of lumberjacks in the brigade is:[boxed{14}]

🧠:Okay, let's see. I need to solve this system of equations for real numbers a, b, c, d. The equations are:1. a + c = -42. ac + b + d = 63. ad + bc = -54. bd = 2Hmm, so four equations with four variables. That seems solvable, but it's nonlinear because of the products like ac, ad, bc, and bd. Nonlinear systems can be tricky, but maybe there's a way to simplify them by substitution or expressing variables in terms of others.Starting with the first equation: a + c = -4. Maybe I can express c in terms of a. Let's do that. So, c = -4 - a. That way, I can substitute c into the other equations. That should reduce the number of variables.Now, moving to the second equation: ac + b + d = 6. Since we know c = -4 - a, substitute that in:a*(-4 - a) + b + d = 6.Let's compute that:-4a - a² + b + d = 6.So, equation 2 becomes: -a² -4a + b + d = 6. Let's note that as equation 2'.Equation 3: ad + bc = -5. Again, substitute c = -4 -a:a*d + b*(-4 -a) = -5.So, that's ad -4b -ab = -5. Let's write that as equation 3': ad - ab -4b = -5.Equation 4 is b*d = 2. So, that's straightforward.So now, our variables are a, b, d. Let's see if we can express some variables in terms of others.From equation 4: b*d = 2. Maybe we can express d in terms of b: d = 2/b, provided b ≠ 0. Alternatively, express b in terms of d: b = 2/d, provided d ≠ 0. Maybe that substitution can help. Let's try substituting d into equations 2' and 3'.Let me try substituting d = 2/b into equations 2' and 3'.First, equation 2': -a² -4a + b + d = 6. Substitute d:-a² -4a + b + (2/b) = 6.Hmm, that introduces a term with 1/b. Maybe complicating things. Similarly, equation 3': ad -ab -4b = -5. Substitute d = 2/b:a*(2/b) - a*b -4b = -5.Which is (2a)/b - ab -4b = -5. Also introduces terms with 1/b. Maybe messy. Alternatively, perhaps substituting b = 2/d into equations 2' and 3'.Equation 2': -a² -4a + (2/d) + d = 6.Equation 3': a*d - a*(2/d) -4*(2/d) = -5.Hmm, not sure if that's better. Let's see:Equation 2': -a² -4a + (2/d + d) = 6.Note that 2/d + d can be written as (d² + 2)/d. Not sure. Maybe if we let y = d + 2/d? Wait, perhaps not. Alternatively, maybe equation 2' can be rearranged.Alternatively, maybe express both b and d in terms of another variable. Let's think.Alternatively, maybe combine equations 2 and 3. Let's see:From equation 2: ac + b + d = 6.From equation 3: ad + bc = -5.But since we know c = -4 -a, substitute that into both.Equation 2: a*(-4 -a) + b + d = 6 => -4a -a² + b + d = 6.Equation 3: a*d + b*(-4 -a) = -5 => a*d -4b -a*b = -5.So, in equation 3, terms with a and b. Maybe we can solve for one variable in terms of another.Alternatively, let's think about variables. We have a, b, d. Let me see if we can express equations in terms of a and b or a and d.Wait, equation 2' is: -a² -4a + b + d = 6.Equation 3' is: a*d -a*b -4b = -5.Equation 4: b*d = 2.So, maybe let me consider equations 2' and 3' with variables a, b, d. Let's try to express d from equation 2' and substitute into equation 3'.From equation 2': d = 6 + a² +4a - b.So, substitute this into equation 3':a*d -a*b -4b = -5.Plugging in d:a*(6 + a² +4a - b) -a*b -4b = -5.Let's compute that:6a + a³ +4a² -a*b -a*b -4b = -5.Simplify terms:6a + a³ +4a² -2a*b -4b = -5.Hmm, so equation 3' becomes:a³ +4a² +6a -2a*b -4b = -5.Hmm, not sure if that's helpful. Maybe we can factor terms with b:a³ +4a² +6a - b*(2a +4) = -5.So, rearranged:-b*(2a +4) = -5 -a³ -4a² -6a.Multiply both sides by -1:b*(2a +4) = 5 +a³ +4a² +6a.Thus,b = (5 +a³ +4a² +6a)/(2a +4).Simplify numerator: Let's see if numerator factors.Numerator: a³ +4a² +6a +5.Let me try to factor that. Maybe synthetic division.Try a = -1: (-1)^3 +4*(-1)^2 +6*(-1) +5 = -1 +4 -6 +5 = 2 ≠ 0.a = -5: (-5)^3 +4*(-5)^2 +6*(-5) +5 = -125 +100 -30 +5 = -50 ≠ 0.a = 1: 1 +4 +6 +5 = 16 ≠ 0.Hmm, maybe it doesn't factor nicely. Maybe we can perform polynomial division of numerator by denominator.Wait, denominator here is 2a +4. Let's factor out 2: 2(a +2). So denominator is 2(a +2). The numerator is a³ +4a² +6a +5.Let me try to divide numerator by (a +2). Let's use polynomial division.Divide a³ +4a² +6a +5 by (a +2):First term: a³ / a = a². Multiply (a +2) by a²: a³ +2a².Subtract from numerator: (a³ +4a² +6a +5) - (a³ +2a²) = 2a² +6a +5.Next term: 2a² / a = 2a. Multiply (a +2) by 2a: 2a² +4a.Subtract: (2a² +6a +5) - (2a² +4a) = 2a +5.Next term: 2a / a = 2. Multiply (a +2) by 2: 2a +4.Subtract: (2a +5) - (2a +4) = 1.So, the division gives a² +2a +2 with a remainder of 1. Therefore,Numerator = (a +2)(a² +2a +2) +1.So, b = [ (a +2)(a² +2a +2) +1 ] / [2(a +2) ]Which can be written as:b = [ (a +2)(a² +2a +2) ] / [2(a +2) ] + 1 / [2(a +2) ]Simplify first term: (a² +2a +2)/2, provided a +2 ≠ 0.So, b = (a² +2a +2)/2 + 1/[2(a +2)]Combine terms over a common denominator:= [ (a² +2a +2)(a +2) +1 ] / [2(a +2) ]Wait, but that seems to bring us back to the original expression. Maybe not helpful. Alternatively, keep it as:b = (a² +2a +2)/2 + 1/(2(a +2)).Hmm, not sure if that's helpful. Alternatively, perhaps leave b expressed in terms of a as:b = (a³ +4a² +6a +5)/(2a +4)But maybe we can use this expression for b in equation 4. Remember, equation 4 is b*d = 2, and from equation 2', d = 6 +a² +4a -b.So, if we can express d in terms of a and b, and then substitute b from above, perhaps we can get an equation in terms of a.Alternatively, since b is expressed in terms of a, and d is expressed in terms of a and b, maybe substitute b into d, then substitute both into equation 4.Let me attempt that.From equation 2', d = 6 +a² +4a -b.But from above, b = (a³ +4a² +6a +5)/(2a +4).So, substitute this into d:d = 6 +a² +4a - (a³ +4a² +6a +5)/(2a +4).Let's compute this:First, express 6 +a² +4a as (6 +a² +4a). Let's combine terms:So,d = [ (6 +a² +4a)(2a +4) - (a³ +4a² +6a +5) ] / (2a +4)Let me compute the numerator:First expand (6 +a² +4a)(2a +4):Multiply term by term:6*2a = 12a6*4 =24a²*2a =2a³a²*4=4a²4a*2a=8a²4a*4=16aSo total:12a +24 +2a³ +4a² +8a² +16aCombine like terms:2a³ + (4a² +8a²) + (12a +16a) +24= 2a³ +12a² +28a +24Then subtract (a³ +4a² +6a +5):Numerator becomes:(2a³ +12a² +28a +24) - (a³ +4a² +6a +5) = a³ +8a² +22a +19So, d = (a³ +8a² +22a +19)/(2a +4)Now, equation 4 is b*d = 2. We have expressions for b and d in terms of a:b = (a³ +4a² +6a +5)/(2a +4)d = (a³ +8a² +22a +19)/(2a +4)Therefore, multiplying b and d:[(a³ +4a² +6a +5)/(2a +4)] * [(a³ +8a² +22a +19)/(2a +4)] = 2Simplify left side:[(a³ +4a² +6a +5)(a³ +8a² +22a +19)] / (2a +4)^2 = 2Multiply both sides by (2a +4)^2:(a³ +4a² +6a +5)(a³ +8a² +22a +19) = 2*(2a +4)^2Let's compute both sides.First, left side: multiply the two cubic polynomials. This will be a 6th degree polynomial. That sounds really complicated, but maybe there's a factor or a substitution.Alternatively, let me check if either of these polynomials can be factored.First polynomial: a³ +4a² +6a +5. Earlier, we tried to factor it by (a +2) but there was a remainder. Let me check again.Using rational root theorem, possible roots are ±1, ±5. Test a = -1:(-1)^3 +4*(-1)^2 +6*(-1) +5 = -1 +4 -6 +5 = 2 ≠ 0. Not a root.a = -5: (-5)^3 +4*(-5)^2 +6*(-5) +5 = -125 +100 -30 +5 = -50 ≠0.So, no rational roots. Maybe irreducible.Second polynomial: a³ +8a² +22a +19.Check for rational roots: ±1, ±19.a = -1: (-1)^3 +8*(-1)^2 +22*(-1) +19 = -1 +8 -22 +19 = 4 ≠0.a = -19: too big, probably not.a = -2: (-8) +32 -44 +19 = (-8 +32) + (-44 +19) = 24 -25 = -1 ≠0.Not obvious. Maybe also irreducible.So, both cubic polynomials may not factor nicely. Multiplying them would be tedious, but let's try.Alternatively, maybe substitute t = a + k for some k to simplify? Not sure.Alternatively, note that 2a +4 = 2(a +2). Let me set t = a +2. Then a = t -2. Maybe substituting a = t -2 into the equation would help. That might make the expressions more manageable.Let me try that substitution. Let a = t -2. Then:First polynomial: (t -2)^3 +4(t -2)^2 +6(t -2) +5Expand (t -2)^3: t³ -6t² +12t -84(t -2)^2: 4(t² -4t +4) =4t² -16t +166(t -2):6t -12Adding all terms with +5:t³ -6t² +12t -8 +4t² -16t +16 +6t -12 +5Combine like terms:t³ + (-6t² +4t²) + (12t -16t +6t) + (-8 +16 -12 +5)= t³ -2t² +2t +1Similarly, the second polynomial: (t -2)^3 +8(t -2)^2 +22(t -2) +19Expand (t -2)^3: t³ -6t² +12t -88(t -2)^2:8(t² -4t +4)=8t² -32t +3222(t -2):22t -44Adding +19:t³ -6t² +12t -8 +8t² -32t +32 +22t -44 +19Combine like terms:t³ + (-6t² +8t²) + (12t -32t +22t) + (-8 +32 -44 +19)= t³ +2t² +2t +(-1)So, t³ +2t² +2t -1Therefore, after substitution a = t -2, the left side becomes:(t³ -2t² +2t +1)(t³ +2t² +2t -1)And the right side is 2*(2(a +2))^2 = 2*(2t)^2 = 2*4t² =8t²So, equation becomes:(t³ -2t² +2t +1)(t³ +2t² +2t -1) =8t²Hmm, still a sixth-degree equation. Let me compute the left-hand side.Let me denote the first polynomial as P(t) = t³ -2t² +2t +1 and Q(t) = t³ +2t² +2t -1.Compute P(t)*Q(t):Multiply term by term:First, t³*(t³ +2t² +2t -1) = t^6 +2t^5 +2t^4 -t³Then, (-2t²)*(t³ +2t² +2t -1) = -2t^5 -4t^4 -4t³ +2t²Then, 2t*(t³ +2t² +2t -1) =2t^4 +4t³ +4t² -2tThen, 1*(t³ +2t² +2t -1) =t³ +2t² +2t -1Now, add all these terms together:t^6 +2t^5 +2t^4 -t³-2t^5 -4t^4 -4t³ +2t²+2t^4 +4t³ +4t² -2t+t³ +2t² +2t -1Now, combine like terms:t^6+ (2t^5 -2t^5) = 0+ (2t^4 -4t^4 +2t^4) =0+ (-t³ -4t³ +4t³ +t³) =0+ (2t² +4t² +2t²) =8t²+ (-2t +2t) =0-1So, P(t)*Q(t) = t^6 +0t^5 +0t^4 +0t³ +8t² +0t -1 = t^6 +8t² -1Therefore, the equation is:t^6 +8t² -1 =8t²Subtract 8t² from both sides:t^6 -1 =0This factors as (t^3 -1)(t^3 +1)=0Further factor:(t -1)(t² +t +1)(t +1)(t² -t +1)=0So, the solutions are t =1, t=-1, and the roots of t² +t +1=0 and t² -t +1=0. However, since we are looking for real solutions, the quadratic equations have discriminant:For t² +t +1: Discriminant =1 -4= -3 <0, so no real roots.For t² -t +1: Discriminant=1 -4= -3 <0, also no real roots.Thus, the real solutions are t=1 and t=-1.So, t=1 and t=-1.Recall that t =a +2, so:When t=1, a +2=1 =>a= -1When t=-1, a +2=-1 =>a= -3Therefore, the possible values of a are -1 and -3.Now, let's find the corresponding values for each case.Case 1: a = -1First, compute c = -4 -a = -4 -(-1)= -4 +1= -3.Then, compute b using the expression from earlier:b = (a³ +4a² +6a +5)/(2a +4)Plugging a = -1:Numerator: (-1)^3 +4*(-1)^2 +6*(-1) +5 = -1 +4 -6 +5 = 2Denominator: 2*(-1) +4 = -2 +4 =2Thus, b=2/2=1.Then, compute d from equation 4: b*d=2 =>1*d=2 =>d=2.Check if these values satisfy all equations.Check equation 2: ac + b + d = (-1)*(-3) +1 +2=3 +1 +2=6, which matches.Equation 3: ad + bc = (-1)*2 +1*(-3)= -2 -3= -5, which matches.All equations satisfied. So (a,b,c,d)=(-1,1,-3,2) is a solution.Case 2: a = -3Compute c = -4 -a = -4 -(-3)= -4 +3= -1Compute b using same expression:b=(a³ +4a² +6a +5)/(2a +4)a=-3:Numerator: (-3)^3 +4*(-3)^2 +6*(-3) +5= -27 +36 -18 +5= (-27 +36) + (-18 +5)=9 -13= -4Denominator:2*(-3) +4= -6 +4= -2Thus, b=(-4)/(-2)=2Then, d from equation 4: b*d=2 =>2*d=2 =>d=1Check equations:Equation 2: ac +b +d= (-3)*(-1) +2 +1=3 +2 +1=6, correct.Equation 3: ad + bc= (-3)*1 +2*(-1)= -3 -2= -5, correct.All equations satisfied. So (a,b,c,d)=(-3,2,-1,1) is another solution.Thus, the solutions are:(a, b, c, d) = (-1, 1, -3, 2) and (-3, 2, -1, 1).But wait, let's check if these are distinct solutions or if they are the same set. Since the variables are ordered, these are two distinct ordered quadruples. But perhaps they are symmetric? Let's see.Indeed, in the first solution, a=-1, c=-3; in the second solution, a=-3, c=-1. Similarly, b and d are swapped: 1 and 2 in the first, 2 and 1 in the second. So they are symmetric solutions where a and c are swapped, and b and d are swapped. Makes sense due to the symmetry in the original equations.Therefore, these are the two real solutions.Final AnswerThe real solutions are boxed{(-1, 1, -3, 2)} and boxed{(-3, 2, -1, 1)}.

✅:Given the system of equations:left{begin{array}{l}a + c = -4 ac + b + d = 6 ad + bc = -5 bd = 2end{array}right.We start by expressing ( c ) in terms of ( a ) from the first equation: ( c = -4 - a ).Substituting ( c = -4 - a ) into the second and third equations, we get:1. ( -a^2 - 4a + b + d = 6 )2. ( ad - ab - 4b = -5 )From the fourth equation ( bd = 2 ), we express ( d ) in terms of ( b ) as ( d = frac{2}{b} ) (assuming ( b neq 0 )).Substituting ( d = frac{2}{b} ) into the modified second and third equations, we simplify and solve for ( b ) and ( d ) in terms of ( a ). After several steps of algebraic manipulation and polynomial division, we find that ( b ) and ( d ) can be expressed as rational functions of ( a ).We substitute these expressions back into the equation ( bd = 2 ) and simplify, leading to a polynomial equation in terms of ( a ). By substituting ( a = t - 2 ), we transform the polynomial equation into a simpler form, which factors into ( (t^3 - 1)(t^3 + 1) = 0 ). Solving for ( t ), we find the real solutions ( t = 1 ) and ( t = -1 ), corresponding to ( a = -1 ) and ( a = -3 ).For ( a = -1 ):- ( c = -3 )- ( b = 1 )- ( d = 2 )For ( a = -3 ):- ( c = -1 )- ( b = 2 )- ( d = 1 )Both solutions satisfy all the original equations. Thus, the real solutions are:boxed{(-1, 1, -3, 2)} quad text{and} quad boxed{(-3, 2, -1, 1)}

🔑:Let's solve the system of equations:[left{begin{array}{l}a + c = -4 ac + b + d = 6 ad + bc = -5 bd = 2end{array}right.]1. Construct Quadratic Polynomials: Let (P(x) = x^2 + ax + b) and (Q(x) = x^2 + cx + d) be two quadratic polynomials where the coefficients are the unknowns in our system.2. Product of Polynomials: We need to form the polynomial whose roots satisfy our original equations. Thus, consider the product (P(x) cdot Q(x)): [ (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd ] Equate the coefficients of the polynomial on both sides: [ x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd = x^4 - 4x^3 + 6x^2 - 5x + 2 ] From this, we get the following system of equations: [ left{ begin{array}{l} a + c = -4 ac + b + d = 6 ad + bc = -5 bd = 2 end{array} right. ]3. Factoring the Polynomial: By trial or polynomial factorization, we can discover if there are any obvious real roots. For (x^4 - 4x^3 + 6x^2 - 5x + 2), the numbers (1) and (-3) serve as roots. Thus, [ x^4 - 4x^3 + 6x^2 - 5x + 2 = (x^2 - 3x + 2)(x^2 - x + 1) ]4. Analyze Possible Scenarios: The quadratic polynomials can be equated to the factors of the quartic polynomial: Case 1: [ begin{array}{l} x^2 + ax + b = x^2 - 3x + 2 x^2 + cx + d = x^2 - x + 1 end{array} ] From this, we get: [ a = -3, quad b = 2, quad c = -1, quad d = 1 ] Case 2: [ begin{array}{l} x^2 + ax + b = x^2 - x + 1 x^2 + cx + d = x^2 - 3x + 2 end{array} ] From this, we get: [ a = -1, quad b = 1, quad c = -3, quad d = 2 ]5. Validate Each Solution: Substitute (a, b, c, d) from each case back into the original system to ensure they satisfy all equations: - For ((a, b, c, d) = (-3, 2, -1, 1)): [ begin{array}{l} a + c = -3 + (-1) = -4 quad (text{True}) ac + b + d = (-3)(-1) + 2 + 1 = 3 + 2 + 1 = 6 quad (text{True}) ad + bc = (-3)(1) + (2)(-1) = -3 - 2 = -5 quad (text{True}) bd = (2)(1) = 2 quad (text{True}) end{array} ] - For ((a, b, c, d) = (-1, 1, -3, 2)): [ begin{array}{l} a + c = -1 + (-3) = -4 quad (text{True}) ac + b + d = (-1)(-3) + 1 + 2 = 3 + 1 + 2 = 6 quad (text{True}) ad + bc = (-1)(2) + (1)(-3) = -2 - 3 = -5 quad (text{True}) bd = (1)(2) = 2 quad (text{True}) end{array} ] Conclusion:The solutions to the system are:[boxed{(-3, 2, -1, 1) text{ and } (-1, 1, -3, 2)}]

🧠:Okay, so I need to find all positive integer solutions (x, y, z, n) to the equation ( x^{2n+1} - y^{2n+1} = x y z + 2^{2n+1} ) where n is at least 2 and z is bounded by ( 5 cdot 2^{2n} ). Hmm, let me start by understanding the equation and the constraints.First, the equation is ( x^{2n+1} - y^{2n+1} = x y z + 2^{2n+1} ). Since n is a positive integer greater than or equal to 2, 2n+1 is at least 5. So we are dealing with exponents that are odd and at least 5. The right-hand side (RHS) is a product of x, y, z, plus a term that's a power of 2. The left-hand side (LHS) is a difference of two (2n+1)th powers. Since x and y are positive integers, x must be greater than y, otherwise the LHS would be non-positive, while the RHS is positive because all terms are positive. So x > y.Let me think about factoring the LHS. The difference of two odd powers can be factored as:( x^{2n+1} - y^{2n+1} = (x - y)(x^{2n} + x^{2n-1}y + dots + y^{2n}) )But I'm not sure if that helps directly. However, maybe I can consider the equation modulo some number to get constraints on x, y, z, or n.Alternatively, since z is bounded by ( 5 cdot 2^{2n} ), maybe z isn't too large relative to n, so perhaps for each n ≥ 2, we can look for possible x, y, z with z ≤ 5·2^{2n}. But n is also a variable here, so we need to consider all n ≥ 2. Hmm.Let me see if there are obvious solutions. Maybe try small n first. Let's start with n=2. Then 2n+1=5, so the equation becomes:( x^5 - y^5 = xyz + 32 )And z ≤ 5·2^{4} = 5·16 = 80.So for n=2, z ≤ 80.Let me see if I can find x, y, z with x > y ≥1, z ≥1, such that x^5 - y^5 - xyz = 32.Trying small y first. Let's try y=1. Then equation becomes x^5 - 1 - x*z = 32 → x^5 - x*z = 33. Then z = (x^5 -33)/x. Since z must be integer, x must divide x^5 -33, which is equivalent to x dividing 33. Because x^5 ≡ 0 mod x, so x divides -33. But x is positive, so x divides 33. Divisors of 33 are 1,3,11,33. But x > y=1, so x ≥2. So possible x:3,11,33.Check x=3: z=(243 -33)/3 = 210/3=70. 70 ≤80? Yes. So z=70. So (x,y,z,n)=(3,1,70,2) is a solution? Let me check the original equation:3^5 -1^5 =243 -1=242. RHS:3*1*70 +32=210 +32=242. Yes, 242=242. So that works.x=11: z=(161051 -33)/11 = (161018)/11. Let me compute 161018 ÷11. 11*14638=160, 11*14638=160, 14638*11= 160, 14638*11=160, 14638*10=146380, plus 14638=160, 146380+160, so 14638*11=160, 14638*11=160, sorry, maybe better to compute 11*14638. 11*14000=154000, 11*638= 7018, total 154000+7018=161,018. So 161018 ÷11=14638. So z=14638. But z must be ≤80 for n=2. 14638 is way larger than 80. So that's invalid. Similarly x=33 would give z even larger. So only x=3, y=1, z=70 is valid for y=1.Next, y=2. Then equation becomes x^5 -32 -2x z =32 → x^5 -2x z =64. Then z=(x^5 -64)/(2x). Must be integer. So x divides x^5 -64. So x divides 64. Because x^5 ≡0 mod x, so x divides -64. So x is a divisor of 64. Divisors of 64 are 1,2,4,8,16,32,64. Since x>y=2, x≥3. So possible x:4,8,16,32,64.Check x=4: z=(1024 -64)/(8)=960/8=120. But z must be ≤80. So 120>80. Not allowed.x=8: z=(32768 -64)/16=32704/16=2044. Way too big.Similarly x=16,32,64 give z even larger. So y=2, no solutions.y=3: Equation is x^5 -243 -3x z =32 → x^5 -3x z =275. Then z=(x^5 -275)/(3x). So x divides x^5 -275, so x divides 275. Divisors of 275:1,5,11,25,55,275. x>3, so x≥4. Possible x:5,11,25,55,275.Check x=5: z=(3125 -275)/15=2850/15=190. z=190 >80. Not allowed.x=11: z=(161051 -275)/33=160,776/33 ≈4872.0... 33*4872=160, 33*4800=158400, 33*72=2376, total 158400+2376=160,776. So z=4872>80. No good.Similarly larger x will give larger z. So y=3 gives no solutions.y=4: Equation x^5 -1024 -4x z =32 → x^5 -4x z =1056. Then z=(x^5 -1056)/(4x). So x divides x^5 -1056 → x divides 1056. Divisors of 1056: Let's factorize 1056. 1056=2^5 *3*11. So divisors are numerous. x>4, so x≥5. Let's check small x.x=5: z=(3125 -1056)/20=2069/20=103.45. Not integer.x=6: z=(7776 -1056)/24=6720/24=280. z=280>80. Not allowed.x=8: z=(32768 -1056)/32=31712/32=991.25. Not integer.x=11: z=(161051 -1056)/44=160, 161051-1056=160, 161051-1000=160051, minus 56=159,995. So 159,995/44 ≈3636.25. Not integer.x=12: z=(248832 -1056)/48=247776/48=5162. Not allowed.Not seeing any valid z here. So y=4 no solutions.y=5: Let's check. x^5 -3125 -5x z =32 →x^5 -5x z =3157. Then z=(x^5 -3157)/(5x). So x divides x^5 -3157 → x divides 3157. Let's factor 3157. 3157 ÷7=451. 451 is 11*41. So 3157=7*11*41. So divisors are 1,7,11,41,77,287,451,3157. x>5, so possible x=7,11,41, etc.x=7: z=(16807 -3157)/35=13650/35=390. z=390>80. No.x=11: z=(161051 -3157)/55=157,894/55≈2870.8. Not integer.x=41: z would be huge. So no solution here.Hmm, this seems tedious. Maybe there's a pattern here. For y=1, we found a solution. Maybe trying higher y is not helpful. Let's check y=1 for other n.Wait, maybe the solution we found for n=2 is part of a general pattern. Let me think. Suppose y=1 for general n. Then the equation becomes:( x^{2n+1} -1 = x cdot 1 cdot z + 2^{2n+1} )So ( x^{2n+1} - x z = 2^{2n+1} +1 )Therefore, z = (x^{2n+1} - 2^{2n+1} -1)/xWe need z to be a positive integer, so x must divide (x^{2n+1} - 2^{2n+1} -1). Since x divides x^{2n+1}, x must divide (-2^{2n+1} -1). Therefore, x divides (2^{2n+1} +1). Let's note that.So x is a divisor of 2^{2n+1} +1. Let's denote D = 2^{2n+1} +1. Then x divides D, so x is a positive divisor of D, and x >1 since y=1 and x > y.But 2^{2n+1} is 2^(2n+1), which is even, so D is odd. Therefore, all divisors of D are odd. So possible x is an odd divisor of D greater than 1.Moreover, since D = 2^{2n+1} +1, which is 2*4^n +1. For n ≥2, 4^n is 16, 64, etc. So D is a large number. However, 2^{2n+1} +1 can sometimes be a prime. For example, if n=2, D=2^5 +1=33=3*11. Which factors as 3*11.So for n=2, D=33, which has divisors 1,3,11,33. x must be 3,11,33. For x=3, z=(3^5 -33 -1)/3=(243 -34)/3=209/3≈69.666, wait, no. Wait, let's recast:Wait, for n=2, D=33, so x must divide 33, and z=(x^5 - 33 -1)/x=(x^5 -34)/x. Wait, earlier for y=1 and n=2, we had x^5 -1 -x z=32, so x^5 - x z=33, so z=(x^5 -33)/x. So in that case, x divides x^5 -33, which implies x divides 33, which is correct.But in the general case, for y=1, D=2^{2n+1} +1. Then x divides D, so x divides 2^{2n+1} +1. Then z=(x^{2n+1} - D)/x. Since D=2^{2n+1} +1, z=(x^{2n+1} -2^{2n+1} -1)/x.Wait, let me check with n=2:z=(x^5 - 32 -1)/x=(x^5 -33)/x. So when x=3, (243-33)/3=210/3=70. Which works. Similarly, if x=11, (161051 -33)/11=161018/11=14638, which is too big. So for n=2, x=3 is the only feasible solution.So, for general n, if we set y=1, then x must be a divisor of 2^{2n+1} +1. But 2^{2n+1} +1 is 2^{2n+1} +1, which factors as (2 +1)(2^{2n} -2^{2n-1} + ... +1) only when 2n+1 is a multiple of 3, but 2n+1 is odd. Wait, actually, 2^{2n+1} +1 factors as (2 +1)(2^{2n} -2^{2n-1} + ... -2 +1) only when 2n+1 is divisible by 3? Wait, maybe not. Let's test with n=2, 2n+1=5: 2^5 +1=33=3*11, which is (2 +1)(2^4 -2^3 +2^2 -2 +1)=3*(16 -8 +4 -2 +1)=3*11=33. So that works. For n=3, 2n+1=7: 2^7 +1=129= 3*43. Let's check the same formula: (2 +1)(2^6 -2^5 +2^4 -2^3 +2^2 -2 +1)=3*(64 -32 +16 -8 +4 -2 +1)=3*(64-32=32; 32+16=48; 48-8=40; 40+4=44; 44-2=42; 42+1=43)=3*43=129. So that works as well. So in general, 2^{2n+1} +1 factors as 3 times something. Specifically, 2^{k} +1 where k is odd factors into (2 +1)(2^{k-1} -2^{k-2} + ... +1). So yes, for k=2n+1, 2^{2n+1} +1=3 * Q, where Q=(2^{2n} -2^{2n-1} + ... +1). Therefore, D=3*Q, so the divisors of D are 1,3,Q,3Q. Therefore, possible x is 3, Q, 3Q. But since x must be greater than y=1, and x is a positive integer, x can be 3, Q, or 3Q.But Q is (2^{2n} -2^{2n -1} + ... +1). Let's compute Q for n=2: 2^4 -2^3 +2^2 -2 +1=16 -8 +4 -2 +1=11. So Q=11. For n=3, Q=2^6 -2^5 +2^4 -2^3 +2^2 -2 +1=64 -32 +16 -8 +4 -2 +1=43. So Q=43. For n=4, Q=2^8 -2^7 +2^6 -2^5 +2^4 -2^3 +2^2 -2 +1=256 -128 +64 -32 +16 -8 +4 -2 +1. Let's compute step by step:256 -128 =128128 +64=192192 -32=160160 +16=176176 -8=168168 +4=172172 -2=170170 +1=171. So Q=171=9*19. Wait, 171=9*19. So not prime. Hmm, so Q is not necessarily prime. So D=3*Q, where Q may be composite. Thus, D can have more divisors. For example, when n=4, D=3*171=513=3^3*19. So divisors of 513 are 1,3,9,19,27,57,171,513. So x could be 3,9,19,27,57,171,513. So for each n, x can be 3 or other divisors. However, when x=3, let's check if that gives a valid z.Suppose x=3, then z=(3^{2n+1} -2^{2n+1} -1)/3. Let's compute this for n=2: (3^5 -2^5 -1)/3=(243 -32 -1)/3=210/3=70. Which works. For n=3: (3^7 -2^7 -1)/3=(2187 -128 -1)/3=2058/3=686. Check z≤5*2^{2n}=5*64=320. 686>320. So invalid. So for n=3, x=3 gives z=686 which is over the limit. Similarly, for n=4: (3^9 -2^9 -1)/3=(19683 -512 -1)/3=19170/3=6390. But z≤5*2^{8}=5*256=1280. 6390>1280. So invalid.Thus, only for n=2, x=3 gives z=70 which is within the limit. For n≥3, x=3 would give z exceeding the maximum allowed. What about x=Q? For example, n=2, x=11. Then z=(11^5 -32 -1)/11=(161051 -33)/11=161018/11=14638, which is too large. Similarly, for n=3, x=43, z=(43^7 -128 -1)/43, which is massive. So x=Q or x=3Q would give z values way beyond the z limit.Therefore, the only solution with y=1 is when n=2, x=3, z=70.Now, perhaps trying other y values for n=2. Wait, we already checked y=1,2,3,4,5 and only y=1 gives a solution. Maybe higher y? Let's try y=6 for n=2. Then equation becomes x^5 - 7776 -6x z =32 →x^5 -6x z =7808. Then z=(x^5 -7808)/(6x). x must divide x^5 -7808, so x divides 7808. 7808 ÷16=488. 7808=16*488=16*8*61=128*61. So divisors are 1,2,4,8,16,32,64,61,122,244,488,976,1952,3904,7808. x>6, so x≥7. Let's try x=8: z=(32768 -7808)/48=24960/48=520. 520>80. No. x=16: z=(1048576 -7808)/96=1,040,768/96≈10,841.33. Not integer. x=61: z would be huge. Not helpful. So no solutions here.Alternatively, maybe for n=2, the only solution is (3,1,70,2). Let's check if there are other solutions with different n.Trying n=3. Then 2n+1=7. The equation becomes:x^7 - y^7 = xyz + 2^7= xyz +128And z ≤5*2^{6}=5*64=320.Looking for x > y ≥1, z ≥1 integers. Let's try y=1 again. Then equation:x^7 -1 =x*z +128 →x^7 -x*z=129 →z=(x^7 -129)/x. So x divides x^7 -129 →x divides 129. 129=3*43. So divisors are 1,3,43,129. x>1, so x=3,43,129.x=3: z=(2187 -129)/3=2058/3=686. But z ≤320. 686>320. Invalid.x=43: z=(43^7 -129)/43. 43^7 is a huge number, but z would be (43^7 -129)/43=43^6 - 129/43. But 129 is divisible by 43? 43*3=129. So 129/43=3. So z=43^6 -3. Which is a gigantic number, way over 320. So invalid.Similarly x=129: z even larger. So no solution for y=1, n=3.Trying y=2: Equation x^7 -128 -2x z=128 →x^7 -2x z=256. Then z=(x^7 -256)/(2x). So x divides x^7 -256, so x divides 256. 256=2^8. Divisors are 1,2,4,8,16,32,64,128,256. x>2, so x≥3. Possible x=4,8,16, etc.x=4: z=(16384 -256)/8=16128/8=2016. z=2016>320. No.x=8: z=(2097152 -256)/16=2096896/16=131056. Too big. So no solutions.Similarly, higher x give higher z. No luck.y=3: x^7 -2187 -3x z=128 →x^7 -3x z=2315. Then z=(x^7 -2315)/(3x). So x divides x^7 -2315 →x divides 2315. 2315=5*463. 463 is prime. So divisors are 1,5,463,2315. x>3, so x=5,463,2315.x=5: z=(78125 -2315)/15=75810/15=5054. 5054>320. No.x=463: z is huge. So invalid.Similarly, y=4: x^7 - 16384 -4x z=128 →x^7 -4x z=16512. z=(x^7 -16512)/(4x). x divides x^7 -16512 →x divides 16512. 16512=16512. Let's factorize. 16512 ÷16=1032. 1032 ÷8=129. 129=3*43. So 16512=16*8*3*43=2^7 *3*43. Divisors are multiple. x>4, possible x=6,8, etc. Trying x=8: z=(2097152 -16512)/32=2080640/32=65020. Too big. So no.This seems not promising. Maybe for n=3, no solutions.Check n=4. 2n+1=9. Equation x^9 - y^9 =xyz +2^9= xyz +512. z ≤5*2^{8}=5*256=1280.Again, try y=1: x^9 -1 =x z +512 →x^9 -x z=513. Then z=(x^9 -513)/x. x divides 513. 513=3^3*19. Divisors:1,3,9,19,27,57,171,513. x>1, so x=3,9,19, etc.x=3: z=(19683 -513)/3=19170/3=6390. z=6390>1280. No.x=9: z=(387,420,489 -513)/9=(387,419,976)/9≈43,046,664. Too big.No solution here. Similarly for y=2:x^9 -512 -2x z=512 →x^9 -2x z=1024. z=(x^9 -1024)/(2x). x divides 1024. 1024=2^10. Divisors x=4,8,16,... x=4: z=(262,144 -1024)/8=261,120/8=32,640>1280. No.So, seems like for n≥3, even if we try y=1, the z value becomes too large. The only solution found so far is for n=2, (3,1,70,2).Wait, perhaps there is another approach. Let's consider modulo 2^{2n+1}. The RHS is xyz +2^{2n+1}, so modulo 2^{2n+1}, the RHS is congruent to xyz mod 2^{2n+1}. The LHS is x^{2n+1} - y^{2n+1}.If x and y are both even, then x=2a, y=2b. Then LHS=(2a)^{2n+1} - (2b)^{2n+1}=2^{2n+1}(a^{2n+1} -b^{2n+1}). RHS=2a*2b*z +2^{2n+1}=4ab z +2^{2n+1}. Then equate:2^{2n+1}(a^{2n+1} -b^{2n+1}) =4ab z +2^{2n+1}Divide both sides by 2^{2n+1}:a^{2n+1} -b^{2n+1} = (4ab z)/2^{2n+1} +1But (4ab z)/2^{2n+1}= (ab z)/2^{2n -1}. Since 2n+1 ≥5, 2n -1 ≥3. For the RHS to be integer, ab z must be divisible by 2^{2n -1}. But since a and b are integers, unless a or b has factors of 2, but x and y are both even, so a and b are integers, but not necessarily even. If a or b is even, but this complicates. Maybe this path isn't helpful.Alternatively, if one of x or y is even and the other is odd. Let's suppose x is even and y is odd. Then LHS=x^{2n+1} - y^{2n+1}. Since x is even, x^{2n+1} is even, and y is odd, so y^{2n+1} is odd. So LHS is even - odd = odd. RHS is xyz +2^{2n+1}. x is even, so xyz is even, 2^{2n+1} is even. So even + even = even. But LHS is odd, which is a contradiction. Similarly, if x is odd and y is even, then LHS is odd - even = odd, RHS is odd*even*z + even = even + even = even. Again, contradiction. Therefore, x and y must be both even or both odd.If both are even, as above, but then the equation may have higher divisibility. If both are odd, then LHS is even (odd - odd = even), RHS is odd*odd*z + even. So odd*z + even. If z is even, RHS is even. If z is odd, RHS is odd + even = odd. But LHS is even, so RHS must be even. Therefore, z must be even. So in the case where x and y are both odd, z must be even.So possible scenarios:1. Both x and y even: then x=2a, y=2b. Then equation becomes 2^{2n+1}(a^{2n+1} -b^{2n+1})=4ab z +2^{2n+1}. Divide by 2^{2n+1}:a^{2n+1} -b^{2n+1} = (4ab z)/2^{2n+1} +1. So (4ab z)/2^{2n+1} must be integer, so 2^{2n+1} divides 4ab z. Since 4=2^2, this implies 2^{2n-1} divides ab z. Since a and b are integers (x=2a, y=2b), and z is positive integer. So ab z must be divisible by 2^{2n -1}. Since 2n -1 ≥3 (as n≥2), so ab z must be divisible by 8,16, etc., depending on n. This might require a or b to be even, but since x and y are both even, a and b could be any integers.Alternatively, perhaps we can factor out 2 from x and y. Let’s set x=2a, y=2b. Then z must be such that:Original equation:(2a)^{2n+1} - (2b)^{2n+1} = (2a)(2b) z +2^{2n+1}Simplify:2^{2n+1}(a^{2n+1} - b^{2n+1}) = 4ab z +2^{2n+1}Divide both sides by 2^{2n+1}:a^{2n+1} - b^{2n+1} = (4ab z)/2^{2n+1} +1Which simplifies to:a^{2n+1} - b^{2n+1} = (ab z)/2^{2n -1} +1Since the LHS is an integer, the RHS must be an integer. Hence, (ab z)/2^{2n -1} must be an integer. Therefore, ab z must be divisible by 2^{2n -1}.Given that z ≤5·2^{2n}, so ab z ≤5·2^{2n} *ab. But since ab z must be divisible by 2^{2n -1}, perhaps ab needs to compensate. However, unless a or b themselves have factors of 2, but since x=2a and y=2b, a and b can be any positive integers. If we set a=1, b=1, then ab z=1*1*z=z. So z must be divisible by 2^{2n -1}. But z ≤5·2^{2n}, so possible z=2^{2n -1}, 2*2^{2n -1}, ..., up to 5·2^{2n}. For example, z= k*2^{2n -1} where k=1,2,...,10. Then:Equation becomes:1 -1 = (1*1*z)/2^{2n -1} +1 →0 = k +1 →k=-1. Impossible. So no solution with a=1,b=1.Alternatively, try a=2, b=1. Then ab=2*1=2. So ab z=2 z must be divisible by 2^{2n -1}, so z must be divisible by 2^{2n -2}. Then z= k*2^{2n -2}, k≥1. Then z ≤5*2^{2n} ⇒k*2^{2n -2} ≤5*2^{2n} ⇒k ≤5*2^{2n}/2^{2n -2}=5*4=20. So k=1,...,20.Plug into equation:a^{2n+1} -b^{2n+1}= (ab z)/2^{2n -1} +1 →2^{2n+1} -1= (2*1*z)/2^{2n -1} +1= (2 z)/2^{2n -1} +1= z/2^{2n -2} +1.But a=2, b=1. So LHS=2^{2n+1} -1. RHS= z/2^{2n -2} +1. Therefore:2^{2n+1} -1 -1= z/2^{2n -2} →2^{2n+1} -2= z/2^{2n -2} →z=(2^{2n+1} -2)*2^{2n -2}=2^{2n -2}(2^{2n+1} -2)=2^{2n -2}*2*(2^{2n} -1)=2^{2n -1}(2^{2n} -1)But z= k*2^{2n -2}, so:k*2^{2n -2}=2^{2n -1}(2^{2n} -1) ⇒k=2*(2^{2n} -1). But k must be ≤20. So 2*(2^{2n} -1) ≤20 →2^{2n} -1 ≤10 →2^{2n} ≤11. Since n≥2, 2n≥4, so 2^{2n}≥16. 16>11. No solution. Hence, no solutions with a=2, b=1.This seems complicated. Maybe there's another way. Let's consider the original equation:x^{2n+1} - y^{2n+1} =xyz +2^{2n+1}If we think of this as x^{2n+1} - y^{2n+1} -2^{2n+1}=xyz.Perhaps, to bound x and y. Since x > y, let's set x = y + k, where k ≥1. Then substitute into the equation:(y + k)^{2n+1} - y^{2n+1} = (y + k)y z +2^{2n+1}Expanding the left-hand side using the binomial theorem:= y^{2n+1} + (2n+1)y^{2n}k + ... + k^{2n+1} - y^{2n+1}= (2n+1)y^{2n}k + ... +k^{2n+1}Therefore,(2n+1)y^{2n}k + ... +k^{2n+1} = (y^2 +ky) z +2^{2n+1}This seems messy, but maybe for small k we can find solutions.Let's try k=1. Then x=y+1. Then LHS becomes:(y +1)^{2n+1} - y^{2n+1} = (2n+1)y^{2n} + ... +1.And the equation becomes:(2n+1)y^{2n} + ... +1 = y(y +1) z +2^{2n+1}But even for k=1, the LHS is a polynomial in y of degree 2n, while the RHS is quadratic in y. For large y, LHS grows much faster than RHS, so maybe only small y are possible. Let's try with n=2, k=1.For n=2, k=1:LHS=5y^4 + 10y^3 +10y^2 +5y +1RHS= y(y+1) z +32We need 5y^4 +10y^3 +10y^2 +5y +1 -32= y(y+1) zSo 5y^4 +10y^3 +10y^2 +5y -31= y(y+1) zLet's compute for y=1:LHS=5 +10 +10 +5 -31= (5+10=15; 15+10=25;25+5=30;30-31=-1). So -1=1*2*z →-1=2z. No solution.y=2:LHS=5*16 +10*8 +10*4 +5*2 -31=80+80+40+10-31=80+80=160;160+40=200;200+10=210;210-31=179.RHS=2*3*z=6z. So 6z=179. 179 prime. Not divisible by 6. No.y=3:LHS=5*81 +10*27 +10*9 +15 -31=405 +270 +90 +15 -31=405+270=675;675+90=765;765+15=780;780-31=749.RHS=3*4*z=12z. 749=12z→z≈62.416. Not integer.y=4:LHS=5*256 +10*64 +10*16 +20 -31=1280 +640 +160 +20 -31=1280+640=1920;1920+160=2080;2080+20=2100;2100-31=2069.RHS=4*5*z=20z. 2069=20z→z=103.45. Not integer.y=5:LHS=5*625 +10*125 +10*25 +25 -31=3125 +1250 +250 +25 -31=3125+1250=4375;4375+250=4625;4625+25=4650;4650-31=4619.RHS=5*6*z=30z. 4619/30≈153.966. Not integer.Not working. So k=1, n=2 gives no solution.Try k=2, n=2. Then x=y+2.LHS=(y+2)^5 -y^5=32 +80y +80y^2 +40y^3 +10y^4 +y^5 -y^5=10y^4 +40y^3 +80y^2 +80y +32.Equation:10y^4 +40y^3 +80y^2 +80y +32 = y(y+2) z +32. Subtract 32:10y^4 +40y^3 +80y^2 +80y = y(y+2) z.Divide both sides by y:10y^3 +40y^2 +80y +80 = (y +2) z.So z=(10y^3 +40y^2 +80y +80)/(y +2). Let's perform polynomial division.Divide numerator by y+2.10y^3 +40y^2 +80y +80 ÷ (y +2).Using synthetic division:-2 | 10 40 80 80 -20 -40 -80 10 20 40 0So quotient is 10y^2 +20y +40. So z=10y^2 +20y +40.Since z must be positive integer, this works for any y≥1. However, z must also satisfy z ≤5*2^{4}=80.So 10y^2 +20y +40 ≤80 →10y^2 +20y +40 ≤80 →10y^2 +20y ≤40 →Divide by 10: y^2 +2y ≤4 →y^2 +2y -4 ≤0.Solve quadratic equation y^2 +2y -4=0. Discriminant=4 +16=20. Roots=(-2 ±√20)/2=(-2 ±2√5)/2=-1 ±√5≈-1 ±2.236. So positive root≈1.236. So y ≤1.236. Since y is positive integer, y=1.Check y=1:z=10*1 +20*1 +40=10+20+40=70. Which is ≤80. So z=70. Then x=y+2=3. So (x,y,z,n)=(3,1,70,2). Which is the solution we found earlier.For y=2: z=10*4 +20*2 +40=40+40+40=120>80. Not allowed.So only y=1 gives valid z. So this is the same solution as before.For k=2, n=2, we recover the solution. So perhaps this is the only solution with k=2.Wait, but this seems specific to n=2. Let's check for general n with k=2. Suppose x=y+2. Then:(y +2)^{2n+1} - y^{2n+1} = y(y +2) z +2^{2n+1}Expand the left-hand side:= Σ_{i=0}^{2n+1} C(2n+1,i)y^{2n+1 -i}2^i - y^{2n+1}= Σ_{i=1}^{2n+1} C(2n+1,i)y^{2n+1 -i}2^i= 2^{2n+1} + C(2n+1,1)y^{2n}2 + C(2n+1,2)y^{2n -1}2^2 + ... + C(2n+1,2n)y*2^{2n}Therefore, LHS=2^{2n+1} +2(2n+1)y^{2n} +4C(2n+1,2)y^{2n -1} + ... +2^{2n+1}yThen equation becomes:2^{2n+1} +2(2n+1)y^{2n} +4C(2n+1,2)y^{2n -1} + ... +2^{2n+1}y = y(y +2) z +2^{2n+1}Subtract 2^{2n+1} from both sides:2(2n+1)y^{2n} +4C(2n+1,2)y^{2n -1} + ... +2^{2n+1}y = y(y +2) zDivide both sides by y:2(2n+1)y^{2n -1} +4C(2n+1,2)y^{2n -2} + ... +2^{2n+1} = (y +2) zThis is similar to the n=2 case. If we can factor the left-hand side by y +2, then z would be an integer. Let's check for y=1.Set y=1:Left-hand side=2(2n+1)(1)^{2n -1} +4C(2n+1,2)(1)^{2n -2} + ... +2^{2n+1}This is a sum of terms. For y=1, each term becomes a coefficient times 1. Let's compute this sum.But for y=1, LHS=2(2n+1) +4C(2n+1,2) +8C(2n+1,3)+...+2^{2n+1}But notice that in the case n=2, this sum was:2(5) +4C(5,2) +8C(5,3) +16C(5,4) +32C(5,5)=10 +4*10 +8*10 +16*5 +32*1=10 +40 +80 +80 +32=242. Then RHS=(1 +2) z=3 z. So 242=3 z → z≈80.666, but in reality, when y=1, we have:Wait, no. Wait, when y=1 and x=3 (k=2), for general n, the equation becomes:(3)^{2n+1} -1^{2n+1} =3*1*z +2^{2n+1}So 3^{2n+1} -1 -2^{2n+1} =3 z → z=(3^{2n+1} -2^{2n+1} -1)/3Which is exactly what we had earlier. For n=2, this gives z=(243 -32 -1)/3=210/3=70, which works. For n=3, z=(3^7 -2^7 -1)/3=(2187 -128 -1)/3=2058/3=686>320. Not allowed.Therefore, only for n=2, this gives a valid z. For n≥3, z exceeds the limit. Hence, the only solution in this family is n=2, x=3, y=1, z=70.Now, check for other k values. Suppose k=3. Then x=y+3. Let's try n=2. Then:(y +3)^5 -y^5=243 + 405y + 270y^2 +90y^3 +15y^4 +y^5 -y^5=15y^4 +90y^3 +270y^2 +405y +243.Equation:15y^4 +90y^3 +270y^2 +405y +243= y(y+3) z +32.So 15y^4 +90y^3 +270y^2 +405y +211= y(y+3) z.This seems difficult to solve. Try small y.y=1:LHS=15 +90 +270 +405 +211= 15+90=105;105+270=375;375+405=780;780+211=991.RHS=1*4*z=4z. 991=4z→z=247.75. Not integer.y=2:LHS=15*16 +90*8 +270*4 +405*2 +211=240 +720 +1080 +810 +211=240+720=960;960+1080=2040;2040+810=2850;2850+211=3061.RHS=2*5*z=10z. 3061/10=306.1. Not integer.Not promising. Similarly, higher y gives larger LHS. So k=3 seems not useful.Another approach: Assume that x and y are both powers of 2. Let’s suppose x=2^a, y=2^b, with a > b ≥1. Then:x^{2n+1} - y^{2n+1}=2^{a(2n+1)} -2^{b(2n+1)}=2^{b(2n+1)}(2^{(a-b)(2n+1)} -1)The RHS is xyz +2^{2n+1}=2^a*2^b*z +2^{2n+1}=2^{a+b}z +2^{2n+1}So equation becomes:2^{b(2n+1)}(2^{(a-b)(2n+1)} -1)=2^{a+b}z +2^{2n+1}Factor out 2^{min(b(2n+1), 2n+1, a+b)}. Let’s suppose b(2n+1) ≤2n+1 and b(2n+1) ≤a+b. Since b≥1, b(2n+1)≥2n+1. Equality when b=1. So if b=1, then left side is 2^{2n+1}(2^{(a-1)(2n+1)} -1). RHS=2^{a+1}z +2^{2n+1}. Let’s factor out 2^{2n+1}:2^{2n+1}[2^{(a-1)(2n+1) - (2n+1)} -1/(2^{2n+1})] =2^{a+1}z +2^{2n+1}Wait, not sure. Let's set b=1. Then equation:2^{2n+1}(2^{(a-1)(2n+1)} -1)=2^{a+1}z +2^{2n+1}Divide both sides by 2^{2n+1}:2^{(a-1)(2n+1)} -1 =2^{a+1 - (2n+1)} z +1Wait, 2^{a+1}/2^{2n+1}=2^{a+1 - (2n+1)}=2^{a -2n}. So:2^{(a-1)(2n+1)} -1 =2^{a -2n} z +1Rearrange:2^{(a-1)(2n+1)} -2 =2^{a -2n} zTherefore, z=(2^{(a-1)(2n+1)} -2)/2^{a -2n}=2^{2n}*(2^{(a-1)(2n+1) -1} -1)/2^{a -2n -1}?Wait, perhaps better to write:z=(2^{(a-1)(2n+1)} -2)/2^{a -2n}For z to be integer, denominator must divide numerator. So 2^{a -2n} divides 2^{(a-1)(2n+1)} -2.Note that 2^{(a-1)(2n+1)} -2=2[2^{(a-1)(2n+1) -1} -1]Thus, z=2[2^{(a-1)(2n+1) -1} -1]/2^{a -2n}=[2^{(a-1)(2n+1) -1} -1]/2^{a -2n -1}For this to be integer, the exponent in the denominator must be ≤ the exponent in the numerator's power of 2 minus the exponent from the subtraction. This is getting too abstract. Maybe set specific a and n.Let's try n=2. Then equation for b=1:2^{5}(2^{(a-1)5} -1)=2^{a+1}z +32So 32(2^{5(a-1)} -1)=2^{a+1}z +32Divide both sides by 32:2^{5(a-1)} -1=2^{a+1 -5}z +1/32. Wait, no. Wait:Wait, 32*(2^{5(a-1)} -1)=2^{a+1} z +32 →32*2^{5(a-1)} -32=2^{a+1} z +32 →32*2^{5(a-1)} -64=2^{a+1} z →32*(2^{5(a-1)} -2)=2^{a+1} z →2^{5}*(2^{5(a-1)} -2)=2^{a+1} z →2^{5(a-1) +5} -2^{6}=2^{a+1} z →2^{5a} -2^{6}=2^{a+1} z →2^{a+1}(2^{5a -a -1} -2^{6 -a -1})=z →2^{a+1}(2^{4a -1} -2^{5 -a})=z.But z must be an integer, so 5 -a must be ≥0, so a ≤5. But a > b=1, and a is integer. Let's try a=2:z=2^{3}(2^{7} -2^{3})=8*(128 -8)=8*120=960. But z ≤5*2^{4}=80. 960>80. Invalid.a=3:z=2^{4}(2^{11} -2^{2})=16*(2048 -4)=16*2044=32704>80.a=4:z=2^{5}(2^{15} -2^{1})=32*(32768 -2)=32*32766=1,048,512>80.a=5:z=2^{6}(2^{19} -2^{0})=64*(524288 -1)=64*524287=33,554,368>80.No solutions here. Therefore, setting b=1 and x=2^a doesn't help.Overall, this line of attack isn't fruitful. Let's return to the initial solution and see if it's unique.We found that for n=2, y=1, x=3, z=70 works. For other n, trying y=1 gives z too large. For other y, no solutions found. Maybe this is the only solution.To confirm, let's check if there are solutions with y>1 for n=2. Suppose y=2, x=4.Then x=4, y=2:4^5 -2^5=1024 -32=992.RHS=4*2*z +32=8z +32.Set 8z +32=992 →8z=960→z=120. But z ≤80. So invalid.Similarly, x=5, y=2:5^5 -32=3125 -32=3093.RHS=5*2*z +32=10z +32.10z=3093 -32=3061→z=306.1. Not integer.x=4, y=3:4^5 -3^5=1024 -243=781.RHS=4*3*z +32=12z +32.12z=781 -32=749→z≈62.416. Not integer.x=5, y=4:5^5 -4^5=3125 -1024=2101.RHS=5*4*z +32=20z +32.20z=2101 -32=2069→z=103.45. Not integer.x=6, y=5:6^5 -5^5=7776 -3125=4651.RHS=6*5*z +32=30z +32.30z=4651 -32=4619→z≈153.966. Not integer.Similarly, x=2, y=1:2^5 -1=32 -1=31. RHS=2*1*z +32=2z +32. 2z=31 -32=-1. Invalid.So no other solutions for n=2 except (3,1,70,2).For n≥3, no solutions found with y=1 or small y. Given that the RHS includes 2^{2n+1}, which grows exponentially, and the LHS is a difference of large powers, it's plausible that no other solutions exist.Therefore, the only solution is (x, y, z, n) = (3, 1, 70, 2).

✅:The equation ( x^{2n+1} - y^{2n+1} = xyz + 2^{2n+1} ) with ( n geq 2 ) and ( z leq 5 cdot 2^{2n} ) is analyzed as follows:1. Initial Exploration: - For ( n = 2 ), the equation becomes ( x^5 - y^5 = xyz + 32 ) with ( z leq 80 ). - Testing small values of ( y ): - Case ( y = 1 ): - The equation simplifies to ( x^5 - xz = 33 ). - ( x ) must divide ( 33 ). Possible ( x = 3 ), yielding ( z = 70 ), which satisfies ( z leq 80 ). - Verification: ( 3^5 - 1^5 = 242 ) and ( 3 cdot 1 cdot 70 + 32 = 242 ), confirming the solution ( (3, 1, 70, 2) ).2. General Case Analysis: - For ( n geq 3 ), testing ( y = 1 ) leads to ( z ) values exceeding ( 5 cdot 2^{2n} ). - Other values of ( y ) (e.g., ( y = 2, 3, 4 )) result in non-integer or excessively large ( z ).3. Structural Considerations: - Factoring the difference of odd powers and modular arithmetic constraints indicate that ( x ) and ( y ) must both be odd or even, with ( z ) even if both are odd. - Testing ( x = y + k ) for small ( k ) and specific values of ( n ) confirms that only ( k = 2 ) for ( n = 2 ) yields a valid solution.4. Conclusion: - The only solution satisfying all conditions is when ( n = 2 ), ( x = 3 ), ( y = 1 ), and ( z = 70 ).Final Answer:[boxed{(3, 1, 70, 2)}]

🔑:Given the equation:[x^{2n+1} - y^{2n+1} = x y z + 2^{2n+1}]We need to find all positive integer solutions (x, y, z, n) that satisfy the conditions:[n geq 2 quad text{and} quad z leq 5 cdot 2^{2n}]1. Establishing Constraints on (x) and (y): - By examining the powers, we can infer that (x - y > 0). - Additionally, (x) and (y) must have the same parity (both even or both odd). Thus, (x - y geq 2).2. Testing Specific Values: - If (y = 1) and (x = 3), we substitute into the original equation to find: [ 3^{2n+1} - 1^{2n+1} = 3 cdot 1 cdot z + 2^{2n+1} ] Simplifying, we get: [ 3^{2n+1} - 1 = 3z + 2^{2n+1} ] Rearrange to solve for (z): [ z = frac{3^{2n+1} - 1 - 2^{2n+1}}{3} ]3. Finding (z) and Validating the Condition (z leq 5 cdot 2^{2n}): - To ensure (z) remains an integer and meets the condition, the inequality must hold: [ 3^{2n+1} leq 5 cdot 2^{2n} + frac{1 + 2^{2n+1}}{3} ] - Simplify, [ 3^{2n+1} leq left(5 + frac{2}{3}right)2^{2n} + frac{1}{3} ] Since (5 + frac{2}{3} approx 5.67), [ 3^{2n+1} leq 6 cdot 2^{2n} ] By comparing powers of the base 3 and 2: [ left(frac{3}{2}right)^{2n+1} leq 6 ]4. Solving Inequality for (n): - Approximate to first satisfy: [ left(frac{3}{2}right)^{5} approx 7.59375 > 6 quad text{(for } n = 2) ] - Since (n=2) marginally satisfies the inequality, having further: [ n = 2 ]5. Substituting (n = 2) Back into the Equation: [ z = frac{3^{2 times 2 + 1} - 1 - 2^{2 times 2 + 1}}{3} ] [ z = frac{3^5 - 1 - 2^5}{3} ] [ z = frac{243 - 1 - 32}{3} = frac{210}{3} = 70 ] Thus, for (n = 2), the quadruple ((x, y, z, n) = (3, 1, 70, 2)) is a solution.6. Confirming Uniqueness: - For (x geq 4) or (y geq 2), it is established through comparison calculations (outstretched from solution steps) that all other configurations exceed the boundaries and fail to meet (n geq 2) or the initial equation conditions. Conclusion:The only set of positive integer solutions is:[boxed{(3, 1, 70, 2)}]

🧠:Okay, so I need to find the symmetric point of M(1,1,1) with respect to the given line. The line is given by the equation (x-2)/1 = (y + 1.5)/(-2) = (z - 1)/1. Hmm, symmetric point with respect to a line in 3D. Let me recall how to do this.First, symmetry with respect to a line means that the line is the perpendicular bisector of the segment joining the original point M and its symmetric counterpart M'. So, M' should be such that the line is perpendicular to the line segment MM' and intersects it at its midpoint. Therefore, the steps I need to take are probably:1. Find the projection of point M onto the given line. Let's call this projection point P.2. Once I have P, since P is the midpoint between M and M', I can solve for M' using the midpoint formula.Yes, that sounds right. So the key here is to find the projection P of M onto the line. Then M' is just 2P - M, right?So, how do I find the projection of a point onto a line in 3D?The line is given in symmetric equations. Let me write it in parametric form. The symmetric equations are (x - 2)/1 = (y + 1.5)/(-2) = (z - 1)/1. So the direction vector of the line is <1, -2, 1>. A point on the line is (2, -1.5, 1). So parametric equations for the line can be written as:x = 2 + ty = -1.5 - 2tz = 1 + twhere t is a parameter.So any point on the line can be represented as (2 + t, -1.5 - 2t, 1 + t).Now, the projection of M onto this line is the point P on the line such that the vector MP is perpendicular to the direction vector of the line.Let me denote the direction vector as v = <1, -2, 1>.The vector from a general point on the line (let's say point Q(2 + t, -1.5 - 2t, 1 + t)) to M(1, 1, 1) is QM = <1 - (2 + t), 1 - (-1.5 - 2t), 1 - (1 + t)> = <-1 - t, 2.5 + 2t, -t>.For QM to be perpendicular to v, their dot product should be zero.So, QM • v = (-1 - t)(1) + (2.5 + 2t)(-2) + (-t)(1) = 0.Let me compute that:First term: (-1 - t)(1) = -1 - tSecond term: (2.5 + 2t)(-2) = -5 -4tThird term: (-t)(1) = -tAdding them up: (-1 - t) + (-5 -4t) + (-t) = -1 - t -5 -4t - t = -6 -6tSet equal to zero: -6 -6t = 0 => -6t = 6 => t = -1So t = -1. Therefore, the projection point P is:x = 2 + (-1) = 1y = -1.5 -2*(-1) = -1.5 + 2 = 0.5z = 1 + (-1) = 0So P is (1, 0.5, 0). Wait, let me check that again.Wait, for t = -1:x = 2 + t = 2 + (-1) = 1y = -1.5 -2t = -1.5 -2*(-1) = -1.5 + 2 = 0.5z = 1 + t = 1 + (-1) = 0Yes, that's correct. So P is (1, 0.5, 0).Now, since P is the midpoint between M and M', then:Coordinates of P = ( (M_x + M'_x)/2, (M_y + M'_y)/2, (M_z + M'_z)/2 )Therefore, solving for M':M'_x = 2*P_x - M_x = 2*1 -1 = 1M'_y = 2*P_y - M_y = 2*0.5 -1 = 1 -1 = 0M'_z = 2*P_z - M_z = 2*0 -1 = -1Therefore, M' is (1, 0, -1). Wait, is that right? Let me verify.Wait, original point M is (1,1,1). Midpoint P is (1, 0.5, 0). So for x-coordinate: (1 + M'_x)/2 = 1 => 1 + M'_x = 2 => M'_x = 1. Okay.For y-coordinate: (1 + M'_y)/2 = 0.5 => 1 + M'_y = 1 => M'_y = 0.For z-coordinate: (1 + M'_z)/2 = 0 => 1 + M'_z = 0 => M'_z = -1.Yes, that's correct. So M' is (1, 0, -1). Let me check if this makes sense.Wait, maybe I should check if the line is indeed the perpendicular bisector. Let me check the vector from M to M'. The vector MM' is (1-1, 0-1, -1 -1) = (0, -1, -2). The direction vector of the line is <1, -2, 1>. The dot product between MM' and the direction vector should be zero if they are perpendicular.Dot product: 0*1 + (-1)*(-2) + (-2)*1 = 0 + 2 -2 = 0. Perfect, so they are perpendicular. That checks out.Also, the midpoint between M and M' is ((1+1)/2, (1+0)/2, (1 + (-1))/2) = (1, 0.5, 0), which is P. So that's correct. Therefore, M' is indeed (1, 0, -1). Hmm, seems good.Wait, but let me think again. Is there another way to approach this problem? Maybe using formula for reflection over a line in 3D?Alternatively, another method. Let's see.Suppose we have a line L: given by a point A and direction vector v. To find the reflection of a point M over L, we can use the formula:M' = 2 proj_L(M - A) + A - MWait, maybe not exactly. Wait, maybe reflection can be found by moving M to the line, projecting, then moving back.Alternatively, here's another approach. The reflection of M over the line is such that the line is the perpendicular bisector between M and M'. Therefore, as we did before, the midpoint is P, the projection of M on the line. So the same result.Alternatively, maybe using vector projection.Given point M, vector from a point on the line to M, then projecting that vector onto the direction vector, then scaling appropriately.But let's see. Let me formalize.Let A be a point on the line, here A is (2, -1.5, 1). The direction vector is v = <1, -2, 1>.Vector AM = M - A = (1 - 2, 1 - (-1.5), 1 - 1) = (-1, 2.5, 0).The projection of AM onto v is [(AM • v) / ||v||²] * vCompute AM • v: (-1)(1) + (2.5)(-2) + (0)(1) = -1 -5 + 0 = -6||v||² = 1² + (-2)² + 1² = 1 +4 +1 =6Therefore, the projection vector is (-6 /6)* v = -v = <-1, 2, -1>Therefore, the projection of vector AM onto v is the vector <-1, 2, -1>Therefore, the projection point P is A + projection vector = (2, -1.5, 1) + (-1, 2, -1) = (1, 0.5, 0). Which matches the previous result. So that's consistent.Therefore, the reflection point M' is such that vector from M to M' is twice the projection vector? Wait, not exactly. Wait, since P is the midpoint between M and M', then vector from M to P is P - M, and vector from P to M' is also P - M. Therefore, M' = P + (P - M) = 2P - M.Which gives the same result as before. So M' = 2*(1, 0.5, 0) - (1,1,1) = (2 -1, 1 -1, 0 -1) = (1, 0, -1). Same answer.Therefore, after two different methods, I still get M'(1, 0, -1). So seems correct.Let me just visualize a bit. The original point is (1,1,1). The line passes through (2, -1.5, 1) and has direction (1, -2, 1). The projection of M onto the line is (1, 0.5, 0). Then reflecting over that gives (1, 0, -1). Since the z-coordinate goes from 1 to 0 to -1, which makes sense as reflection over the midpoint. Similarly, y-coordinate from 1 to 0.5 to 0. That seems okay.Another check: compute the distance from M to the line and from M' to the line. They should be equal.The distance from a point to a line in 3D can be calculated by |AM × v| / ||v||, where AM is the vector from point A on the line to M.We already have AM = (-1, 2.5, 0). v = <1, -2, 1>Compute AM × v:|i   j   k||-1 2.5 0||1  -2  1|= i*(2.5*1 - 0*(-2)) - j*(-1*1 - 0*1) + k*(-1*(-2) - 2.5*1)= i*(2.5) - j*(-1) + k*(2 - 2.5)= (2.5, 1, -0.5)The magnitude of this cross product is sqrt(2.5² +1² + (-0.5)^2) = sqrt(6.25 +1 +0.25) = sqrt(7.5) ≈ 2.7386The magnitude of v is sqrt(1 +4 +1) = sqrt(6) ≈ 2.4495Therefore, distance from M to line is |AM × v| / ||v|| = sqrt(7.5)/sqrt(6) = sqrt(7.5/6) = sqrt(1.25) ≈ 1.1180Now, let's compute the distance from M' to the line. Let's do the same.Vector AM' = M' - A = (1 -2, 0 - (-1.5), -1 -1) = (-1, 1.5, -2)Compute AM' × v:|i   j   k||-1 1.5 -2||1  -2  1|= i*(1.5*1 - (-2)*(-2)) - j*(-1*1 - (-2)*1) + k*(-1*(-2) -1.5*1)= i*(1.5 -4) - j*(-1 +2) + k*(2 -1.5)= (-2.5, -1, 0.5)The magnitude of this cross product is sqrt((-2.5)^2 + (-1)^2 +0.5^2) = sqrt(6.25 +1 +0.25) = sqrt(7.5) ≈2.7386Divide by ||v|| = sqrt(6): same as before, so distance is sqrt(7.5)/sqrt(6) ≈1.1180. So same distance. Correct.Therefore, distances are equal, which is consistent with M' being the reflection of M over the line.Another check: direction vector of line is <1,-2,1>. The line passes through (2, -1.5, 1). Let's see if the midpoint P(1, 0.5, 0) is on the line. Wait, when t = -1, yes, as we found earlier.Also, the vector MP is (1-1, 0.5 -1, 0 -1) = (0, -0.5, -1). Wait, but earlier we had the vector QM was <-1 -t, 2.5 +2t, -t> which when t=-1, gives (0, 0.5, 1). Wait, no, perhaps I need to check again.Wait, vector from M(1,1,1) to P(1,0.5,0) is (0, -0.5, -1). The direction vector of the line is <1, -2,1>. Dot product is 0*1 + (-0.5)*(-2) + (-1)*1 = 0 +1 -1 =0. So they are perpendicular. Correct. Hence, the segment MP is perpendicular to the line, as required.Therefore, all checks out. So the final answer is M'(1, 0, -1).Final AnswerThe symmetric point ( M^{prime} ) is boxed{(1, 0, -1)}.

✅:To find the point ( M^{prime} ) that is symmetric to the point ( M(1, 1, 1) ) with respect to the given line (frac{x-2}{1} = frac{y+1.5}{-2} = frac{z-1}{1}), we follow these steps:1. Convert the line equation to parametric form: - The line has a direction vector (mathbf{v} = langle 1, -2, 1 rangle) and passes through the point (A(2, -1.5, 1)). - Parametric equations: (x = 2 + t), (y = -1.5 - 2t), (z = 1 + t).2. Find the projection of (M) onto the line: - The vector (overrightarrow{AM}) from (A) to (M) is (langle -1, 2.5, 0 rangle). - The projection parameter (t) is found by solving the dot product (overrightarrow{QM} cdot mathbf{v} = 0), where (overrightarrow{QM}) is the vector from a point (Q) on the line to (M). - Solving (-6 - 6t = 0) gives (t = -1). - The projection point (P) is ((1, 0.5, 0)).3. Determine the symmetric point (M^{prime}): - Using the midpoint formula, (M^{prime}) is calculated as (2P - M). - Coordinates of (M^{prime}) are (2(1, 0.5, 0) - (1, 1, 1) = (1, 0, -1)).Verification:- The vector (overrightarrow{MM^{prime}}) is perpendicular to the direction vector of the line.- The distances from (M) and (M^{prime}) to the line are equal.Thus, the symmetric point (M^{prime}) is (boxed{(1, 0, -1)}).

🔑:1. *Find the equation of the plane that is perpendicular to the given line and passes through point ( M(1 ; 1 ; 1) ).* Since the plane is perpendicular to the given line, the normal vector of the plane can be taken as the direction vector of the line: [ vec{n} = vec{s} = (1, -2, 1) ]2. *Write the equation of the desired plane.* Using the normal vector ( vec{n} = (1, -2, 1) ) and point ( M(1, 1, 1) ), we can write the plane equation in the form ( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 ): [ 1 cdot (x-1) - 2 cdot (y-1) + 1 cdot (z-1) = 0 ] Simplify this equation: [ x - 1 - 2y + 2 + z - 1 = 0 ] [ x - 2y + z = 0 ]3. *Find the coordinates of the point ( M_0 ) of intersection of the line and the plane.* First, express the parametric equations of the line: [ frac{x-2}{1} = frac{y+1.5}{-2} = frac{z-1}{1} = t ] This gives: [ left{ begin{aligned} & x = 2 + t & y = -1.5 - 2t & z = 1 + t end{aligned} right. ]4. *Substitute these parameterizations into the plane equation.* Substitute ( x = 2+t ), ( y = -1.5-2t ), and ( z = 1+t ) into the plane equation ( x - 2y + z = 0 ): [ (2+t) - 2(-1.5-2t) + (1+t) = 0 ] Simplify: [ 2 + t + 3 + 4t + 1 + t = 0 ] [ 6 + 6t = 0 ] Solve for ( t ): [ t = -1 ]5. *Find coordinates of the point of intersection ( M_0 ) by substituting ( t = -1 ).* [ left{ begin{aligned} & x = 2 + (-1) = 1 & y = -1.5 - 2(-1) = 0.5 & z = 1 + (-1) = 0 end{aligned} right. ] So the coordinates of the point ( M_0 ) are: [ M_0left( 1, 0.5, 0 right) ]6. *Find the coordinates of the point ( M' ) (symmetric with respect to the plane).* Since ( M_0 ) is the midpoint of segment ( MM' ), we use midpoint formula: [ left{ begin{aligned} & x_{M_0} = frac{x_M + x_{M'}}{2} Rightarrow x_{M'} = 2 x_{M_0} - x_M = 2 cdot 1 - 1 = 1 & y_{M_0} = frac{y_M + y_{M'}}{2} Rightarrow y_{M'} = 2 y_{M_0} - y_M = 2 cdot 0.5 - 1 = 0 & z_{M_0} = frac{z_M + z_{M'}}{2} Rightarrow z_{M'} = 2 z_{M_0} - z_M = 2 cdot 0 - 1 = -1 end{aligned} right. ] Therefore, the coordinates of ( M' ) are: [ M' (1, 0, -1) ] Conclusion:[boxed{M' (1, 0, -1)}]