🧠:Okay, so I have this problem: Find all integers ( n > 1 ) such that ( n ) divides ( 1^n + 2^n + cdots + (n-1)^n ). Hmm, interesting. Let me try to figure this out step by step.First, I need to understand what the problem is asking. It wants all integers greater than 1 where n divides the sum of each number from 1 to n-1 raised to the nth power. So, mathematically, we can write this as ( n ) divides ( S(n) ), where ( S(n) = 1^n + 2^n + cdots + (n-1)^n ). That means ( S(n) equiv 0 mod n ).Let me start by testing small values of n to see if I can find a pattern or get some intuition.Starting with n = 2:( S(2) = 1^2 = 1 ). Then check if 2 divides 1. 1 mod 2 is 1, so no. Therefore, n=2 is not a solution.n = 3:( S(3) = 1^3 + 2^3 = 1 + 8 = 9 ). Check if 3 divides 9. Yes, 9/3 = 3. So n=3 is a solution.n = 4:( S(4) = 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98 ). Check if 4 divides 98. 98 divided by 4 is 24.5, which is not an integer. So 98 mod 4 is 2. Therefore, n=4 is not a solution.n = 5:( S(5) = 1^5 + 2^5 + 3^5 + 4^5 = 1 + 32 + 243 + 1024 = 1400 ). Check if 5 divides 1400. 1400 / 5 = 280, which is an integer. So n=5 is a solution.n = 6:Compute S(6) = 1^6 + 2^6 + 3^6 + 4^6 + 5^6. Let's compute each term:1^6 = 12^6 = 643^6 = 7294^6 = 40965^6 = 15625Sum: 1 + 64 = 65; 65 + 729 = 794; 794 + 4096 = 4890; 4890 + 15625 = 20515. Now check if 6 divides 20515. Divide by 6: 20515 / 6 ≈ 3419.166..., so remainder 1. 20515 mod 6 is 1. So n=6 is not a solution.n=7:Hmm, this might take a while. Maybe there's a smarter way than computing each term. Let me see if I can find some properties or theorems that might help here.I recall that for prime numbers, Fermat's Little Theorem states that for any integer a, ( a^{p} equiv a mod p ). So if n is prime, then for each term ( k^n equiv k mod n ). Therefore, the sum ( S(n) = 1 + 2 + cdots + (n-1) ) modulo n. The sum of 1 to n-1 is ( frac{n(n-1)}{2} ). So modulo n, that's ( frac{n(n-1)}{2} mod n ). Since n divides ( n(n-1)/2 ), the term ( frac{n(n-1)}{2} ) is congruent to 0 modulo n if and only if n divides ( frac{n(n-1)}{2} ). Wait, but ( frac{n(n-1)}{2} ) divided by n is ( frac{n-1}{2} ). So, for this to be an integer, ( frac{n-1}{2} ) must be an integer. Therefore, n-1 must be even, which means n is odd. But if n is prime and odd (since 2 is the only even prime), then for primes p > 2 (which are odd), ( frac{p-1}{2} ) is an integer, so ( frac{p(p-1)}{2} ) is divisible by p. Therefore, the sum ( S(p) equiv 0 mod p ).But wait, for n=2, which is prime, the sum S(2) = 1^2 = 1, which is not divisible by 2. So maybe primes p > 2 satisfy the condition. Let's check n=3 and n=5, which are primes. For n=3, we saw S(3) = 9, which is divisible by 3. For n=5, S(5) = 1400, which is divisible by 5. So maybe all odd primes satisfy this. But what about composite numbers?Let me check n=9, which is composite and odd.n=9: Compute S(9) = 1^9 + 2^9 + ... + 8^9. This is going to be a huge number. Maybe instead of computing the actual sum, compute modulo 9.Using Fermat's Little Theorem, since 9 is composite, but perhaps we can use Euler's theorem. Wait, Euler's theorem requires that the base and modulus are coprime. For numbers k from 1 to 8, all are coprime with 9 except 3, 6, 9. But since we're only going up to 8, the numbers not coprime with 9 are 3 and 6. So for k=1,2,4,5,7,8, we have ( k^{phi(9)} = k^6 equiv 1 mod 9 ). Therefore, ( k^6 equiv 1 mod 9 ), so ( k^9 = k^{6+3} = (k^6)(k^3) equiv 1 cdot k^3 mod 9 ). So ( k^9 equiv k^3 mod 9 ) for k coprime to 9. For k=3 and 6, since they are multiples of 3, we can compute their powers modulo 9.Let's compute each term modulo 9:1^9 mod 9: 1 mod 9 =12^9 mod 9: Since 2 and 9 are coprime, 2^6 ≡1 mod 9, so 2^9 =2^(6+3)= (2^6)(2^3) ≡1*8=8 mod93^9 mod9: 3 and 9 are not coprime. 3^1=3 mod9, 3^2=0 mod9, so all higher powers of 3 are 0 mod9. So 3^9 ≡0 mod94^9 mod9: 4 and 9 coprime. 4^6 ≡1, so 4^9=4^(6+3)=4^6*4^3≡1*64≡64 mod9. 64/9=7*9=63, 64-63=1, so 64≡1 mod9. Therefore, 4^9≡1 mod95^9 mod9: 5 and 9 coprime. 5^6≡1, so 5^9=5^6*5^3≡1*125 mod9. 125/9=13*9=117, 125-117=8, so 125≡8 mod9. Thus, 5^9≡8 mod96^9 mod9: 6 and 9 not coprime. 6^1=6 mod9, 6^2=36≡0 mod9, so higher powers are 0. So 6^9≡0 mod97^9 mod9: 7 and 9 coprime. 7^6≡1, so 7^9=7^6*7^3≡1*343 mod9. 343 divided by 9: 9*38=342, so 343≡1 mod9. So 7^9≡1 mod98^9 mod9: 8 and 9 coprime. 8^6≡1, so 8^9=8^6*8^3≡1*512 mod9. 512/9: 9*56=504, 512-504=8, so 512≡8 mod9. Thus, 8^9≡8 mod9So adding all these up modulo 9:1 (from 1^9) +8 (from2^9)+0 (3^9)+1 (4^9)+8 (5^9)+0 (6^9)+1 (7^9)+8 (8^9) =1+8=9; 9+0=9; 9+1=10; 10+8=18; 18+0=18; 18+1=19; 19+8=27.27 mod9=0. So S(9) ≡0 mod9. Therefore, n=9 is a solution.Wait, but 9 is composite. So the initial thought that primes might be the only solutions is not correct. So n=3,5 (primes), n=9 (composite) are solutions. What about n=7? Let me check n=7.n=7: Since 7 is prime. Using Fermat's Little Theorem, each term k^7 ≡k mod7. So S(7)=1+2+3+4+5+6=21. 21 mod7=0. So yes, 21 is divisible by7. Therefore, n=7 is a solution. So primes are solutions, but also n=9. Hmm, interesting. So perhaps there are more solutions.Wait, n=15? Let me check n=15. Wait, but computing S(15) modulo15 would be tedious. Maybe there is a pattern here.Wait, n=3,5,7,9 are solutions. Let me check n=15. But before that, maybe there's a general approach.Let me consider the problem again: Find all integers n>1 such that n divides the sum S(n) = 1^n +2^n +...+(n-1)^n.First, let's consider the case when n is prime.If n is an odd prime p, then by Fermat's Little Theorem, each k^{p} ≡k mod p. Therefore, S(p) =1 +2 + ... + (p-1) = p(p-1)/2. So S(p) ≡0 mod p if and only if p divides p(p-1)/2, which is equivalent to p divides (p-1)/2. But since p and p-1 are coprime, p divides (p-1)/2 only if p divides 2. But p is an odd prime, so p does not divide 2. Wait, this is conflicting with our earlier conclusion. Wait, hold on, maybe I made a mistake here.Wait, S(p) = sum_{k=1}^{p-1} k^{p} ≡ sum_{k=1}^{p-1} k mod p. Therefore, S(p) ≡ sum_{k=1}^{p-1}k = (p-1)p/2 mod p. Then, (p-1)p/2 mod p is equal to ( (p-1)/2 ) * p mod p. Since p mod p is 0, the entire expression is 0 mod p. Therefore, S(p) ≡0 mod p regardless. Wait, so for any prime p, S(p) ≡0 mod p. Therefore, all primes should satisfy the condition. But when p=2, S(2)=1^2=1, which is 1 mod2=1≠0. So the exception is p=2. Therefore, all primes p≥3 satisfy the condition.Wait, but according to our calculation for n=3,5,7, they do. So primes except 2 work. So maybe all odd primes are solutions. Then for composite numbers, like n=9, we saw that n=9 works. So perhaps composite numbers can also work. So need to find composite numbers where S(n) ≡0 modn.Let me check n=15. Let's compute S(15) modulo15. Instead of computing the entire sum, let's compute each term modulo15.Note that 15=3*5. For each k from1 to14, compute k^{15} mod15.First, note that Euler's theorem: φ(15)=φ(3*5)=φ(3)*φ(5)=2*4=8. So for any k coprime to15, k^8 ≡1 mod15. Therefore, k^{15}=k^{8*1 +7}= (k^8)^1 *k^7 ≡1 *k^7 mod15. So for k coprime to15, k^{15}≡k^7 mod15.But for k not coprime to15, i.e., multiples of3 or5:If k is a multiple of3 but not5, then k=3m, where m from1 to4 (since up to14). Similarly, multiples of5 not3 are5,10. Multiples of15 would be15, but we are only going up to14.So, let's compute each term:1^15 mod15: 1^k=1, so 1.2^15 mod15: 2 and15 coprime. 2^8≡1 mod15, so 2^15=2^(8+7)=2^8*2^7≡1*(128 mod15). 128/15=8*15=120, 8, so 128≡8 mod15. So 2^15≡8.3^15 mod15: 3 and15 not coprime. 3^1=3, 3^2=9, 3^3=27≡12 mod15, 3^4=36≡6 mod15, 3^5=18≡3 mod15. The cycle repeats every4. 15 divided by4 is3 with remainder3. So 3^15≡3^(4*3 +3)= (3^4)^3 *3^3≡6^3 *12 mod15. 6^3=216≡6 mod15. 6*12=72≡12 mod15. So 3^15≡12.4^15: 4 and15 coprime. 4^8≡1 mod15. So 4^15≡4^7 mod15. Compute 4^2=16≡1, 4^4=(4^2)^2≡1^2=1, 4^7=4^4*4^2*4^1≡1*1*4=4 mod15. So 4^15≡4.5^15: 5 and15 not coprime. 5^1=5, 5^2=25≡10 mod15, 5^3=50≡5 mod15. Cycle repeats every2. 15 is odd. 15 divided by2 is7 with remainder1. So 5^15≡5^(2*7 +1)= (5^2)^7 *5^1≡10^7 *5 mod15. 10^2=100≡10 mod15, 10^4=(10^2)^2≡10^2≡10, 10^7=10^4 *10^2 *10^1≡10*10*10=1000≡10 mod15. So 10^7≡10. Therefore, 5^15≡10*5=50≡5 mod15.6^15: 6 and15 not coprime. 6=2*3. Compute 6^1=6, 6^2=36≡6 mod15. So cycle is 6,6,6,... So 6^15≡6 mod15.7^15: 7 and15 coprime. 7^8≡1, so 7^15≡7^7 mod15. Let's compute 7^2=49≡4, 7^4=(7^2)^2≡4^2=16≡1 mod15. 7^7=7^4 *7^2 *7^1≡1*4*7=28≡13 mod15. So 7^15≡13.8^15: 8 and15 coprime. 8^8≡1, so 8^15≡8^7 mod15. Compute 8^2=64≡4, 8^4=16≡1 mod15. 8^7=8^4*8^2*8^1≡1*4*8=32≡2 mod15. So 8^15≡2.9^15:9 and15 not coprime. 9=3^2. 9 mod15=9. Let's compute 9^1=9, 9^2=81≡6 mod15, 9^3=54≡9 mod15, cycle repeats every2. 15 is odd. 15 divided by2 is7 with remainder1. So 9^15≡9^(2*7 +1)= (9^2)^7 *9^1≡6^7 *9 mod15. 6^2=36≡6, so 6^7≡6 mod15. Thus, 6*9=54≡9 mod15. So 9^15≡9.10^15:10 and15 not coprime. 10=5*2. 10^1=10, 10^2=100≡10 mod15. So cycle is10,10,... So 10^15≡10.11^15:11 and15 coprime. 11≡-4 mod15. So 11^15≡(-4)^15 mod15. Since exponent is odd, (-4)^15≡-4^15. As before, 4^8≡1 mod15, so 4^15≡4^7≡4 mod15. Thus, -4 mod15=11. So 11^15≡11 mod15. Wait, maybe better to compute directly. 11^2=121≡1 mod15. Then 11^4=(11^2)^2≡1^2=1 mod15. 11^8≡1. Then 11^15=11^8 *11^4 *11^2 *11^1≡1*1*1*11≡11 mod15.12^15:12 and15 not coprime. 12=3*4. Compute 12^1=12, 12^2=144≡9 mod15, 12^3=108≡3 mod15, 12^4=36≡6 mod15, 12^5=72≡12 mod15. Cycle repeats every4. 15 divided by4 is3 with remainder3. So 12^15≡12^(4*3 +3)= (12^4)^3 *12^3≡6^3 *3 mod15. 6^3=216≡6 mod15. 6*3=18≡3 mod15. So 12^15≡3.13^15:13 and15 coprime. 13≡-2 mod15. So 13^15≡(-2)^15=-2^15 mod15. 2^8≡1, so 2^15=2^8*2^7≡1*8=8 mod15. So -8 mod15=7. Therefore, 13^15≡7 mod15.14^15:14 and15 coprime. 14≡-1 mod15. So 14^15≡(-1)^15≡-1 mod15≡14 mod15.Now sum all these up modulo15:1 (k=1) +8 (k=2) +12 (k=3) +4 (k=4) +5 (k=5) +6 (k=6) +13 (k=7) +2 (k=8) +9 (k=9) +10 (k=10) +11 (k=11) +3 (k=12) +7 (k=13) +14 (k=14)Let's add step by step:Start with 1.1 +8 =99 +12=21→21 mod15=66 +4=1010 +5=15→15 mod15=00 +6=66 +13=19→19 mod15=44 +2=66 +9=15→00 +10=1010 +11=21→66 +3=99 +7=16→11 +14=15→0So total sum modulo15 is0. Therefore, S(15)≡0 mod15. Therefore, n=15 is a solution.Hmm, so n=15 also works. So now we have n=3,5,7,9,15 as solutions. Wait, what's the pattern here?Looking at these numbers: 3,5,7 (primes), 9=3^2, 15=3*5. So perhaps numbers that are odd and have all prime factors dividing their Carmichael function or something?Alternatively, maybe numbers where n is odd. Wait, n=2, which is even, does not work. n=4, even, does not work. n=6, even, does not work. n=8? Let's check n=8.n=8: Compute S(8) modulo8. Since n=8, even.Compute each term:1^8 mod8=12^8=256 mod8=03^8. Since 3 and8 coprime. φ(8)=4, so 3^4≡1 mod8. Thus, 3^8=(3^4)^2≡1^2=1 mod8.4^8. 4 is not coprime with8. 4^2=16≡0 mod8. So higher powers also 0. So 4^8≡0.5^8. 5 and8 coprime. 5^4≡1 mod8 (since φ(8)=4). So 5^8=(5^4)^2≡1 mod8.6^8. 6 and8 not coprime. 6=2*3. 6^2=36≡4 mod8, 6^4=16≡0 mod8. So 6^8=(6^4)^2≡0^2=0 mod8.7^8. 7 and8 coprime. 7≡-1 mod8. So (-1)^8=1 mod8.So sum:1 (k=1) +0 (k=2) +1 (k=3) +0 (k=4) +1 (k=5) +0 (k=6) +1 (k=7) =1+0+1+0+1+0+1=4 mod8. Therefore, S(8)=4 mod8≠0. So n=8 does not work.Therefore, even numbers don't work. So maybe only odd numbers?Wait, n=9 (odd) works, n=15 (odd) works. Let me check n=21. Maybe n=21 is a solution. But computing S(21) modulo21 would take time. Alternatively, perhaps there's a theorem.Wait, I remember that if n is an odd integer, then the sum ( 1^n + 2^n + cdots + (n-1)^n ) might have some properties modulo n. Let's see.For odd n, consider the sum S(n) = sum_{k=1}^{n-1} k^n.Notice that for each k in 1 to n-1, pair k with n - k. So k + (n -k) = n. Let's consider the sum of k^n + (n -k)^n.Since n is odd, we can write (n -k)^n ≡ (-k)^n = -k^n modn, because n is odd, so (-k)^n = -k^n. Therefore, k^n + (n -k)^n ≡ k^n + (-k)^n ≡ k^n -k^n ≡0 modn.Wait, so if n is odd, then the terms can be paired such that each pair sums to 0 modn. However, if n is odd, then n-1 is even, so the number of terms is even. Therefore, the entire sum would be 0 modn. Wait, is that correct?Wait, let me check with n=3. Then pairs are (1,2). 1^3 +2^3=1+8=9≡0 mod3. Yes. For n=5: pairs (1,4), (2,3). 1^5 +4^5=1 +1024=1025≡1025 mod5. 1025/5=205, so 0 mod5. Similarly 2^5 +3^5=32 +243=275≡0 mod5. So sum is 0+0=0 mod5. For n=9: pairs (1,8), (2,7), (3,6), (4,5). Each pair's sum is 0 mod9, as we saw earlier. Therefore, the entire sum is 0 mod9.So if n is odd, then the sum S(n) ≡0 modn. Therefore, all odd integers n>1 satisfy the condition. Wait, is that true? Then why did n=15 work? Let me check n=7 again. For n=7, sum S(7)=1+2+3+4+5+6=21≡0 mod7. And in general, for any odd n, the sum S(n) ≡0 modn.Wait, let me formalize this. For any odd n>1, the sum ( sum_{k=1}^{n-1} k^n equiv0 modn ).Proof: Since n is odd, for each k in 1 to n-1, pair k with n -k. Then, k^n + (n -k)^n ≡k^n + (-k)^n =k^n -k^n=0 modn, since n is odd. Since there are (n-1)/2 pairs, and each pair sums to 0 modn, the total sum is 0 modn. Hence, S(n) ≡0 modn.Therefore, all odd integers n>1 satisfy the condition. Conversely, if n is even, then does S(n) ≡0 modn?For even n, let's test n=2: sum=1≡1 mod2≠0.n=4: sum=1^4 +2^4 +3^4=1+16+81=98≡2 mod4≠0.n=6: sum=20515≡1 mod6≠0.n=8: sum≡4 mod8≠0.So even n do not satisfy the condition. Therefore, the solution is all odd integers n>1.But wait, the problem says "Find all integers n >1 such that n divides...". So according to this, all odd integers n>1 work. However, let me confirm with n=9 and n=15 which we checked earlier, and they worked. Also n=1, but n>1. So the answer is all odd integers greater than1.But wait, let me check n=1, which is excluded. For n=3,5,7,9,15, all worked. So in general, for any odd n>1, the sum is divisible by n.Therefore, the conclusion is that all odd integers greater than1 satisfy the condition. Therefore, the answer is all odd integers n >1.But wait, in the case of n=1, which is excluded, but n=1 is odd. But the problem states n>1.Therefore, the solution is all odd integers n >1. So the final answer is all odd integers greater than1. So the answer is boxed{n text{ is an odd integer greater than }1} or in boxed notation, boxed{text{All odd integers } n > 1}.But let me verify this with another odd composite number, say n=21.Compute S(21) modulo21. Since n=21 is odd, according to the earlier logic, it should work. Let's confirm.For n=21, which is 3*7. For each k from1 to20, pair k with21 -k. Then each pair's sum is k^{21} + (21 -k)^{21}. Since21 is odd, (21 -k)^{21} ≡(-k)^{21}≡-k^{21} mod21. Therefore, each pair contributes 0 mod21. Since there are10 pairs, sum is0 mod21. Therefore, S(21)≡0 mod21. Therefore, n=21 is a solution.Therefore, the pattern holds. Thus, all odd integers n>1 satisfy the condition. So the answer is all odd integers greater than1.Final AnswerThe integers ( n > 1 ) are all odd numbers. Thus, the solution is boxed{text{All odd integers } n > 1}.

✅:To find all integers ( n > 1 ) such that ( n ) divides ( 1^n + 2^n + cdots + (n-1)^n ), we start by testing small values and analyzing properties of the sum modulo ( n ).1. Testing Small Values: - For ( n = 2 ), the sum ( 1^2 = 1 ) is not divisible by 2. - For ( n = 3 ), the sum ( 1^3 + 2^3 = 9 ) is divisible by 3. - For ( n = 4 ), the sum ( 1^4 + 2^4 + 3^4 = 98 ) is not divisible by 4. - For ( n = 5 ), the sum ( 1^5 + 2^5 + 3^5 + 4^5 = 1400 ) is divisible by 5. - For ( n = 6 ), the sum ( 1^6 + 2^6 + cdots + 5^6 = 20515 ) is not divisible by 6.2. Using Fermat's Little Theorem for Primes: - For a prime ( p ), Fermat's Little Theorem implies ( k^p equiv k mod p ). Thus, the sum ( 1 + 2 + cdots + (p-1) equiv 0 mod p ).3. Exploring Composite Numbers: - For composite ( n ), such as ( n = 9 ) and ( n = 15 ), we compute the sum modulo ( n ) and find it divisible by ( n ).4. Generalizing for Odd Numbers: - For odd ( n ), pair each term ( k ) with ( n - k ). Since ( n ) is odd, ( (n - k)^n equiv (-k)^n equiv -k^n mod n ). Each pair sums to 0 modulo ( n ), leading the entire sum to be 0 modulo ( n ).5. Conclusion: - All odd integers ( n > 1 ) satisfy the condition, as shown by the pairing argument and verification with specific examples. Even integers do not satisfy the condition due to non-zero remainders.Thus, the solution is (boxed{text{All odd integers } n > 1}).

🔑:To solve the problem of finding all integers ( n>1 ) such that ( n ) divides the sum ( 1^{n} + 2^{n} + 3^{n} + cdots + (n-1)^{n} ), we will break it into two cases based on whether ( n ) is odd or even.1. Case 1: ( n ) is odd If ( n ) is odd, let's analyze the behavior of the terms modulo ( n ). For any integer ( k ), we consider the term ( (n - k)^n ): [ (n - k)^n equiv (-k)^n pmod{n} ] Since ( n ) is odd, raising (-k) to the power ( n ) gives: [ (-k)^n = -k^n ] Therefore: [ (n - k)^n equiv -k^n equiv -k^n pmod{n} ] Consequently, ( k^n + (n - k)^n ) is divisible by ( n ): [ k^n + (n - k)^n equiv k^n - k^n equiv 0 pmod{n} ] We can then pair the terms in the sum ( 1^n + 2^n + cdots + (n-1)^n ) as follows: [ sum_{k=1}^{n-1} k^n = sum_{k=1}^{frac{n-1}{2}} left(k^n + (n-k)^n right) ] Each pair sums to zero modulo ( n ), hence the whole sum is divisible by ( n ). Therefore, when ( n ) is odd, the condition holds that ( n ) divides ( 1^n + 2^n + cdots + (n-1)^n ).2. Case 2: ( n ) is even Let ( n = 2^alpha m ), where ( m ) is an odd number and ( alpha ) is the 2-adic valuation of ( n ). We will use induction on (alpha) to show that for all odd ( k ): [ k^{2^alpha} equiv 1 pmod{2^alpha} ] Base case: For (alpha = 1), ( n = 2 ), and any odd ( k ): [ k^{2} equiv 1 pmod{2} ] This is trivial since ( k equiv 1 pmod{2} ). Inductive step: Assume for some ( alpha ): [ k^{2^alpha} equiv 1 pmod{2^alpha} ] For (alpha + 1): [ k^{2^{alpha + 1}} = (k^{2^alpha})^2 equiv 1^2 = 1 pmod{2^{alpha + 1}} ] This completes the induction. Therefore, for all odd ( k ): [ k^{2^alpha} equiv 1 pmod{2^alpha} ] Given this, we find that in the sum ( 1^n + 2^n + cdots + (n-1)^n ): - ( k^n equiv 0 pmod{2^alpha} ) if ( k ) is even, and ( k^n equiv 1 pmod{2^alpha} ) if ( k ) is odd. Since there are ( 2^{alpha-1} m ) odd terms: [ sum_{k text{ odd}} k^n equiv 2^{alpha-1} m equiv 2^{alpha-1} pmod{2^alpha} ] Thus: [ 1^n + 2^n + cdots + (n-1)^n equiv 2^{alpha-1} notequiv 0 pmod{2^alpha} ] Therefore, the sum is not divisible by ( 2^alpha ), and consequently not by ( n ) when ( n ) is even. Conclusion:Only the odd integers greater than 1 satisfy the condition that ( n ) divides ( 1^n + 2^n + cdots + (n-1)^n ). [boxed{text{Odd } n}]

🧠:Okay, so I need to prove that the sum from k=0 to n of (2n)! divided by (k! squared times (n minus k)! squared) equals the square of the binomial coefficient C(2n, n). Hmm, let me start by understanding what each part of this equation represents.First, the left side of the equation is a summation. For each term in the sum, we have (2n)! divided by (k!)^2 times ((n - k)!)^2. The summation runs from k=0 to k=n. The right side is the square of the binomial coefficient C(2n, n), which is the number of ways to choose n items from 2n, squared.Let me recall that the binomial coefficient C(2n, n) is equal to (2n)! / (n! n!). So, squaring that would give ( (2n)! )^2 / (n!^4 ). But the left side is a sum of terms involving (2n)! divided by (k!^2 (n - k)!^2). Hmm, maybe there's a combinatorial interpretation here?Alternatively, perhaps generating functions could be useful. The square of a binomial coefficient might relate to the convolution of coefficients in a product of generating functions. Let me think.Another approach: Maybe the left side can be rewritten in terms of binomial coefficients. Let's see. Each term in the sum is (2n)! / (k!^2 (n - k)!^2). Let's factor that as [ (2n)! / (k! (n - k)! )^2 ].Wait, the denominator is (k! (n - k)! ) squared. So, each term is (2n)! divided by (k! (n - k)! ) squared. Hmm.Alternatively, note that C(n, k) = n! / (k! (n - k)! ), so (C(n, k))^2 would be (n!^2) / (k!^2 (n - k)!^2 ). So, the term inside the sum is (2n)! / (k!^2 (n - k)!^2 ) = (2n)! / (n!^2 ) * (n!^2 ) / (k!^2 (n - k)!^2 ) = C(2n, n) * (C(n, k))^2.Wait, that seems useful. Let's check:C(2n, n) is (2n)! / (n! n!), so multiplying that by (C(n, k))^2, which is (n! / (k! (n - k)! ))^2, gives:(2n)! / (n! n!) * (n!^2 ) / (k!^2 (n - k)!^2 ) = (2n)! / (k!^2 (n - k)!^2 ). Yes! So each term in the sum is C(2n, n) multiplied by (C(n, k))^2. Therefore, the entire sum is C(2n, n) multiplied by the sum from k=0 to n of (C(n, k))^2.But we know that the sum of (C(n, k))^2 from k=0 to n is equal to C(2n, n). That's a standard combinatorial identity. For example, it can be proven by considering the coefficient of x^n in the expansion of (1 + x)^2n, which is C(2n, n), but also by recognizing that (1 + x)^n * (1 + x)^n = (1 + x)^2n, and the coefficient of x^n in the product is the sum of C(n, k) * C(n, n - k) = sum of (C(n, k))^2, which equals C(2n, n).Therefore, if we substitute back, the sum becomes C(2n, n) * C(2n, n) = (C(2n, n))^2, which is exactly the right-hand side of the equation. So, that proves the identity.Wait, let me verify each step again to make sure I didn't skip anything.Starting with the left-hand side:Sum_{k=0}^n [ (2n)! / (k!^2 (n - k)!^2 ) ]Express each term as (2n)! / (k!^2 (n - k)!^2 )Factor this as [ (2n)! / (n!^2 ) ] * [ n!^2 / (k!^2 (n - k)!^2 ) ]Which is equal to C(2n, n) * (C(n, k))^2, since C(2n, n) = (2n)! / (n! n!) and (C(n, k))^2 = (n! / (k! (n - k)! ))^2.Therefore, the sum becomes C(2n, n) * Sum_{k=0}^n (C(n, k))^2.And as per the known identity, Sum_{k=0}^n (C(n, k))^2 = C(2n, n).Hence, substituting back, the entire sum is C(2n, n) * C(2n, n) = (C(2n, n))^2, which matches the right-hand side. Therefore, the equality holds.Alternatively, maybe there's another way to approach it combinatorially. Let's consider a combinatorial proof.Suppose we have two separate groups of n people each, say Group A and Group B. The total number of people is 2n. The right-hand side, (C(2n, n))^2, can be thought of as choosing n people out of 2n, and then another n people out of the same 2n, allowing overlap. But actually, squaring C(2n, n) would represent choosing two independent committees of size n each from the 2n people. Wait, but committees could overlap. However, the left-hand side is a sum over k from 0 to n of [ (2n)! / (k!^2 (n - k)!^2 ) ].Alternatively, maybe interpret the left-hand side as counting the number of ways to choose k people from Group A and k people from Group B, but since each group has n people, if you choose k from Group A, you need to choose n - k from Group B? Wait, not sure.Wait, perhaps the term (2n)! / (k!^2 (n - k)!^2 ) can be interpreted as follows: Suppose we have two groups, each with n elements. The number of ways to choose k elements from the first group and k elements from the second group is C(n, k) * C(n, k) = (C(n, k))^2. Then, multiplied by C(2n, n), which might be some other selection. Wait, maybe not directly.Alternatively, if we consider the entire sum as summing over all possible k the product (C(n, k))^2, and then scaling by C(2n, n), which as we saw before gives the total sum as (C(2n, n))^2.But perhaps a more direct combinatorial interpretation. Let's think of forming two committees, each of size n, from 2n people. The total number is (C(2n, n))^2. Now, for the left-hand side, perhaps we can partition this count based on the number of overlapping members between the two committees.Suppose we fix k as the number of people common to both committees. Then, to form two committees with k overlapping members, we need to: choose k people to be in both committees, then choose (n - k) people from the remaining (2n - k) to be in the first committee only, and (n - k) people from the remaining (2n - k - (n - k)) = (n) people to be in the second committee only. Wait, that might not be straightforward.Alternatively, think of the committees as two subsets of size n. The number of pairs of committees that intersect in exactly k elements is C(2n, k) * C(2n - k, n - k) * C(2n - n, n - k). Wait, maybe not.Wait, let's break it down. If two committees (A and B) each have n members, and their intersection is k, then:- Choose k elements common to both: C(2n, k).- Then, choose the remaining n - k elements for committee A from the remaining 2n - k: C(2n - k, n - k).- Then, choose the remaining n - k elements for committee B from the remaining 2n - k - (n - k) = n elements: C(n, n - k) = C(n, k).Therefore, the number of such pairs is C(2n, k) * C(2n - k, n - k) * C(n, k).But simplifying this:C(2n, k) * C(2n - k, n - k) * C(n, k) = [ (2n)! / (k! (2n - k)! ) ] * [ (2n - k)! / ( (n - k)! (2n - k - (n - k))! ) ) ] * [ n! / (k! (n - k)! ) ]Simplify step by step:First term: (2n)! / (k! (2n - k)! )Second term: (2n - k)! / ( (n - k)! (n)! ) ), because 2n - k - (n - k) = n.Third term: n! / (k! (n - k)! )Multiplying them together:(2n)! / (k! (2n - k)! ) * (2n - k)! / ( (n - k)! n! ) * n! / (k! (n - k)! )Simplify:The (2n - k)! cancels out.The n! in the denominator cancels with the n! in the numerator.So we have (2n)! / (k! (n - k)! ) * 1 / ( (n - k)! k! )So overall, (2n)! / (k!^2 (n - k)!^2 )Which is exactly the term in the summation on the left-hand side. Therefore, the number of pairs of committees with intersection size k is (2n)! / (k!^2 (n - k)!^2 ). Therefore, summing over all k from 0 to n gives the total number of pairs of committees, which is (C(2n, n))^2. Hence, the equality holds.So both algebraic manipulation using known identities and combinatorial interpretation confirm the equality. Therefore, the proof is complete.Final Answerboxed{left( binom{2n}{n} right)^2}

✅:To prove the equalitysum_{k=0}^{n} frac{(2 n)!}{(k!)^{2}((n-k)!)^{2}} = left( binom{2n}{n} right)^{2}we start by analyzing the left-hand side. Each term in the summation can be rewritten using binomial coefficients. First, note that each term in the sum is:frac{(2n)!}{(k!^2 (n - k)!^2)}This can be factored as:frac{(2n)!}{n!^2} cdot frac{n!^2}{k!^2 (n - k)!^2}Recognizing the binomial coefficients, this becomes:binom{2n}{n} cdot left( binom{n}{k} right)^2Thus, the entire summation can be expressed as:binom{2n}{n} sum_{k=0}^{n} left( binom{n}{k} right)^2We use the known combinatorial identity that the sum of the squares of the binomial coefficients is:sum_{k=0}^{n} left( binom{n}{k} right)^2 = binom{2n}{n}Substituting this identity into our expression, we get:binom{2n}{n} cdot binom{2n}{n} = left( binom{2n}{n} right)^2This confirms that the left-hand side equals the right-hand side. Additionally, a combinatorial interpretation verifies this result by considering pairs of committees of size ( n ) from ( 2n ) people, where the number of such pairs is indeed ( left( binom{2n}{n} right)^2 ).Thus, the equality holds, and the final answer is:boxed{left( binom{2n}{n} right)^2}

🔑:The given problem requires us to show that for any n in mathbb{N}, the following equality holds:[ sum_{k=0}^{n} frac{(2n)!}{(k!)^2((n-k)!)^2} = left( C_{2n}^{n} right)^2 ]We start by using the binomial theorem. Recall the binomial theorem which states:[ (1+x)^{m} = sum_{k=0}^{m} C_{m}^{k} x^{k} ]First, we expand ((1+x)^{2n}) and rewrite it as ((1+x)^n cdot (1+x)^n):[ (1+x)^{2n} = (1+x)^n (1+x)^n ]Using the binomial theorem on both sides, we get:[ sum_{k=0}^{2n} C_{2n}^{k} x^k = left( sum_{i=0}^{n} C_{n}^{i} x^i right) left( sum_{j=0}^{n} C_{n}^{j} x^j right) ]Next, we are particularly interested in the coefficient of (x^n) on both sides of the equation. The left-hand side is simply (C_{2n}^n x^n).On the right-hand side, we need to consider all terms of the form (x^i cdot x^j) such that (i+j = n). Thus, the coefficient of (x^n) is:[ C_{2n}^{n} = sum_{i+j=n} C_{n}^{i} C_{n}^{j} ]Using the identity (C_{n}^{k} = C_{n}^{n-k}) for binomial coefficients, and rewriting sum limits, we get:[ C_{2n}^{n} = sum_{k=0}^{n} (C_{n}^{k})^2 ]Thus, we have shown:[ sum_{k=0}^{n} (C_{n}^{k})^2 = C_{2n}^{n} ]Now, returning to the original problem:[ sum_{k=0}^{n} frac{(2n)!}{(k!)^2((n-k)!)^2} ]Notice that (frac{(2n)!}{(k!)^2((n-k)!)^2}) can be rewritten using binomial coefficients as:[ frac{(2n)!}{n! n!} sum_{k=0}^{n} frac{(n!)^2}{k! k! (n-k)! (n-k)!} ][ = C_{2n}^{n} sum_{k=0}^{n} left( C_{n}^{k} right)^2 ]The inner sum has already been shown to be (C_{2n}^{n}):[ sum_{k=0}^{n} left( C_{n}^{k} right)^2 = C_{2n}^{n} ]Thus:[ C_{2n}^{n} sum_{k=0}^{n} left( C_{n}^{k} right)^2 = C_{2n}^{n} cdot C_{2n}^{n} = left( C_{2n}^{n} right)^2 ]Hence, we conclude:[ sum_{k=0}^{n} frac{(2n)!}{(k!)^2((n-k)!)^2} = left( C_{2n}^{n} right)^2 ][ boxed{left( C_{2n}^{n} right)^2} ]

🧠:Okay, so I need to find the range of the function y = x/(1 + x²). Hmm, the range is all the possible y-values that the function can take. Let me think about how to approach this.First, maybe I can start by considering the behavior of the function as x gets very large. If x approaches positive infinity, then x² dominates the denominator, so the denominator becomes approximately x². Then the function would be roughly x/x² = 1/x, which approaches 0. Similarly, as x approaches negative infinity, the denominator is still x², so the function behaves like x/x² = 1/x, which also approaches 0. So, the function tends to 0 at both ends. That suggests that the maximum and minimum values might be somewhere in the middle.Next, maybe I can take the derivative to find critical points. Since the function is differentiable everywhere (the denominator is always 1 + x², which is never zero), critical points will occur where the derivative is zero. Let's compute the derivative.Using the quotient rule: if y = f/g, then y' = (f'g - fg')/g².Here, f = x, so f' = 1, and g = 1 + x², so g' = 2x.Therefore, y' = [1*(1 + x²) - x*(2x)] / (1 + x²)².Simplify the numerator: (1 + x² - 2x²) = 1 - x².So y' = (1 - x²)/(1 + x²)².Set the derivative equal to zero to find critical points:(1 - x²)/(1 + x²)² = 0.The denominator is always positive, so the numerator must be zero. Thus, 1 - x² = 0 => x² = 1 => x = 1 or x = -1.So the critical points are at x = 1 and x = -1. Let's find the corresponding y-values.At x = 1: y = 1/(1 + 1) = 1/2.At x = -1: y = (-1)/(1 + 1) = -1/2.So the function has a maximum at y = 1/2 and a minimum at y = -1/2. Since the function approaches 0 as x approaches infinity and negative infinity, it seems that the range of the function should be between -1/2 and 1/2. But wait, let me verify this to make sure.Alternatively, maybe I can solve for x in terms of y and find the possible y-values that satisfy the equation. Let's try that.Starting with y = x/(1 + x²). Let's solve for x.Multiply both sides by (1 + x²):y(1 + x²) = xBring all terms to one side:yx² - x + y = 0This is a quadratic equation in terms of x: yx² - x + y = 0.For this equation to have real solutions for x, the discriminant must be non-negative. The discriminant D of a quadratic equation ax² + bx + c = 0 is D = b² - 4ac.Here, a = y, b = -1, c = y.So D = (-1)² - 4*y*y = 1 - 4y².For real solutions, D ≥ 0:1 - 4y² ≥ 0=> 4y² ≤ 1=> y² ≤ 1/4=> |y| ≤ 1/2Therefore, y must be between -1/2 and 1/2. So the range is [-1/2, 1/2].This confirms what I found earlier with the critical points. The maximum value is 1/2 and the minimum is -1/2, so the range is all real numbers from -1/2 to 1/2.Wait, just to be thorough, let me check if there are any restrictions I might have missed. For instance, could there be any y-values outside of this interval that the function can take?Suppose someone claims that y can be 1/2 or -1/2, which we saw occurs at x = 1 and x = -1. What about y = 0? That occurs when x = 0, so y = 0 is included. Then, for values between -1/2 and 1/2, does the equation y = x/(1 + x²) have solutions?Yes, because we showed that for any y in [-1/2, 1/2], the quadratic equation has real solutions. So every y in that interval is attainable by some x.Therefore, the range of the function is all real numbers y such that -1/2 ≤ y ≤ 1/2.Alternatively, if we wanted to present it in interval notation, it's [-1/2, 1/2].I think that's it. Both methods—calculus and solving for x in terms of y with the discriminant—give the same result, so I'm confident the range is from -1/2 to 1/2.

✅:The range of the function ( y = frac{x}{1 + x^2} ) is determined by analyzing critical points and solving for real solutions in terms of ( y ).1. Critical Points Analysis:- Compute the derivative using the quotient rule: [ y' = frac{(1)(1 + x^2) - x(2x)}{(1 + x^2)^2} = frac{1 - x^2}{(1 + x^2)^2}. ]- Set ( y' = 0 ), leading to ( 1 - x^2 = 0 ), which gives critical points at ( x = 1 ) and ( x = -1 ).- Evaluate ( y ) at these points: [ y(1) = frac{1}{2}, quad y(-1) = -frac{1}{2}. ]- As ( x to pminfty ), ( y to 0 ). Thus, the function attains maximum and minimum values of ( frac{1}{2} ) and ( -frac{1}{2} ), respectively.2. Algebraic Verification:- Rearrange ( y = frac{x}{1 + x^2} ) into a quadratic equation in ( x ): [ yx^2 - x + y = 0. ]- For real solutions, the discriminant must be non-negative: [ (-1)^2 - 4(y)(y) geq 0 implies 1 - 4y^2 geq 0 implies y^2 leq frac{1}{4} implies |y| leq frac{1}{2}. ]Conclusion:The range of ( y = frac{x}{1 + x^2} ) is all real numbers between ( -frac{1}{2} ) and ( frac{1}{2} ), inclusive.[boxed{left[ -frac{1}{2}, frac{1}{2} right]}]

🔑:To find the range of the function ( y = frac{x}{1 + x^2} ), we need to determine the values of ( y ) that the function can attain as ( x ) ranges over all real numbers.1. Rewrite the Function: Consider the equation ( y = frac{x}{1 + x^2} ) and rewrite it in terms of ( x ): [ y = frac{x}{1 + x^2} ] Multiply both sides of the equation by ( 1 + x^2 ) to clear the fraction: [ y(1 + x^2) = x ]2. Rearrange the Equation: Rearrange the equation to form a quadratic equation in ( x ): [ yx^2 + y = x Rightarrow yx^2 - x + y = 0 ]3. Solve the Quadratic Equation: To ensure this quadratic equation has real solutions for ( x ), its discriminant must be non-negative. The discriminant ((Delta)) of the quadratic equation ( ax^2 + bx + c = 0 ) is given by: [ Delta = b^2 - 4ac ] In our equation ( yx^2 - x + y = 0 ), we have: [ a = y, quad b = -1, quad c = y ] Thus, the discriminant is: [ Delta = (-1)^2 - 4(y)(y) = 1 - 4y^2 ]4. Ensure the Discriminant is Non-negative: To have real solutions for ( x ), we require: [ Delta geq 0 Rightarrow 1 - 4y^2 geq 0 ] Solve for ( y ): [ 1 geq 4y^2 Rightarrow frac{1}{4} geq y^2 Rightarrow y^2 leq frac{1}{4} ] Taking the square root of both sides: [ -frac{1}{2} leq y leq frac{1}{2} ]5. Conclusion: The range of the function ( y = frac{x}{1 + x^2} ) is: [ boxed{left[-frac{1}{2}, frac{1}{2}right]} ]

🧠:Okay, so I need to prove that if ( z + frac{1}{z} = 2 cos alpha ), then ( z^{n} + frac{1}{z^{n}} = 2 cos n alpha ). Hmm, let's start by understanding what the given equation tells us about ( z ). The equation ( z + frac{1}{z} = 2 cos alpha ) seems familiar. I remember that for complex numbers, expressions like ( z + frac{1}{z} ) often relate to Euler's formula. Maybe I can express ( z ) in terms of ( e^{ialpha} )?Wait, Euler's formula says that ( e^{ialpha} + e^{-ialpha} = 2 cos alpha ). That looks exactly like the right-hand side of the given equation. So maybe ( z ) is ( e^{ialpha} )? Let me check that. If ( z = e^{ialpha} ), then ( frac{1}{z} = e^{-ialpha} ), and adding them gives ( e^{ialpha} + e^{-ialpha} = 2 cos alpha ), which matches the given equation. So perhaps ( z ) is a complex number on the unit circle with angle ( alpha ).But wait, could there be other solutions for ( z )? The equation ( z + frac{1}{z} = 2 cos alpha ) is a quadratic equation in terms of ( z ). Let me solve it to see. Multiplying both sides by ( z ), we get ( z^2 + 1 = 2 z cos alpha ), which rearranges to ( z^2 - 2 z cos alpha + 1 = 0 ). Using the quadratic formula, the solutions are ( z = frac{2 cos alpha pm sqrt{4 cos^2 alpha - 4}}{2} = cos alpha pm i sin alpha ). So ( z = e^{ialpha} ) or ( z = e^{-ialpha} ). Therefore, ( z ) is either ( e^{ialpha} ) or its reciprocal ( e^{-ialpha} ).So regardless of which solution ( z ) is, ( z ) can be written as ( e^{ialpha} ) or ( e^{-ialpha} ). Let's take ( z = e^{ialpha} ) first. Then ( z^n = e^{i n alpha} ), and ( frac{1}{z^n} = e^{-i n alpha} ). Adding these gives ( e^{i n alpha} + e^{-i n alpha} = 2 cos n alpha ), which is exactly what we need to prove. Similarly, if ( z = e^{-ialpha} ), then ( z^n = e^{-i n alpha} ), and ( frac{1}{z^n} = e^{i n alpha} ), so adding them also gives ( 2 cos n alpha ).Therefore, in both cases, ( z^n + frac{1}{z^n} = 2 cos n alpha ). So the proof seems straightforward if we express ( z ) in exponential form. But maybe there's another way to approach this using induction or trigonometric identities? Let me think.If I use mathematical induction, for the base case ( n = 1 ), the equation holds by the given condition. Assume it holds for some integer ( k ), that is ( z^{k} + frac{1}{z^{k}} = 2 cos k alpha ). Then for ( n = k + 1 ), maybe I can express ( z^{k+1} + frac{1}{z^{k+1}} ) in terms of ( z^{k} + frac{1}{z^{k}} ) and ( z^{k-1} + frac{1}{z^{k-1}} ). Let's try that.Consider multiplying ( z + frac{1}{z} = 2 cos alpha ) by ( z^{k} + frac{1}{z^{k}} ):[ left( z + frac{1}{z} right) left( z^{k} + frac{1}{z^{k}} right) = z^{k+1} + z^{k-1} + frac{1}{z^{k-1}} + frac{1}{z^{k+1}} ]This simplifies to:[ z^{k+1} + frac{1}{z^{k+1}} + z^{k-1} + frac{1}{z^{k-1}} ]But the left-hand side is ( 2 cos alpha cdot 2 cos k alpha = 4 cos alpha cos k alpha ). So:[ z^{k+1} + frac{1}{z^{k+1}} = 4 cos alpha cos k alpha - left( z^{k-1} + frac{1}{z^{k-1}} right) ]If the induction hypothesis holds for ( k ) and ( k - 1 ), then ( z^{k-1} + frac{1}{z^{k-1}} = 2 cos (k - 1) alpha ). Therefore:[ z^{k+1} + frac{1}{z^{k+1}} = 4 cos alpha cos k alpha - 2 cos (k - 1) alpha ]Now, using the trigonometric identity for the product of cosines:[ 2 cos alpha cos k alpha = cos (k + 1) alpha + cos (k - 1) alpha ]Therefore, multiplying by 2:[ 4 cos alpha cos k alpha = 2 cos (k + 1) alpha + 2 cos (k - 1) alpha ]Substituting back:[ z^{k+1} + frac{1}{z^{k+1}} = 2 cos (k + 1) alpha + 2 cos (k - 1) alpha - 2 cos (k - 1) alpha ]Simplifying, the ( 2 cos (k - 1) alpha ) terms cancel out:[ z^{k+1} + frac{1}{z^{k+1}} = 2 cos (k + 1) alpha ]Which completes the induction step. Therefore, by mathematical induction, the formula holds for all positive integers ( n ).Alternatively, we can use De Moivre's theorem. If ( z = cos alpha + i sin alpha ), then ( z^n = cos n alpha + i sin n alpha ), and ( frac{1}{z^n} = cos n alpha - i sin n alpha ). Adding them gives ( 2 cos n alpha ). Similarly, if ( z = cos alpha - i sin alpha ), the same result holds.Another approach might involve using recurrence relations. Since ( z^n + frac{1}{z^n} ) can be expressed in terms of ( z^{n-1} + frac{1}{z^{n-1}} ) and ( z^{n-2} + frac{1}{z^{n-2}} ), similar to the induction step. Let me verify that.Assume ( S_n = z^n + frac{1}{z^n} ). Then, from the previous multiplication:( S_1 cdot S_n = S_{n+1} + S_{n-1} ).Given that ( S_1 = 2 cos alpha ), we have:( 2 cos alpha cdot S_n = S_{n+1} + S_{n-1} ).Rearranged:( S_{n+1} = 2 cos alpha cdot S_n - S_{n-1} ).This is the same recurrence relation as the one for ( 2 cos n alpha ). Let's check if ( 2 cos n alpha ) satisfies the same recurrence.Using the trigonometric identity:( 2 cos (n + 1) alpha = 2 cos alpha cos n alpha - 2 sin alpha sin n alpha ).But this doesn't directly match. Wait, perhaps using another identity:We know that ( cos (n + 1) alpha + cos (n - 1) alpha = 2 cos alpha cos n alpha ).Therefore,( 2 cos (n + 1) alpha = 2 cdot 2 cos alpha cos n alpha - 2 cos (n - 1) alpha ).Wait, no. Let's rearrange:From ( cos (n + 1) alpha + cos (n - 1) alpha = 2 cos alpha cos n alpha ), we get:( cos (n + 1) alpha = 2 cos alpha cos n alpha - cos (n - 1) alpha ).Multiplying both sides by 2:( 2 cos (n + 1) alpha = 2 cdot 2 cos alpha cos n alpha - 2 cos (n - 1) alpha ).But this is exactly the same recurrence as ( S_{n+1} = 2 cos alpha S_n - S_{n-1} ). Therefore, if ( S_n = 2 cos n alpha ), then the recurrence holds. Since the base cases ( S_1 = 2 cos alpha ) and ( S_0 = z^0 + frac{1}{z^0} = 1 + 1 = 2 = 2 cos 0 alpha ), which is consistent, the recurrence relation confirms that ( S_n = 2 cos n alpha ) for all non-negative integers ( n ).So there are multiple ways to approach this proof: using Euler's formula and exponential form, mathematical induction, recurrence relations, or trigonometric identities. All these methods converge to the same conclusion. Therefore, the proof is solid.

✅:To prove that ( z^{n} + frac{1}{z^{n}} = 2 cos n alpha ) given ( z + frac{1}{z} = 2 cos alpha ), we proceed as follows: Step 1: Solve for ( z )The equation ( z + frac{1}{z} = 2 cos alpha ) is quadratic in ( z ). Multiplying through by ( z ):[z^2 - 2z cos alpha + 1 = 0.]Using the quadratic formula:[z = frac{2 cos alpha pm sqrt{4 cos^2 alpha - 4}}{2} = cos alpha pm i sin alpha = e^{pm i alpha}.]Thus, ( z = e^{i alpha} ) or ( z = e^{-i alpha} ).--- Step 2: Use Exponential Form to Compute ( z^n + frac{1}{z^n} )For ( z = e^{i alpha} ):[z^n = e^{i n alpha}, quad frac{1}{z^n} = e^{-i n alpha}.]Adding these:[z^n + frac{1}{z^n} = e^{i n alpha} + e^{-i n alpha} = 2 cos n alpha.]For ( z = e^{-i alpha} ):[z^n = e^{-i n alpha}, quad frac{1}{z^n} = e^{i n alpha}.]Similarly:[z^n + frac{1}{z^n} = e^{-i n alpha} + e^{i n alpha} = 2 cos n alpha.]--- Step 3: Alternative Proof via Mathematical InductionBase Case (( n = 1 )): Given ( z + frac{1}{z} = 2 cos alpha ), the base case holds.Inductive Step: Assume ( z^k + frac{1}{z^k} = 2 cos k alpha ) holds for some ( k in mathbb{N} ). Multiply the given equation by ( z^k + frac{1}{z^k} ):[left(z + frac{1}{z}right)left(z^k + frac{1}{z^k}right) = z^{k+1} + z^{k-1} + frac{1}{z^{k-1}} + frac{1}{z^{k+1}}.]This simplifies to:[z^{k+1} + frac{1}{z^{k+1}} + z^{k-1} + frac{1}{z^{k-1}}.]Using ( z + frac{1}{z} = 2 cos alpha ) and the inductive hypothesis:[2 cos alpha cdot 2 cos k alpha = z^{k+1} + frac{1}{z^{k+1}} + 2 cos (k-1)alpha.]By the trigonometric identity ( 2 cos alpha cos k alpha = cos (k+1)alpha + cos (k-1)alpha ):[4 cos alpha cos k alpha = 2 cos (k+1)alpha + 2 cos (k-1)alpha.]Thus:[z^{k+1} + frac{1}{z^{k+1}} = 2 cos (k+1)alpha.]By induction, the result holds for all ( n in mathbb{N} ).--- ConclusionIn all cases, ( z^{n} + frac{1}{z^{n}} = 2 cos n alpha ). This follows from the exponential representation of ( z ), mathematical induction, or recurrence relations derived from trigonometric identities.[boxed{z^{n} + frac{1}{z^{n}} = 2 cos n alpha}]

🔑:We are given the equation:[z + frac{1}{z} = 2 cos alpha]Our goal is to prove that:[z^n + frac{1}{z^n} = 2 cos n alpha]We will use mathematical induction to prove this statement.1. Base Case: For ( n = 1 ): [ z^1 + frac{1}{z^1} = z + frac{1}{z} ] We are given that: [ z + frac{1}{z} = 2 cos alpha ] Therefore, the base case holds true.2. Inductive Step: Assume the statement is true for ( n = k ), i.e., assume: [ z^k + frac{1}{z^k} = 2 cos k alpha ] We need to show that the statement is true for ( n = k+1 ). Consider: [ z^{k+1} + frac{1}{z^{k+1}} ] Using the given equation, we can write the product: [ (z + frac{1}{z})(z^k + frac{1}{z^k}) = 2 cos alpha cdot 2 cos k alpha ] Expanding and rearranging: [ (z + frac{1}{z})(z^k + frac{1}{z^k}) = z^{k+1} + z^{k-1} + frac{1}{z^k z} + frac{1}{z^{k+1}} ] Using: [ z^{k+1} + frac{1}{z^{k+1}} = (z + frac{1}{z})(z^k + frac{1}{z^k}) - z^{k-1} - frac{1}{z^{k-1}} ] Substituting the inductive hypothesis ( z^k + frac{1}{z^k} = 2 cos k alpha ): [ (z + frac{1}{z})(2 cos k alpha) = 2 cos alpha cdot 2 cos k alpha = 4 cos alpha cos k alpha ] Therefore: [ z^{k+1} + frac{1}{z^{k+1}} = 4 cos alpha cos k alpha - (z^{k-1} + frac{1}{z^{k-1}}) ] Using the double-angle identity for cosines: [ 2 cos (k+1)alpha + 2 cos (k-1) alpha ] Rearranging: [ z^{k+1} + frac{1}{z^{k+1}} = 2 cos (k+1) alpha + 2 cos (k-1) alpha - 2 cos (k-1) alpha ] Simplifying the equation: [ z^{k+1} + frac{1}{z^{k+1}} = 2 cos (k+1) alpha ] Therefore, by induction, the assumption holds true for all ( n in ℕ ). Conclusion:[boxed{z^n + frac{1}{z^n} = 2 cos n alpha}]

🧠:Okay, so I need to find the number of ordered pairs of positive integers (x, y) with x and y both less than or equal to 2020 such that the expression 3x² + 10xy + 3y² is a power of a prime number. Hmm, let me break this down step by step.First, let's look at the quadratic form 3x² + 10xy + 3y². Maybe I can factor this expression to simplify it. Let me try factoring:3x² + 10xy + 3y². Let's see if this can be factored into two binomials. The coefficients are 3 and 3 for x² and y², so maybe (3x + ay)(bx + 3y). Let's check:(3x + ay)(bx + 3y) = 3b x² + (9x y + a b x y) + 3a y². Comparing to the original expression 3x² + 10xy + 3y², we need 3b = 3, so b = 1. Similarly, 3a = 3, so a = 1. Then the middle term becomes 9xy + 1*1*xy = 10xy. Perfect! So the expression factors as (3x + y)(x + 3y).So, 3x² + 10xy + 3y² = (3x + y)(x + 3y). Now, the problem states that this product should be a power of a prime. Let's denote that prime by p. So, (3x + y)(x + 3y) = p^k, where p is a prime and k is a positive integer.Since p^k is a prime power, its only divisors are 1, p, p², ..., p^k. Therefore, the two factors (3x + y) and (x + 3y) must themselves be powers of the same prime p. Let me write this down:Let 3x + y = p^m and x + 3y = p^n, where m and n are non-negative integers such that m + n = k. But since both 3x + y and x + 3y are positive integers (as x and y are positive integers), m and n must be at least 1.Now, since 3x + y and x + 3y are both powers of the same prime p, we can set up equations:3x + y = p^m x + 3y = p^n We need to solve these equations for positive integers x, y, given that p is a prime and m, n are positive integers. Also, x and y must satisfy x, y ≤ 2020.To solve this system, let's express x and y in terms of p^m and p^n. Let me write the system of equations:Equation 1: 3x + y = p^m Equation 2: x + 3y = p^n Let me solve this system. Let's multiply Equation 1 by 3: 9x + 3y = 3p^m Then subtract Equation 2: 9x + 3y - (x + 3y) = 3p^m - p^n Which simplifies to: 8x = 3p^m - p^n Therefore, x = (3p^m - p^n)/8 Similarly, let's solve for y. Multiply Equation 2 by 3: 3x + 9y = 3p^n Subtract Equation 1: 3x + 9y - (3x + y) = 3p^n - p^m Which simplifies to: 8y = 3p^n - p^m Thus, y = (3p^n - p^m)/8 Since x and y must be positive integers, both (3p^m - p^n) and (3p^n - p^m) must be divisible by 8, and the results of these divisions must be positive integers. Let's note that.So, the conditions for x and y to be positive integers are:1. 3p^m ≡ p^n mod 8 2. 3p^n ≡ p^m mod 8 3. 3p^m > p^n 4. 3p^n > p^m 5. x = (3p^m - p^n)/8 ≤ 2020 6. y = (3p^n - p^m)/8 ≤ 2020 Additionally, since p is a prime, we need to consider different primes and exponents m and n such that all conditions are satisfied.First, let's analyze the congruences. Let's consider the possible residues of primes modulo 8.Primes can be 2, or odd primes. Odd primes can be 1, 3, 5, or 7 mod 8. Let's handle p=2 first.Case 1: p=2.Then, we have 3*2^m ≡ 2^n mod 8, and 3*2^n ≡ 2^m mod 8.Let's compute 3*2^m mod 8:If m >= 3, 2^m ≡ 0 mod 8, so 3*2^m ≡ 0 mod 8. Similarly, 2^n mod 8 is 0 if n >=3. So for m, n >=3, we have 0 ≡ 0 mod8, which is okay. But let's check for smaller exponents.If m=1: 3*2=6 mod8. So 3*2^m=6 mod8. Then 2^n must be congruent to 6 mod8. But 2^n mod8 is 2,4,0 for n=1,2,>=3. So 6 mod8 is impossible. Hence, m=1 is invalid.m=2: 3*4=12≡4 mod8. So 3*2^m=4 mod8. Then 2^n ≡4 mod8. Which happens when n=2. So possible.Similarly, check 3*2^n ≡2^m mod8. If n=2, then 3*4=12≡4 mod8. So 2^m ≡4 mod8. Thus m=2. So p=2, m=n=2. Then x=(3*2² -2²)/8=(12 -4)/8=8/8=1, y=(3*2² -2²)/8= same=1. So x=1, y=1. Check 3x² +10xy +3y²=3 +10 +3=16=2^4, which is a prime power. So that's valid.Now, check if m and n can be higher. Suppose m >=3, n >=3. Then 3*2^m -2^n must be divisible by8. Since 2^m and 2^n are multiples of8, so 3*2^m -2^n=2^3*(3*2^{m-3} -2^{n-3}). So the difference is divisible by8. Similarly for y. So x= (3*2^m -2^n)/8, y=(3*2^n -2^m)/8.But we need x and y positive. So:3*2^m >2^n and 3*2^n >2^m Let me write these inequalities:From 3*2^m >2^n, take log base2:log2(3) + m > n Similarly, log2(3) + n > m Given that log2(3) ≈1.58496.So n < m +1.58496 and m < n +1.58496 Since m and n are integers, this implies that |m -n| <=1.So possible cases: m =n, m =n +1, or n =m +1.Let's check these.Subcase 1: m =n.Then x=(3*2^m -2^m)/8=(2*2^m)/8=2^{m+1}/8=2^{m-3}.Similarly y=2^{m-3}.Thus x=y=2^{m-3}. Since x and y must be positive integers, m-3 >=0 => m >=3.But x=2^{m-3} <=2020, so 2^{m-3} <=2020. Let's solve for m.Take log2: m-3 <= log2(2020) ≈11. So m-3 <=11 => m <=14. So m can be from 3 to14.Thus, for m=3 to14, p=2, x=y=2^{m-3}. For each m, x and y are equal and equal to 2^{m-3}. Check if x and y <=2020:For m=14, x=2^{11}=2048. But 2048>2020. So m=14 gives x=2048 which is over. So m can be up to13: m=13 gives x=2^{10}=1024<=2020. m=14 is too big.Wait, wait, m-3<=log2(2020). log2(2020)= approx 11. So m-3 <=11, m<=14. But 2^{11}=2048, which is bigger than 2020, so indeed, m=14 would give x=2^{11}=2048>2020. So the maximum m is13, which gives x=2^{10}=1024<=2020.Wait, wait, let me check:If m=13: x=2^{13-3}=2^{10}=1024<=2020. Yes. m=14 would be 2^{11}=2048>2020, which is invalid. So m can be from3 to13. That's 11 values (3,4,...,13). Thus for each m from3 to13, we have x=y=2^{m-3}. So that's 11 ordered pairs (x,y)=(1024,1024), etc. Wait, but x and y are equal here.But wait, in the case m=n, we have x=y. But in this case, each m from3 to13 gives one pair (x,y)=(2^{m-3},2^{m-3}). So 11 pairs here.Subcase 2: m =n +1.So m =n +1. Then let's substitute into x and y.x=(3*2^{n+1} -2^n)/8=(3*2*2^n -2^n)/8=(6*2^n -2^n)/8=5*2^n /8=5*2^{n-3}.Similarly, y=(3*2^n -2^{n+1})/8=(3*2^n -2*2^n)/8=2^n /8=2^{n-3}.So x=5*2^{n-3}, y=2^{n-3}. Since x and y must be positive integers, n-3 >=0 =>n>=3.But x=5*2^{n-3} must be <=2020. Let's solve 5*2^{n-3} <=2020.2^{n-3} <=404. log2(404)= approx8.66. So n-3<=8 =>n<=11. Therefore, n can be3,4,...,11. Thus m=n+1 would be4,...,12.So for each n=3 to11, m=4 to12, x=5*2^{n-3}, y=2^{n-3}. Check x and y:For n=11: x=5*2^8=5*256=1280<=2020, y=2^8=256<=2020. So okay.Similarly, y=2^{n-3} must be <=2020, which for n=11, 2^8=256 okay.Thus, n=3 to11 gives 9 values. Each gives a distinct pair (x,y)=(5*2^{n-3}, 2^{n-3}).Similarly, Subcase3: n =m +1.By symmetry, this will give x=2^{m-3}, y=5*2^{m-3}. So similar to Subcase2, but with x and y swapped. Since the original problem counts ordered pairs, (x,y) and (y,x) are different unless x=y.So for each m=3 to11, n=m+1, we get pairs (2^{m-3},5*2^{m-3}). Similarly, m=3 to11, so 9 pairs. Thus, total in Subcases2 and3: 9 +9=18.Thus, for p=2:Case1: m=n: 11 pairs Case2: m=n+1 and n=m+1: 18 pairs Wait, no, m=n+1 and n=m+1 are different. Wait, actually, Subcase2 is m=n+1, which gives (x,y)=(5*2^{n-3}, 2^{n-3}) Subcase3 is n=m+1, which gives (x,y)=(2^{m-3},5*2^{m-3}) So total pairs here would be 9 (from Subcase2) +9 (from Subcase3)=18. Plus the 11 pairs from m=n.Wait, but for m=n, since x=y, these are ordered pairs where x=y, so each counts once. So total for p=2:11 (from m=n) +9 (from m=n+1) +9 (from n=m+1) = 11 +18=29? Wait, but maybe some overlap? Wait, no, because in m=n, the pairs are (x,x), while in m=n+1 and n=m+1, they are (5k, k) and (k,5k), which are distinct.So total of 29 ordered pairs for p=2.Wait, let's check for n=3 in Subcase2: n=3, m=4. Then x=5*2^{0}=5, y=1. So (5,1). Similarly, for m=3, n=4 in Subcase3: x=1, y=5. So (1,5). So these are distinct. So yes, 9 each for Subcases2 and3.So total 11+9+9=29.But wait, earlier with m=n, starting from m=3 to13: pairs (1,1) when m=3? Wait, no. Wait, when m=3, x=2^{0}=1, y=1. Wait, no, when m=n=3, x=(3*2^3 -2^3)/8=(24 -8)/8=16/8=2. Wait, hold on, let me recast.Wait, earlier, for m=n, x=(3*2^m -2^m)/8=(2*2^m)/8=2^{m+1}/8=2^{m-3}. Wait, when m=3, x=2^{0}=1, so y=1. Then m=4: x=2^{1}=2, y=2; m=5: x=4, y=4, etc. So the pairs when m=n are (1,1), (2,2), (4,4), ..., up to (1024,1024) for m=13. So 11 pairs from m=3 to m=13 (since m=3 gives 1, m=4 gives2,..., m=13 gives 1024). 11 terms.But wait, m=3 gives x=1, which is okay. Then m=3: x=1, y=1. Then 3x² +10xy +3y²=3 +10 +3=16=2^4. Which is correct.Wait, but when m=3 and n=3, the equations are:3x + y =2^3=8 x +3y=8 Solving, 3x + y=8 and x +3y=8. Subtract: 2x -2y=0 =>x=y. Then 4x=8 =>x=2. So x=2, y=2. Wait, but according to the previous formula, x=2^{m-3}=2^{0}=1? Contradiction. Hmm, so I made a mistake here.Wait, wait, perhaps my mistake is in the derivation when m=n. Let me re-examine.If m=n, then 3x + y =p^m and x +3y =p^m. Therefore, the two equations are:3x + y = p^m x + 3y = p^m Subtracting the second equation from the first:2x - 2y =0 =>x=y.Then substituting back into 3x +x=4x= p^m =>x= p^m /4.Since x must be a positive integer, p^m must be divisible by4. If p=2, then 2^m divisible by4 => m >=2. Then x=2^{m}/4=2^{m-2}. So x=2^{m-2}, y=2^{m-2}.Wait, this contradicts the earlier result where I thought x=2^{m-3}. So where did I go wrong?Wait, in the previous steps when I solved for x and y:From the system:3x + y = p^m x + 3y = p^n If m=n, then p^m = p^n. So the equations become:3x + y = p^m x + 3y = p^m Subtracting: 2x -2y=0 =>x=y. Then substituting back, 4x=p^m. So x=p^m /4. Therefore, p^m must be divisible by4, so if p=2, m must be >=2. Then x=2^{m-2}, y=2^{m-2}.But earlier, using the formula x=(3p^m -p^n)/8, when m=n, that would be (3p^m -p^m)/8=(2p^m)/8=p^m /4. Which is correct. So x=2^{m}/4=2^{m-2} when p=2. So my mistake earlier was miscalculating x=2^{m-3}. Actually, x=2^{m-2} when p=2 and m=n.So let's correct this. For p=2 and m=n, x=2^{m-2}. Then x must be a positive integer, so m >=2. Also, x=2^{m-2} <=2020. So 2^{m-2} <=2020. Therefore, m-2 <= log2(2020) ≈11. So m <=13. Therefore, m=2 to13.But wait, in the equations above, m and n are exponents in 3x + y = p^m and x +3y = p^n. If m=n, then m can be from2 upwards. But earlier, when we considered p=2, in the case m=n, we had x=2^{m-2}, so m needs to be at least2, and m<=13 since 2^{13-2}=2^11=2048>2020. Wait, 2^{11}=2048, which is greater than2020, so the maximum m such that x=2^{m-2} <=2020 is m=12: 2^{10}=1024<=2020. m=13:2^{11}=2048>2020. Therefore, m=2 to12. Thus, m=2,3,...,12. That's11 values.Wait, let's check m=2: x=2^{0}=1, y=1. Then 3x + y=4=2², x +3y=4=2². So that's valid.For m=12: x=2^{10}=1024, y=1024. Check if x,y<=2020: yes, 1024 is okay. Then 3x + y=3*1024 +1024=4096=2^12, and x +3y=1024 +3*1024=4096=2^12. So that works.Thus, in the case p=2 and m=n, we have m=2 to12, which is11 pairs (x,y)=(1,1),(2,2),..., (1024,1024).Earlier, I thought m starts at3, but actually, m starts at2. So the earlier calculation was wrong. Let's correct all.So for p=2:Case1: m=n: m=2 to12, x=y=2^{m-2}. So 11 pairs.Case2: m=n+1. Let's recast with the corrected formulas.If m=n+1, then:x=(3*2^{n+1} -2^n)/8=(6*2^n -2^n)/8=5*2^n /8=5*2^{n-3}Similarly, y=(3*2^n -2^{n+1})/8=(3*2^n -2*2^n)/8=2^n /8=2^{n-3}Thus, x=5*2^{n-3}, y=2^{n-3}. For x and y to be positive integers, n-3 >=0 =>n>=3. So n>=3, and x=5*2^{n-3} <=2020.Thus, 5*2^{n-3} <=2020 =>2^{n-3} <=404. As before, log2(404)=~8.66, so n-3 <=8 =>n <=11. So n=3 to11, which is9 values.Similarly, for each n=3 to11, we have x=5*2^{n-3}, y=2^{n-3}. So pairs like (5*1,1), (5*2,2), ..., up to n=11: x=5*2^8=1280, y=256.Similarly, for Subcase3: n=m+1. Then x=2^{m-3}, y=5*2^{m-3}. So m >=3. Then x=2^{m-3} <=2020 and y=5*2^{m-3} <=2020.So for y=5*2^{m-3} <=2020: 5*2^{m-3} <=2020 =>2^{m-3} <=404 =>m-3 <=8 =>m<=11. Thus m=3 to11, which is9 values. So pairs (1,5), (2,10), ..., (256,1280).Therefore, Subcases2 and3 give 9+9=18 ordered pairs.Therefore, total for p=2:11 +18=29 ordered pairs.But wait, let's check if any of these pairs exceed x,y<=2020. For example, when n=11 in Subcase2, x=5*2^{8}=1280<=2020, y=256<=2020. Similarly, when m=11 in Subcase3, x=256, y=1280. Both okay.So total 29 ordered pairs for p=2. Now, let's check if these are all. But wait, also, when m and n are greater, but since x and y can't exceed2020, we can't have m or n beyond certain points. But we already considered up to m=12 for m=n, which gives x=1024. Then m=13 would give x=2048, which is over. So the counts are correct.So p=2 gives29 ordered pairs.Now, let's move to the other primes. So p is an odd prime. Then p is congruent to1,3,5,7 mod8.We need to consider the congruences:From earlier, for the equations:3x + y = p^m x + 3y = p^n We derived:x=(3p^m - p^n)/8 y=(3p^n - p^m)/8 For x and y to be positive integers, the numerators must be divisible by8, and the results positive.Given that p is odd, p is congruent to1,3,5,7 mod8. Let's analyze the congruence conditions.First, 3p^m - p^n ≡0 mod8 and 3p^n - p^m ≡0 mod8.Since p is odd, p ≡1,3,5,7 mod8. Let's compute 3p^m mod8 and p^n mod8.Note that for odd primes, p^k mod8 cycles with period depending on p.For example:If p ≡1 mod8: p^k ≡1 mod8 for any k.If p ≡3 mod8: 3^1=3, 3^2=9≡1 mod8, 3^3=3*1=3, 3^4=1, etc. So cycle length2.Similarly, p≡5 mod8: 5^1=5,5^2=25≡1 mod8, cycle length2.p≡7 mod8:7^1=7,7^2=49≡1 mod8, cycle length2.So regardless of p, if p is odd, p^k mod8 cycles with period1 or2. Let's tabulate:- If p ≡1 mod8: p^k ≡1 mod8 for any k.- If p ≡3 mod8: p^k ≡3 if k odd, 1 if k even.- If p ≡5 mod8: p^k ≡5 if k odd,1 if k even.- If p ≡7 mod8: p^k ≡7 if k odd,1 if k even.Therefore, for odd primes, p^k mod8 is either1,3,5,7 depending on p and k.Given this, let's analyze the congruences 3p^m ≡p^n mod8 and 3p^n ≡p^m mod8.Case2: p is an odd prime.So, we have two congruence equations:1. 3p^m ≡p^n mod8 2. 3p^n ≡p^m mod8 We need to find primes p and exponents m,n such that both congruences hold.Let's consider different cases based on p mod8.Subcase1: p ≡1 mod8.Then p^k ≡1 mod8 for any k. So:1. 3*1 ≡1 mod8 =>3≡1 mod8, which is false. Therefore, no solutions for primes p≡1 mod8.Subcase2: p ≡3 mod8.Then p^k ≡3 if k odd, 1 if k even.So, let's consider the possible parities of m and n.Case1: m odd, n odd.Then p^m ≡3, p^n≡3 mod8.So equation1:3*3=9≡1≡3 mod8? 9 mod8=1, but equation1 requires 3p^m ≡p^n mod8 =>1≡3 mod8? No, not possible.Case2: m odd, n even.Then p^m ≡3, p^n≡1.Equation1:3*3=9≡1≡1 mod8:1≡1 mod8, which holds.Equation2:3*1=3≡3≡p^m mod8. Since p^m=3, equation2:3≡3 mod8 holds. So both congruences hold.Case3: m even, n odd.Similarly:p^m ≡1, p^n≡3.Equation1:3*1=3≡3 mod8, which must equal p^n≡3: yes, holds.Equation2:3*3=9≡1≡p^m≡1 mod8: yes, holds.Case4: m even, n even.p^m≡1, p^n≡1.Equation1:3*1≡1 mod8:3≡1 mod8, no.Thus, for p≡3 mod8, the only valid cases are when one exponent is even and the other is odd. So m and n have opposite parities.Similarly for p≡5 and p≡7 mod8.Let's check p≡5 mod8.Subcase3: p≡5 mod8.Then p^k≡5 if k odd,1 if k even.Equation1:3p^m ≡p^n mod8.Case1: m odd, n odd.3*5=15≡7≡5 mod8? No.Case2: m odd, n even.3*5=15≡7≡1 mod8? 7≡1? No.Case3: m even, n odd.3*1=3≡5 mod8? No.Case4: m even, n even.3*1≡1 mod8? 3≡1? No. So no solutions for p≡5 mod8.Subcase4: p≡7 mod8.Then p^k≡7 if k odd,1 if k even.Equation1:3p^m ≡p^n mod8.Case1: m odd, n odd.3*7=21≡5≡7 mod8? No.Case2: m odd, n even.3*7=21≡5≡1 mod8? No.Case3: m even, n odd.3*1=3≡7 mod8? No.Case4: m even, n even.3*1≡1 mod8? 3≡1? No. So no solutions for p≡7 mod8.Thus, only primes p≡3 mod8 can satisfy the congruences, with exponents m and n of opposite parity.So for odd primes, only p≡3 mod8 can yield solutions, with m and n of opposite parity (one even, one odd).Now, let's focus on primes p≡3 mod8. Let's denote p=8k+3 for some integer k.Given that, we need to solve:3x + y =p^m x +3y =p^n with m and n of opposite parity, and x=(3p^m -p^n)/8, y=(3p^n -p^m)/8 being positive integers.Additionally, x and y must be <=2020.First, we need to find all primes p≡3 mod8, and exponents m,n with opposite parity such that x and y are positive integers and <=2020.Let's analyze the positivity conditions:3p^m -p^n >0 =>3p^m >p^n =>3 >p^{n -m} and 3p^n -p^m >0 =>3p^n >p^m =>3 >p^{m -n}Since p is at least3 (the smallest prime ≡3 mod8 is3), and p^{|m -n|} >=3^{1}=3.But 3 >p^{|m -n|} is only possible if |m -n|=0 and p^{0}=1 <3. But m and n have opposite parity, so |m -n| is at least1. Hence, p^{|m -n|} >=3, which contradicts 3>p^{|m -n|}.Wait, this is a problem. Wait, this suggests that for primes p>=3, which is always true as p is a prime≡3 mod8 (so p=3,11,19,...), and |m -n| >=1 (since m and n have opposite parity), then p^{|m -n|} >=3, hence 3p^m >p^n would require that 3 >p^{n -m}. But if n >m, then n -m >=1, so p^{n -m} >=p >=3, hence 3 >p^{n -m} would require p^{n -m} <3, but p>=3 and exponent>=1 gives p^{n -m} >=3. Thus, impossible.Similarly, if m >n, then m -n >=1, so p^{m -n} >=p >=3, so 3p^n >p^{m} =>3 >p^{m -n} >=3, so 3>3 is false. Hence, impossible.Therefore, there are no solutions for odd primes p≡3 mod8? But this contradicts our earlier congruence analysis.Wait, perhaps I made a mistake here.Wait, let's re-examine the positivity conditions:From x=(3p^m -p^n)/8 >0 =>3p^m >p^n and y=(3p^n -p^m)/8 >0 =>3p^n >p^m So both 3p^m >p^n and 3p^n >p^m must hold.Let me denote a =p^m and b =p^n. Then we have 3a >b and 3b >a. These inequalities imply that a and b are positive real numbers such that a/3 < b <3a.But since a and b are powers of the same prime p, let’s assume that m ≠n. Because if m =n, then a=b, and 3a >a and 3a >a hold, but in our earlier analysis for p≡3 mod8, m and n must have opposite parity, so m ≠n.But since a =p^m and b =p^n, then if m >n, then a =p^{m} >b =p^{n}, so 3a >b is automatic, but 3b >a would require 3p^{n} >p^{m} =>3 >p^{m -n}. Since m >n, m -n is at least1 (since m and n have opposite parity, difference is at least1). Thus, 3 >p^{m -n}. But p is at least3, so p^{m -n} >=3^{1}=3. Hence, 3>3 is false. Similarly, if n >m, then 3p^{m} >p^{n} =>3>p^{n -m} >=p >=3. So 3>3 again false. Hence, no solutions.This is a contradiction. There must be a mistake. Wait, so our earlier congruence analysis suggested that primes p≡3 mod8 with exponents of opposite parity could work, but the positivity conditions rule them out. Therefore, there are no solutions for odd primes?Wait, this can't be. Let me check with p=3.Take p=3, which is ≡3 mod8. Let's try small exponents.Suppose m=1 (odd), n=2 (even). Then:x=(3*3^1 -3^2)/8=(9 -9)/8=0/8=0. Not positive.Alternatively, m=2 (even), n=1 (odd):x=(3*3² -3^1)/8=(27 -3)/8=24/8=3 y=(3*3^1 -3^2)/8=(9 -9)/8=0. Not positive.Another example: m=1, n=0. But n must be at least1. Wait, exponents must be positive integers.Wait, perhaps m=1, n=2:x=(3*3 -9)/8=0, invalid. m=2, n=1: y=(3*3 -9)/8=0, invalid.Another example: m=3 (odd), n=4 (even):x=(3*27 -81)/8=(81 -81)/8=0. Invalid.m=4, n=3:x=(3*81 -27)/8=(243 -27)/8=216/8=27 y=(3*27 -81)/8=(81 -81)/8=0. Invalid.Hmm, so in all cases where one exponent is even and the other odd, either x or y becomes zero, which is invalid since we need positive integers. So this suggests that for odd primes p≡3 mod8, there are no solutions. Therefore, the only solutions are for p=2.Thus, the total number of ordered pairs is just the29 we found earlier for p=2.Wait, but this contradicts my initial thought that there might be solutions for other primes, but according to this analysis, there are none. Let me verify with p=3 and m=0 and n=1, but exponents must be at least1.Alternatively, maybe there's a mistake in the assumption that m and n have opposite parity. Wait, for p≡3 mod8, the congruences required that m and n have opposite parity. But in reality, when we set up the equations:3p^m ≡p^n mod8 3p^n ≡p^m mod8 For p≡3 mod8, and m odd, n even:3p^m ≡3*3≡9≡1 mod8 p^n ≡3^even≡1 mod8 So 1≡1 mod8, which holds.Similarly, 3p^n ≡3*1≡3 mod8 p^m ≡3 mod8 So 3≡3 mod8, holds.But in this case, let's take m=1 (odd), n=2 (even).Then x=(3*3^1 -3^2)/8=(9 -9)/8=0. Not valid.If m=3 (odd), n=4 (even):x=(3*27 -81)/8=0. Also invalid.Similarly, if n >m, let's take m=1, n=2: same as above.If m=2 (even), n=1 (odd):x=(3*9 -3)/8=(27 -3)/8=24/8=3 y=(3*3 -9)/8=0. Again invalid.Thus, regardless of which exponent is larger, one of x or y becomes zero, which is invalid. Hence, no solutions for p≡3 mod8.Therefore, the only valid solutions are for p=2, giving29 ordered pairs.Therefore, the answer should be29.Wait, but let's verify with p=2, the pairs we found.For m=n=2: x=(3*4 -4)/8=(12 -4)/8=1, y=1. Correct.For m=3, n=3: x=(3*8 -8)/8=16/8=2, y=2. So (2,2), which gives 3*4 +10*4 +3*4=12 +40 +12=64=2^6. Yes, a prime power.For m=4, n=4: x=4, y=4. 3*16 +10*16 +3*16=48 +160 +48=256=2^8. Correct.Similarly, for m=2 to12, x=y=2^{m-2}. The exponents of2 in the result are2m. For example, m=2: p^m=4, so 3x + y=4, x +3y=4. x=1,y=1, expression=16=2^4. m=3: x=2,y=2, expression=64=2^6. So each time, the expression is2^{2m}. Indeed, 3x² +10xy +3y²= (3x + y)(x +3y)= p^m * p^m= p^{2m}. Since p=2, it's2^{2m}.Similarly, for the cases where m and n differ by1:For example, m=3, n=4: x=(3*8 -16)/8= (24 -16)/8=1, y=(3*16 -8)/8=(48 -8)/8=40/8=5. So (1,5). Check the expression:3*1 +10*1*5 +3*25=3 +50 +75=128=2^7. Which is correct. Similarly, (5,1) would be the pair when m=4,n=3: (5,1). Check expression:3*25 +10*5*1 +3*1=75 +50 +3=128=2^7.Similarly, m=4,n=5: x=(3*16 -32)/8= (48 -32)/8=16/8=2, y=(3*32 -16)/8= (96 -16)/8=80/8=10. So (2,10). Check:3*4 +10*20 +3*100=12 +200 +300=512=2^9. Correct.So all these pairs are valid. Therefore, the total of29 pairs seems correct.Hence, the final answer is29. So the number of ordered pairs is29.Final Answerboxed{29}

✅:To find the number of ordered pairs of positive integers ((x, y)) with (x, y leq 2020) such that (3x^2 + 10xy + 3y^2) is a power of a prime, we start by factoring the quadratic form:[3x^2 + 10xy + 3y^2 = (3x + y)(x + 3y)]We need this product to be a power of a prime (p). Therefore, we set:[(3x + y) = p^m quad text{and} quad (x + 3y) = p^n]where (m) and (n) are non-negative integers. Solving these equations for (x) and (y), we get:[x = frac{3p^m - p^n}{8} quad text{and} quad y = frac{3p^n - p^m}{8}]For (x) and (y) to be positive integers, the numerators must be divisible by 8 and the results must be positive. We analyze the cases for different primes (p).Case 1: (p = 2)For (p = 2), the equations become:[3x + y = 2^m quad text{and} quad x + 3y = 2^n]Solving these, we find:1. (m = n): This gives (x = y = 2^{m-2}). The valid range for (m) is 2 to 12, yielding 11 pairs.2. (m = n + 1): This gives (x = 5 cdot 2^{n-3}) and (y = 2^{n-3}). The valid range for (n) is 3 to 11, yielding 9 pairs.3. (n = m + 1): This gives (x = 2^{m-3}) and (y = 5 cdot 2^{m-3}). The valid range for (m) is 3 to 11, yielding 9 pairs.Total pairs for (p = 2): (11 + 9 + 9 = 29).Case 2: Odd primes (p equiv 3 mod 8)For odd primes (p equiv 3 mod 8), we find that the conditions for (x) and (y) to be positive integers cannot be satisfied due to the constraints on the exponents (m) and (n). Thus, there are no valid solutions for odd primes.Thus, the only valid solutions are for (p = 2), resulting in 29 ordered pairs.[boxed{29}]

🔑:Given:[ 3x^2 + 10xy + 3y^2 ]We need to find the number of ordered pairs ((x, y)) such that this expression evaluates to a power of a prime number. To proceed, we observe that the given quadratic form can be factorized. Consider:[ 3x^2 + 10xy + 3y^2 ]Let us factorize it:[ 3x^2 + 10xy + 3y^2 = (3x + y)(x + 3y) ]Now, we need ((3x + y)(x + 3y)) to be a power of some prime. Let's analyze this under two possible scenarios based on the integer values of (x) and (y): Case 1: When (x geq y)We begin by considering ( frac{3x + y}{x + 3y} ) as an integer:[ frac{3x + y}{x + 3y} ]We need this ratio to be an integer, implying ( 3x + y = k(x + 3y) ) for some integer (k). Let’s analyze possible values of (k):# Subcase 1.1: (k = 1)[ 3x + y = x + 3y ]Solving for (x) and (y):[ 2x = 2y ]So, we have:[ x = y ]For (x leq 2020), both (x) and (y) need to be powers of 2. Within the constraint (x, y leq 2020), we find the following powers of 2 that are less than or equal to 2020:[ 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 ]This results in:[ 11 text{possible values of } x text{ and } y ]# Subcase 1.2: (k = 2)[ 3x + y = 2(x + 3y) ]Solving for (x) and (y):[ 3x + y = 2x + 6y ][ x = 5y ]Here, ( y ) must be a power of 2 for (x = 5y) and (x leq 2020). For the powers of 2:[ y = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 ]These values ensure (5y leq 2020), giving us:[ 9 text{possible values of } y text{ (and hence } x text{)} Case 2: When (y geq x)By symmetry and considering the similar steps as before, we find the corresponding results by swapping (x) and (y):# Subcase 2.1: (k = 1)[ y = x ]Using symmetry, ( 11 ) possible pairs as before since (y = x) is identical to ( x = y ).# Subcase 2.2: (k = 2)[ y = 5x ]Using similar logic, yielding:[ 9 ] ConclusionAdding all possible pairs from both cases,[ 11 + 9 + 9 = 29 ]Thus, the total number of ordered pairs ((x, y)) such that (3x^2 + 10xy + 3y^2) is a power of some prime, while (x, y leq 2020) is[ boxed{29} ]

🧠:Okay, so I need to prove that for all natural numbers n, A_{n+1} is greater than B_n, where A_n is a tower of n 3s and B_n is a tower of n 8s. Let me start by understanding what exactly these towers mean. First, let's clarify the notation. A tower of exponents with n layers means that each number is stacked as an exponent. For example, A_1 would be 3, A_2 is 3^3, A_3 is 3^{3^3}, and so on. Similarly, B_1 is 8, B_2 is 8^8, B_3 is 8^{8^8}, etc. So, for each subsequent n, we add another layer to the exponent tower.The problem states that for any natural number n, A_{n+1} (which is a tower of n+1 3s) is greater than B_n (a tower of n 8s). So even though 8 is larger than 3, because we're adding an extra layer to the A tower, it should surpass the B tower. That makes intuitive sense because exponent towers grow extremely quickly, so an additional layer can compensate for a smaller base. But I need to formalize this into a proof.Let me think about how to approach this. Maybe induction? Mathematical induction is a common technique for proving statements about natural numbers. Let's try that.Base case: n=1. Then A_{2} = 3^3 = 27, and B_1 = 8. So 27 > 8, which holds true. That's straightforward.Now, let's check n=2. A_{3} = 3^{3^3} = 3^27 ≈ 7.6 trillion. B_2 = 8^8 = 16,777,216. So 7.6 trillion is definitely bigger than ~16 million. So base case holds for n=2 as well. Maybe check n=3? A_4 is 3^{3^{3^3}} which is 3^(7.6 trillion), an astronomically huge number. B_3 is 8^{8^8}. 8^8 is 16,777,216, so B_3 is 8^{16,777,216}. Now, 3^{7.6 trillion} vs 8^{16 million}. Which is bigger? Wait, 8^{16 million} can be written as (2^3)^{16 million} = 2^{48 million}. On the other hand, 3^{7.6 trillion} is 3^{7,600,000,000,000}. Now, 3^1 = 3, 2^2 = 4, so 2^2 > 3^1. But exponentials with different bases: when comparing a^b vs c^d, it's not straightforward. But perhaps taking logarithms?Wait, but even 3^{7.6 trillion} is way bigger than 2^{48 million}. Let's see: Let's convert both to the same base. Let's take natural logarithm.ln(3^{7.6 trillion}) = 7.6 trillion * ln 3 ≈ 7.6 * 10^12 * 1.0986 ≈ 8.35 * 10^12.ln(8^{16 million}) = 16 million * ln 8 ≈ 16 * 10^6 * 2.0794 ≈ 33.27 * 10^6 ≈ 3.327 * 10^7.So 8.35 * 10^12 is way larger than 3.327 * 10^7, so 3^{7.6 trillion} is much larger than 8^{16 million}. Therefore, A_4 > B_3. So the pattern seems to hold for n=3 as well. But of course, checking specific cases isn't a proof. We need a general inductive step.Assume that for some k ≥ 1, A_{k+1} > B_k. We need to show that A_{k+2} > B_{k+1}.So, A_{k+2} = 3^{A_{k+1}}} and B_{k+1} = 8^{B_k}}.By the inductive hypothesis, A_{k+1} > B_k. Therefore, 3^{A_{k+1}}} > 3^{B_k}}.So if we can show that 3^{B_k} > 8^{B_k}}, then since A_{k+2} = 3^{A_{k+1}}} > 3^{B_k} > 8^{B_k}} = B_{k+1}, that would complete the inductive step.But wait, 3^{B_k} vs 8^{B_k}? Since 8 > 3, but both are exponents with the same exponent B_k. So 8^{B_k} is larger than 3^{B_k}. Wait, that's the opposite of what I need. So this approach might not work.Hmm, so perhaps my initial thought is wrong. If I have 3^{A_{k+1}}} compared to 8^{B_k}}, but since A_{k+1} > B_k, then 3^{A_{k+1}}} > 3^{B_k}, but 3^{B_k} is less than 8^{B_k}, so this doesn't directly help. So the inductive step is not straightforward.Wait, maybe I need a different approach. Let's consider logarithms to compare the growth rates.Alternatively, perhaps instead of induction, we can use a direct comparison by analyzing the towers.Let me think. For any n, A_{n+1} is a tower of n+1 3s, so it's 3^{A_n}. Similarly, B_n is a tower of n 8s, which is 8^{B_{n-1}}}. So we need to show that 3^{A_n} > 8^{B_{n-1}}} for all n ≥ 1 (since n is natural, starting at 1). Wait, but the original statement is for all natural numbers n, A_{n+1} > B_n. So indeed, for each n, compare 3^{A_n} vs 8^{B_{n-1}}}. Hmm.Alternatively, maybe compare the logarithms. Let's take the logarithm of both sides multiple times.But given that these are towers, taking logarithms successively might be a way to reduce the problem.But perhaps another approach is to note that each time, adding a layer of 3 on top of A_n vs a layer of 8 on top of B_{n-1}. Since 3^3^...^3 (n+1 times) vs 8^8^...^8 (n times). The key is that even though 8 is bigger, the fact that A has one more layer can compensate.Alternatively, maybe we can find that for each n, A_{n} > 8^{8^{...^{8}}} with (n-1) layers. Then A_{n+1} = 3^{A_n} > 3^{8^{8^{...^{8}}}} with (n-1) layers. But we need to show that 3^{8^{8^{...^{8}}}} > 8^{8^{...^{8}}} with n layers.Wait, this seems similar to comparing 3^{C} vs 8^{C}, where C is some tower. But 3^C vs 8^C, since 8^C = (2^3)^C = 2^{3C}. So 3^C vs 2^{3C}. Which is larger? Take log base 2: log2(3^C) = C log2(3) ≈ C * 1.58496, whereas log2(2^{3C}) = 3C. Since 1.58496 < 3, 3^C < 2^{3C}, so 3^C < 8^C. Therefore, 3^{C} < 8^{C} for any C > 0. So that approach might not work.Wait, but in our case, C is itself a tower. For example, if C is 8^{8^{...^{8}}} with n-1 layers, then 3^C vs 8^C. Since 8^C is larger, but we need to have 3^{A_n} > 8^{B_{n-1}}. But A_n is a tower of n 3s, which is different from C. So maybe not directly applicable.Wait, let's think recursively. Let's try to see how A_{n+1} and B_n relate in terms of their recursive definitions.Given:A_{n+1} = 3^{A_n}B_n = 8^{B_{n-1}}}We need to show A_{n+1} > B_n for all n ≥ 1.Base case n=1: A_2 = 3^3 =27, B_1 =8. 27>8.Assume for some n ≥1, A_{n+1} > B_n. Then we need to show A_{n+2} > B_{n+1}.Compute A_{n+2} = 3^{A_{n+1}}} and B_{n+1} =8^{B_n}.By the inductive hypothesis, A_{n+1} > B_n. Therefore, 3^{A_{n+1}}} > 3^{B_n}.But we need to compare 3^{B_n} vs 8^{B_n}. As previously established, 8^{B_n} = (2^3)^{B_n} = 2^{3 B_n}. So, 3^{B_n} vs 2^{3 B_n}.Which is larger? Let's take log base 2:log2(3^{B_n}) = B_n * log2(3) ≈ B_n * 1.58496log2(2^{3 B_n}) = 3 B_nSince 1.58496 < 3, we have log2(3^{B_n}) < log2(8^{B_n}), so 3^{B_n} < 8^{B_n}But from the inductive step, A_{n+2} =3^{A_{n+1}} > 3^{B_n} (since A_{n+1} > B_n), but 3^{B_n} < 8^{B_n} = B_{n+1}Therefore, we have A_{n+2} > 3^{B_n} < B_{n+1}, but this doesn't establish A_{n+2} > B_{n+1}. The problem is that 3^{B_n} < 8^{B_n}, so even though A_{n+2} > 3^{B_n}, it's not necessarily greater than 8^{B_n}.Therefore, the inductive step as initially set up fails because the inequality goes the wrong way when comparing 3^{B_n} to 8^{B_n}. So, my initial approach is flawed.Hmm. Maybe induction isn't the right approach here, or perhaps I need a stronger inductive hypothesis. Alternatively, maybe there's a different way to relate A_{n} and B_{n}.Let me try to analyze the towers layer by layer. Let's define A_n and B_n recursively:A_1 =3, A_{k+1}=3^{A_k}B_1=8, B_{k+1}=8^{B_k}We need to show that A_{n+1} > B_n for all n.Let me compute the first few terms to see the pattern:n=1: A2=27, B1=8 → 27>8n=2: A3=3^27 ≈7.6*10^12, B2=8^8=16,777,216 ≈1.6*10^7 → 7.6*10^12 >1.6*10^7n=3: A4=3^(7.6*10^12), B3=8^(1.6*10^7). Even though 8^(1.6e7) is huge, 3^(7.6e12) is unimaginably larger.So, the pattern seems to hold. But how to formalize this?Perhaps instead of induction, we can compare the towers directly by considering the number of layers.Since A_{n+1} has n+1 layers of 3s, and B_n has n layers of 8s. Maybe we can show that each additional layer in A compensates for the larger base in B.Alternatively, consider that each 3 in the tower for A_{n+1} can be paired with an 8 in B_n, but since A has one more layer, the last layer of A is a 3, which is exponentiated over a much larger tower than the corresponding layer in B.Alternatively, perhaps use the concept of hyperoperators or Knuth's up-arrow notation, but I'm not sure.Wait, another idea: Let's compare A_{n+1} and B_n by taking logarithms multiple times. Since both are towers of exponents, taking the logarithm repeatedly reduces the problem by one layer each time.Define for a tower X with k layers, log(log(...log(X)...)) with k-1 logs would give the bottom element. But since the towers have different numbers of layers, maybe this approach can help.Let me try for n=1: A2=3^3, B1=8. Taking log once: log(A2)=3 log3 ≈ 3*1.0986≈3.2958, which is greater than log(B1)=log8≈2.0794. So A2 > B1.For n=2: A3=3^{3^3}, B2=8^8. Take log twice:log(log(A3)) = log(3^3 * log3) = log(27 * 1.0986) ≈ log(29.66) ≈3.39log(log(B2))= log(8^8 * log8) = log(16,777,216 * 2.0794) ≈ log(34,952,582) ≈17.37Wait, 3.39 < 17.37. But A3=3^{27} ≈7.6*10^12 and B2=8^8≈1.6*10^7, so A3 > B2. But the double log of A3 is less than the double log of B2. Hmm, that's contradictory. Wait, maybe I made a mistake in the calculation.Wait, actually, the logarithm of A3 is log(3^{27}) = 27 log3 ≈27*1.0986≈29.66Then the log of that is log(29.66)≈3.39Similarly, log(B2)=log(8^8)=8 log8≈8*2.0794≈16.635Then log of that is log(16.635)≈2.81Wait, wait, hold on. Let's recast:For A3: log(A3) = log(3^{3^3}) = 3^3 * log3 =27 *1.0986≈29.66Then log(log(A3))=log(29.66)≈3.39For B2: log(B2)=log(8^8)=8 * log8≈8*2.079≈16.635Then log(log(B2))=log(16.635)≈2.81Wait, so 3.39 >2.81, which would mean that A3 > B2, which is correct. Wait, so even though the first log of A3 (≈29.66) is greater than the first log of B2 (≈16.635), the second log of A3 (≈3.39) is greater than the second log of B2 (≈2.81). So, each subsequent logarithm of A remains larger than that of B. But since A has more layers, when we take logs n times for B, which has n layers, we end up comparing a number from A after n logs with the base of B's tower.Wait, maybe this is a way to proceed. Let's formalize it.Suppose we have A_{n+1} which is a tower of n+1 3s, and B_n which is a tower of n 8s. To compare A_{n+1} and B_n, take the logarithm (base whatever) n times of both. For B_n, after n logarithms, we get 8. For A_{n+1}, after n logarithms, we get a tower of 1 3, which is 3. Wait, that can't be. Wait, no.Wait, let's take natural logarithms for example. Let's define recursively:For a tower X with k layers, define L(X, m) as the result of applying m logarithms to X. So, L(X,0)=X, L(X,1)=log(X), L(X,2)=log(log(X)), etc.For B_n, which is a tower of n 8s, L(B_n, n-1) = 8. Because each logarithm peels off one layer. For example, B_2=8^8, L(B_2,1)=log(8^8)=8 log8, L(B_2,2)=log(8 log8) which is not 8. Wait, maybe my initial thought was wrong.Wait, no. Let's see. Let's take B_n = 8^{B_{n-1}}. So, L(B_n,1)=log(B_n)=B_{n-1}*log8. Then L(B_n,2)=log(B_{n-1}*log8)=log(B_{n-1}) + log(log8). Which is not the same as B_{n-2}*log8. Hmm, so each logarithm doesn't just peel off a layer; it complicates it with additive terms.Similarly, for A_{n+1}=3^{A_n}, L(A_{n+1},1)=log(3^{A_n})=A_n * log3. Then L(A_{n+1},2)=log(A_n * log3)=log(A_n) + log(log3). If we compare this to L(B_n,2)=log(B_{n-1}*log8)=log(B_{n-1}) + log(log8).So, unless the logs of the previous terms dominate, which they might, but this seems messy.Alternatively, perhaps instead of natural logarithms, use logarithms with base 3 or 8 to make the expressions simpler.Let me try base 3 for A_{n+1} and base 8 for B_n.For A_{n+1}=3^{A_n}, so log3(A_{n+1})=A_n.For B_n=8^{B_{n-1}}, so log8(B_n)=B_{n-1}.But how does this help? Maybe by comparing A_n and B_{n-1}.If we can show that A_n > B_{n-1}, then log3(A_{n+1})=A_n > B_{n-1}=log8(B_n). But how does that relate?Wait, if A_n > B_{n-1}, then A_{n+1}=3^{A_n} > 3^{B_{n-1}}. But we need to compare 3^{B_{n-1}} with 8^{B_{n-2}}} (since B_n=8^{B_{n-1}} and B_{n-1}=8^{B_{n-2}}}). Wait, perhaps not directly.Alternatively, let's try to see if we can establish a relationship where each A_k is greater than some function of B_{k-1}.Wait, maybe another angle. Let's consider that 3^3 =27 >8= B_1. So A_2 > B_1. Then A_3=3^{27} which is way bigger than B_2=8^8. So perhaps each A_{k+1} is significantly larger than B_k. The question is, how to generalize this.Alternatively, maybe find a function f(k) such that A_{k} > f(k) and f(k+1) > B_{k}, establishing a chain. But I need to find such a function.Alternatively, consider that each layer of 3s grows faster than each layer of 8s. Wait, but 8 is bigger than 3. However, exponentiation is not linear. So a tower with more layers can surpass a tower with fewer layers even with a smaller base.In general, for any two towers, if one has more layers, even with a smaller base, it will eventually surpass a taller tower with a larger base. But we need to prove that even with just one extra layer, the 3s tower surpasses the 8s tower.This seems similar to the concept of hyperoperators where the number of operations (here, layers) has a more significant impact on growth rate than the base numbers. For example, 3^^(n+1) > 8^^n, where ^^ denotes tetration.There's a known result in mathematics that for any bases a and b, if you have sufficiently many layers, the tower with the smaller base but more layers will exceed the tower with the larger base. But here, the difference is fixed at one layer. So we need to show that adding one layer of 3s is enough to overcome the larger base of 8s.Perhaps we can use the fact that 3^{3^{3^{...}}} with n+1 layers is greater than 8^{8^{8^{...}}} with n layers by showing that each additional layer contributes a factor that bridges the gap.Alternatively, let's consider that for any n, 3^{A_n} > 8^{B_{n-1}}}. If we can show that A_n > B_{n-1} * log_3 8, since 3^{A_n} > 8^{B_{n-1}}} is equivalent to A_n > B_{n-1} * log_3 8.Given that log_3 8 is approximately 1.8928, so if we can show that A_n > 1.8928 * B_{n-1}, then 3^{A_n} > 8^{B_{n-1}}}.If we can establish this inequality, then combining it with the inductive hypothesis could work.Let me try this approach.We need to show that A_{n} > c * B_{n-1}, where c = log_3 8 ≈1.8928.If this holds, then A_{n+1} =3^{A_n} >3^{c * B_{n-1}}}.But we need to compare this to B_n =8^{B_{n-1}}} =3^{c * B_{n-1}}}.Therefore, if A_{n} > c * B_{n-1}, then 3^{A_n} >3^{c * B_{n-1}}} =8^{B_{n-1}}}=B_n. Thus, A_{n+1} > B_n.Therefore, if we can show that A_n > c * B_{n-1} for all n ≥1, with c=log_3 8, then the result follows.So now, the problem reduces to proving that A_n > c * B_{n-1} for all n ≥1, with c≈1.8928.Let's check the base case n=1. A_1=3, B_0=? Wait, B_n is defined for natural numbers n, so B_0 is undefined. Wait, original problem states n is a natural number, so n ≥1. Then B_{n-1} for n=1 would be B_0, which isn't defined. Hmm, perhaps my indexing is off.Wait, let's re-express the problem. We need to show for all natural numbers n (n ≥1), A_{n+1} > B_n. If we want to use the inequality A_n > c * B_{n-1}, then for n ≥1, we need to consider A_1 > c * B_0. But B_0 is not defined. So maybe this approach needs adjustment.Alternatively, perhaps start the induction from n=1, where A_2=27 > B_1=8. Then assume for n=k, A_{k+1} > B_k, and try to show A_{k+2} > B_{k+1}.But as before, A_{k+2}=3^{A_{k+1}}} and B_{k+1}=8^{B_k}}.From the inductive hypothesis, A_{k+1} > B_k. So, 3^{A_{k+1}}} >3^{B_k}. But we need to compare 3^{B_k} vs 8^{B_k}. As before, 3^{B_k} <8^{B_k}, so this doesn't help.Alternatively, perhaps find a multiplicative factor between A_{k+1} and B_k that can be leveraged. For example, if A_{k+1} > 2 B_k, then 3^{A_{k+1}}} >3^{2 B_k}=9^{B_k}. Since 9>8, 9^{B_k} >8^{B_k}=B_{k+1}. Hence, if A_{k+1} >2 B_k, then A_{k+2}=3^{A_{k+1}}} >9^{B_k} >8^{B_k}=B_{k+1}. Therefore, if we can show that A_{k+1} >2 B_k for all k ≥1, then the inductive step holds.Let's check if this is true.For n=1: A_2=27, B_1=8. 27>2*8=16. Yes.For n=2: A_3=3^27≈7.6*10^12, B_2=8^8=16,777,216. 7.6*10^12 >2*16,777,216≈33 million. Yes.For n=3: A_4=3^{7.6*10^12}, B_3=8^{16,777,216}. Even 3^{7.6*10^12} is obviously way larger than 2*8^{16 million}.So, it seems that A_{k+1} is not just greater than B_k, but much greater, possibly by a factor exceeding 2. If we can establish that A_{k+1} >2 B_k for all k ≥1, then we can use this to drive the induction.So, let's try this approach.Base case n=1: A_2=27 >2*8=16. True.Assume that A_{k+1} >2 B_k for some k ≥1. Then we need to show A_{k+2} >2 B_{k+1}}.Compute A_{k+2}=3^{A_{k+1}}} >3^{2 B_k}}.And B_{k+1}=8^{B_k}}.We need to show that 3^{2 B_k} >2 *8^{B_k}}.Which is equivalent to (3^2)^{B_k} >2*8^{B_k} →9^{B_k} >2*8^{B_k}.Divide both sides by8^{B_k}:(9/8)^{B_k} >2.Since 9/8=1.125, and (1.125)^{B_k} >2.Given that B_k is at least 8 for k≥1 (since B_1=8, B_2=8^8, etc.), and (1.125)^8 ≈2.297>2. So for k≥1, B_k ≥8, hence (9/8)^{B_k} ≥(9/8)^8≈2.297>2. Therefore, 9^{B_k} >2*8^{B_k}, hence 3^{2 B_k} >2*8^{B_k}.Thus, A_{k+2}=3^{A_{k+1}}} >3^{2 B_k} >2*8^{B_k}=2 B_{k+1}}.Therefore, the inductive step holds. Hence, by induction, A_{n+1} >2 B_n for all n≥1. Since 2 B_n > B_n, it follows that A_{n+1} > B_n for all n≥1.Therefore, we've proven the statement.Final Answerboxed{A_{n+1} > B_{n} text{ for all natural numbers } n}

✅:To prove that ( A_{n+1} > B_n ) for all natural numbers ( n ), we use mathematical induction and analyze the growth rates of the towers of exponents.Base Cases:- For ( n = 1 ): [ A_2 = 3^3 = 27 quad text{and} quad B_1 = 8 ] Clearly, ( 27 > 8 ).- For ( n = 2 ): [ A_3 = 3^{3^3} = 3^{27} approx 7.6 times 10^{12} quad text{and} quad B_2 = 8^8 = 16,777,216 ] Clearly, ( 3^{27} gg 16,777,216 ).- For ( n = 3 ): [ A_4 = 3^{3^{3^3}} quad text{and} quad B_3 = 8^{8^8} ] Even though ( 8^{8^8} ) is extremely large, ( 3^{3^{3^3}} ) is much larger due to the additional layer.Inductive Step:Assume for some ( k geq 1 ), ( A_{k+1} > B_k ). We need to show ( A_{k+2} > B_{k+1} ).- By the inductive hypothesis, ( A_{k+1} > B_k ).- Therefore, ( A_{k+2} = 3^{A_{k+1}} > 3^{B_k} ).We need to show ( 3^{B_k} > 8^{B_k} ). However, this is not directly true since ( 8^{B_k} = (2^3)^{B_k} = 2^{3B_k} ), and ( 3^{B_k} < 2^{3B_k} ).Instead, we show a stronger result: ( A_{k+1} > 2B_k ). This allows us to use:- ( 3^{A_{k+1}} > 3^{2B_k} = 9^{B_k} )- Since ( 9^{B_k} > 2 cdot 8^{B_k} ) for ( B_k geq 8 ), we have ( 9^{B_k} > 2 cdot 8^{B_k} ).Thus, ( A_{k+2} = 3^{A_{k+1}} > 9^{B_k} > 2 cdot 8^{B_k} ). Since ( 2 cdot 8^{B_k} > 8^{B_k} = B_{k+1} ), it follows that ( A_{k+2} > B_{k+1} ).By induction, we conclude that ( A_{n+1} > B_n ) for all natural numbers ( n ).[boxed{A_{n+1} > B_{n} text{ for all natural numbers } n}]

🔑:```markdown Problem:设 A_{n}=3^{3^{3}} (共 n 重3), B_{n}=8^{8^{8}} (共 n 重 8). 求证: 对一切自然数 n, 都有 A_{n+1} > B_{n}.1. 分析: 如果试着证明命题本身，我们发现这并不是一件易事，但如果我们把命题加强，就好证明。我们先将命题加强为: 对一切自然数 n, 都有 [ A_{n+1} > 3 B_{n} quad cdots left( * right) ]2. 我们先验证 n=1 时的情况: [ A_{2} = 3^{3} = 27 > 24 = 3 cdot 8 = 3 B_{1} ] 所以 (*) 成立。3. 假设当 n=k 时，有 A_{k+1} > 3 B_{k} 成立。4. 我们需要证明当 n=k+1 时, 也有 A_{k+1+1} > 3 B_{k+1}。 令: [ A_{k+2} = 3^{A_{k+1}} = 3^{3^{3}}^{3} > 3^{3} cdot B_{k} = 27 cdot B_{k} ] 显然: [ 27 cdot B_{k} > 24 cdot B_{k} = 3 cdot B_{k} cdot 8 cdot B_{k} ] 所以: [ A_{k+2} = 3^{A_{k+1}}^{3} > 3^{B_{k}} > 8 B_{k} = 3 B_{k+1} ]5. 综上所述，由数学归纳法原理知，对一切自然数 n, 都有 A_{n+1} > 3 B_{n} 成立。6. 从而更有 [ A_{n+1} > 3 B_{n} > B_{n} ] 成立。结论:[boxed{A_{n+1} > B_{n}}]```