🧠:Okay, let me start by trying to figure out these chromatic number questions. I remember that the chromatic number is the smallest number of colors needed to color the vertices of a graph so that no two adjacent vertices share the same color. Let me take each question one by one.1. Chromatic number of K₄ and K₏:First, for the complete graph K₄. A complete graph is one where every vertex is connected to every other vertex. So in K₄, each vertex has degree 3, right? Because there are three other vertices it's connected to. Now, since every vertex is adjacent to every other vertex, no two vertices can share the same color. That means each vertex needs a unique color. So, for Kₙ, the chromatic number should be n, because you need one color per vertex. Wait, but is that always the case? Let me think. For example, K₂ is just an edge, so chromatic number 2. K₃ is a triangle, which can't be colored with two colors because it's an odd cycle, so needs three. Similarly, K₄ would need four colors. So yes, the chromatic number of Kₙ is n. Therefore, K₄ has chromatic number 4, and generally, Kₙ has chromatic number n.2. Chromatic number of a square:Hmm, a square. So the graph of a square is a cycle with four vertices, right? C₄. For cycles, the chromatic number depends on whether they have an even or odd number of vertices. Since 4 is even, C₄ is bipartite. Bipartite graphs are 2-colorable. Let me verify. If you color alternate vertices with two colors, say red and blue, then no two adjacent vertices share the same color. So yes, a square can be colored with two colors. Is that optimal? Well, since it's not a single edge, you can't do it with one color. So two is the minimum. An optimal coloring would be alternating colors around the square. For example, vertices A, B, C, D: A-red, B-blue, C-red, D-blue.3. Chromatic numbers of trees from exercise 3:Wait, the user mentions "the trees obtained in the solution of exercise 3." But I don't have access to exercise 3. Maybe there was a previous problem where they constructed some trees? Since I don't have that context, maybe I need to recall that all trees are bipartite. Because a tree is an acyclic connected graph, and any tree is bipartite since you can color it with two colors by doing a level order traversal, alternating colors. So regardless of the tree structure, the chromatic number should be 2. Unless the tree is a single vertex, which would have chromatic number 1. So if the trees in question have at least one edge, their chromatic number is 2. If it's just an isolated vertex, then 1. But since the question mentions "trees," which usually have at least two vertices, so chromatic number 2. Let me confirm. A tree with two vertices is an edge, which needs two colors. A tree with three vertices is a path, which is also 2-colorable. Indeed, all trees are 2-colorable because they contain no cycles, especially no odd cycles. So their chromatic number is 2.Wait, but let me think again. If a tree has a root and layers, you can alternate colors between layers. So yes, two colors suffice. Therefore, unless the tree is a single node, it's 2. So unless exercise 3 had some special trees, but given the information here, I think it's safe to say that the chromatic number of any tree (with at least one edge) is 2.Double-checking:For the complete graph, since every vertex is adjacent to every other, you can't reuse colors. So Kₙ needs n colors. Makes sense.For the square (C₄), bipartition works, so 2 colors. If it were an odd cycle like C₅, it would need 3, but since 4 is even, 2 is enough.For trees, since they're bipartite, 2 colors. Only exception is a single vertex tree, which is 1 color. But unless specified otherwise, trees with edges need 2.I think that covers all the questions. Let me just make sure I didn't mix up any concepts. Chromatic number is about vertex coloring, not edges. Yes. So all the answers should be consistent.

✅:1. Chromatic Number of Complete Graphs The chromatic number of ( K_4 ) is 4, and for any complete graph ( K_n ), the chromatic number is ( n ). Reasoning: In a complete graph, every pair of vertices is adjacent. Thus, each vertex requires a distinct color to ensure no two adjacent vertices share the same color. This necessitates ( n ) colors for ( K_n ).---2. Chromatic Number of a Square (Cycle ( C_4 )) The chromatic number of a square (cycle ( C_4 )) is 2. Optimal Coloring: Alternate colors around the cycle (e.g., red, blue, red, blue). This ensures adjacent vertices have distinct colors while using the minimum number of colors. Reasoning: Even-length cycles (( C_{2k} )) are bipartite and thus 2-colorable. No fewer than 2 colors suffice since edges exist.---3. Chromatic Number of Trees The chromatic number of any tree (with at least one edge) is 2. Reasoning: Trees are acyclic and connected, making them bipartite graphs. A bipartition allows coloring vertices with two colors such that no adjacent vertices share the same color. Single-vertex trees (trivially) have chromatic number 1, but trees with edges require 2 colors. Summary of Solutions: 1. ( chi(K_n) = n ). 2. ( chi(C_4) = 2 ); optimal coloring alternates two colors. 3. ( chi(text{Tree}) = 2 ) (for trees with edges). (boxed{4}), (boxed{n}), (boxed{2}), (boxed{2})

🔑:# Problème 1 1. Déterminer le nombre chromatique du graphe complet mathrm{K}_{4}Le graphe complet mathrm{K}_{4} est un graphe où chaque sommet est connecté à tous les autres sommets. Il est important de rappeler que dans un graphe complet K_n, chaque sommet doit avoir une couleur distincte puisqu'il est adjacente à tous les autres sommets. Pour mathrm{K}_{4} étant donné que nous avons 4 sommets et chaque sommet est connecté à tous les autres sommets, chaque sommet doit avoir une couleur distincte. Par conséquent, le nombre chromatique de mathrm{K}_{4} est:[chi(mathrm{K}_{4}) = 4]2. Déterminer le nombre chromatique du graphe complet mathrm{K}_{n}On peut généraliser l’argument précédent. Si K_n est un graphe complet avec n sommets, chaque sommet doit être colorié différemment de tous les autres sommets, car chaque sommet est directement connecté à tous les autres sommets. Par conséquent, le nombre chromatique de K_n est:[chi(mathrm{K}_{n}) = n]# Problème 2 Déterminer le nombre chromatique d'un carréUn carré contient 4 sommets et 4 arêtes. Nous voulons colorier les sommets du carré de manière que deux sommets adjacents n'aient pas la même couleur. Considérons le processus de coloriage suivant:1. Colorions deux sommets opposés en jaune.2. Colorons les deux autres sommets opposés en rouge.Par ce processus, nous ne voyons pas deux sommets adjacents ayant la même couleur.Ainsi, le nombre chromatique d'un carré est :[chi(text{carré}) = 2]# Problème 3 Déterminer les nombres chromatiques des arbres obtenus dans l'exercice 3Un arbre est un type spécial de graphe sans cycles. Chaque arbre avec au moins deux sommets peut être coloré avec deux couleurs : une couleur pour les sommets de niveau pair et une autre couleur pour les sommets de niveau impair.Par la solution précédente à l'exercice 3, nous savons que deux des arbres peuvent être dessinés sous une forme particulière qui permet une coloration avec deux couleurs sans que deux sommets adjacents aient la même couleur. Ce processus s’applique à tous les arbres parce que les arbres n'ont pas de cycles.Ainsi, chaque arbre obtenu peut être colorié avec deux couleurs, ce qui signifie que le nombre chromatique de chacun de ces arbres est :[chi(text{arbre}) = 2] Conclusion:[boxed{ 4, , n, , 2, , 2 }]

🧠:Okay, let's try to tackle this problem. So, we have a right-angled triangle where the sum of the lengths of the three sides is 16 cm, and the sum of the squares of the three sides is 98 cm². We need to find the area of the triangle, which is (base × height)/2, right? Since it's a right-angled triangle, the legs are the base and height.First, let's recall some basics. In a right-angled triangle, the Pythagorean theorem holds: if the legs are 'a' and 'b', and the hypotenuse is 'c', then a² + b² = c². Also, the perimeter is a + b + c = 16 cm. The sum of the squares is given as a² + b² + c² = 98 cm². Hmm, but we know from the Pythagorean theorem that a² + b² = c², so substituting that into the sum of squares gives c² + c² = 2c² = 98? Wait, that would mean c² = 49, so c = 7 cm. Let me check that again.Wait, if a² + b² + c² = 98, and since a² + b² = c², then substituting that in, we get c² + c² = 98, so 2c² = 98, which leads to c² = 49, so c = 7 cm. That seems straightforward. So the hypotenuse is 7 cm. Then the perimeter is a + b + c = 16, so a + b = 16 - c = 16 - 7 = 9 cm. So a + b = 9. Also, from the Pythagorean theorem, a² + b² = 49.Now, we need to find the area, which is (a × b)/2. So if we can find the product ab, we can compute the area. Let's see. We have two equations: a + b = 9 and a² + b² = 49. Let me think. I remember that (a + b)² = a² + 2ab + b². So if we square the sum a + b, we get 9² = 81. But a² + b² is 49. So substituting into the equation: 81 = 49 + 2ab. Then, solving for ab: 81 - 49 = 2ab => 32 = 2ab => ab = 16. Therefore, the area is ab/2 = 16/2 = 8 cm². So the answer should be A) 8.Wait, but let me verify again to make sure I didn't make any mistakes. Let me go through each step again.1. Given that it's a right-angled triangle, so a² + b² = c².2. Sum of sides: a + b + c = 16.3. Sum of squares: a² + b² + c² = 98.4. Substitute a² + b² with c² in the sum of squares: c² + c² = 98 => 2c² = 98 => c² = 49 => c = 7 cm. That's correct.5. Then, a + b = 16 - 7 = 9 cm.6. We also know a² + b² = 49.7. Using (a + b)² = a² + 2ab + b². So 9² = 49 + 2ab => 81 = 49 + 2ab => 2ab = 32 => ab = 16. Therefore, area is 16/2 = 8. So yes, A) 8.But wait, the answer options include A) 8, which matches. Let me check if there's another way to approach this problem in case I missed something. Maybe considering that the sum of the squares is given, but that's already accounted for. Let me see if the numbers make sense. If the hypotenuse is 7, and the other sides sum to 9, and their squares sum to 49, then their product is 16, which gives the area. Let's test actual numbers. Suppose a and b are integers. Let's see. If a + b = 9 and ab = 16, then solving for a and b.The quadratic equation would be x² - 9x + 16 = 0. Let's compute the discriminant: 81 - 64 = 17. So the roots are (9 ± √17)/2. Which are not integers, so the sides aren't integers. But the problem doesn't state they need to be integers. So that's okay. The area is still 8. So seems correct.Alternatively, maybe there's a mistake in the sum of squares? Let me check the substitution again. Sum of squares is a² + b² + c². But since a² + b² = c², substituting gives c² + c² = 2c² = 98. So yes, c² = 49, c =7. That part is correct. So the rest follows. Therefore, I think the answer is A) 8. But let me check the answer choices again. The options are A) 8, B)10, C)12, D)14, E)16. So 8 is an option. Hmm. Wait, but the problem says "the sum of the squares of the lengths of the three sides of the triangle is 98 cm²". So sum of squares is 98, which is a² + b² + c² = 98, but in a right-angled triangle, a² + b² = c², so 98 = c² + c² = 2c², so c² =49, c=7. That's correct. Then perimeter is 16, so a + b +7 =16 => a +b=9. Then (a +b)^2 =81 = a² +2ab +b² =49 + 2ab => 81-49=32=2ab => ab=16, area=8. So yes. Therefore, the answer is A)8.But wait, let me just check once more. Suppose c=7. Then a + b=9. Let's pick some numbers where a + b=9 and a² + b²=49. Let me try a=5 and b=4. Then a + b=9. But 5² +4²=25+16=41, which is not 49. So that doesn't work. How about a= sqrt( (49 - b²) ). Let's set a =9 - b. Then substitute into a² + b²=49: (9 - b)^2 + b²=49. Expand: 81 -18b + b² + b²=49 => 2b² -18b +81=49 =>2b² -18b +32=0. Divide by 2: b² -9b +16=0. The discriminant is 81 -64=17. So b=(9 ±√17)/2. So the sides are irrational, but that's okay. The product ab would be (9 -b)b=9b -b². From the equation b² -9b +16=0, so 9b -b²=16. Therefore, ab=16. So the area is 8. That's correct. So regardless of the specific values of a and b, the product ab is 16, so area is 8. Therefore, the answer is A)8. Seems solid. I don't see any errors in the reasoning. So I'll go with option A.

✅:The problem involves a right-angled triangle with a perimeter of (16 , text{cm}) and a sum of the squares of its sides equal to (98 , text{cm}^2). To find the area:1. Key Relationships: - Let (a) and (b) be the legs, and (c) the hypotenuse. - From the Pythagorean theorem: (a^2 + b^2 = c^2). - Perimeter: (a + b + c = 16). - Sum of squares: (a^2 + b^2 + c^2 = 98).2. Solve for (c): - Substitute (a^2 + b^2 = c^2) into the sum of squares equation: [ c^2 + c^2 = 98 implies 2c^2 = 98 implies c^2 = 49 implies c = 7 , text{cm}. ]3. Find (a + b): - Using the perimeter: (a + b + 7 = 16 implies a + b = 9).4. Determine (ab): - Square the sum (a + b): [ (a + b)^2 = a^2 + 2ab + b^2. ] - Substitute (a^2 + b^2 = 49) and (a + b = 9): [ 9^2 = 49 + 2ab implies 81 = 49 + 2ab implies 2ab = 32 implies ab = 16. ]5. Calculate the Area: - The area is (frac{1}{2}ab = frac{16}{2} = 8 , text{cm}^2).Answer: (boxed{A})

🔑:1. Setup the Problem: Let the lengths of the sides of the right-angled triangle be ( p ), ( q ), and ( r ), where ( r ) is the hypotenuse. We are given two pieces of information: - The sum of the lengths of the sides: ( p + q + r = 16 ) cm. - The sum of the squares of the lengths of the sides: ( p^2 + q^2 + r^2 = 98 ) cm(^2).2. Apply Pythagoras' Theorem: Since it is a right-angled triangle, according to the Pythagorean theorem: [ p^2 + q^2 = r^2 ] 3. Derive Equation for ( r^2 ): Substituting ( r^2 ) into the sum of the squares formula: [ p^2 + q^2 + r^2 = 98 implies r^2 + r^2 = 98 implies 2r^2 = 98 ] [ r^2 = frac{98}{2} = 49 ] [ r = sqrt{49} = 7 ] Thus, the hypotenuse ( r ) is 7 cm.4. Find the Sum of the Other Two Sides: From the sum of the lengths of the sides: [ p + q + r = 16 implies p + q + 7 = 16 ] [ p + q = 16 - 7 = 9 ]5. Formulate an Expression for ( (p + q)^2 ): [ (p + q)^2 = 9^2 = 81 ]6. Rewrite and Simplify: From the expansion of ( (p + q)^2 ): [ (p + q)^2 = p^2 + 2pq + q^2 ] Using ( p^2 + q^2 = r^2 = 49 ): [ p^2 + 2pq + q^2 = 81 ] [ 49 + 2pq = 81 ] [ 2pq = 81 - 49 = 32 ] [ pq = frac{32}{2} = 16 ]7. Calculate the Area: The area of a right-angled triangle is given by: [ text{Area} = frac{1}{2}pq ] [ text{Area} = frac{1}{2} times 16 = 8 ] Conclusion:The area of the right-angled triangle is ( boxed{8} ) cm(^2).

🧠:Alright, let's tackle this problem step by step. So, we need to find prime numbers ( p ) that satisfy the system of equations:[p + 1 = 2x^2 quad text{(1)}][p^2 + 1 = 2y^2 quad text{(2)}]where ( x ) and ( y ) are integers. Hmm, okay. Let's break this down.First, since ( p ) is a prime number, it has to be greater than 1. So, possible primes are 2, 3, 5, 7, 11, etc. Let's start by checking small primes to see if they fit the equations. Maybe we can find a pattern or figure out a general approach from there.Starting with the smallest prime, ( p = 2 ):Plugging into equation (1): ( 2 + 1 = 3 = 2x^2 ). So, ( x^2 = 3/2 ). That's not an integer since 3/2 is 1.5, which isn't a perfect square. So, ( p = 2 ) is out.Next prime, ( p = 3 ):Equation (1): ( 3 + 1 = 4 = 2x^2 ) → ( x^2 = 2 ). Again, ( x ) isn't an integer here. So, ( p = 3 ) doesn't work.Next, ( p = 5 ):Equation (1): ( 5 + 1 = 6 = 2x^2 ) → ( x^2 = 3 ). Still not a perfect square. Nope.Next prime, ( p = 7 ):Equation (1): ( 7 + 1 = 8 = 2x^2 ) → ( x^2 = 4 ). Ah, ( x = 2 ) or ( x = -2 ). Since ( x ) is an integer, this works. So, equation (1) is satisfied. Let's check equation (2):( p^2 + 1 = 49 + 1 = 50 = 2y^2 ) → ( y^2 = 25 ). So, ( y = 5 ) or ( y = -5 ). That works too! So, ( p = 7 ) is a valid solution. Cool, found one.Let's check the next prime, ( p = 11 ):Equation (1): ( 11 + 1 = 12 = 2x^2 ) → ( x^2 = 6 ). Not a perfect square. Doesn't work.Next prime, ( p = 13 ):Equation (1): ( 13 + 1 = 14 = 2x^2 ) → ( x^2 = 7 ). Not a square. Nope.Next, ( p = 17 ):Equation (1): ( 17 + 1 = 18 = 2x^2 ) → ( x^2 = 9 ). So, ( x = 3 ) or ( x = -3 ). That works. Now check equation (2):( 17^2 + 1 = 289 + 1 = 290 = 2y^2 ) → ( y^2 = 145 ). 145 isn't a perfect square (12^2=144, 13^2=169). So, no good.Hmm, okay. So, ( p = 17 ) fails equation (2). Next prime, ( p = 19 ):Equation (1): ( 19 + 1 = 20 = 2x^2 ) → ( x^2 = 10 ). Not a square. Moving on.( p = 23 ):Equation (1): 24 = 2x² → x²=12. Not a square.( p = 29 ):Equation (1): 30 = 2x² → x²=15. Nope.( p = 31 ):Equation (1): 32 = 2x² → x²=16. So, x=4 or -4. Now check equation (2):( 31² + 1 = 961 + 1 = 962 = 2y² → y²=481 ). 481 is between 21²=441 and 22²=484. Not a square. So, no.So, up to ( p = 31 ), only ( p = 7 ) works. Let's check a few more primes just to be thorough.( p = 37 ):Equation (1): 38 = 2x² → x²=19. Not a square.( p = 43 ):Equation (1): 44 = 2x² → x²=22. Nope.( p = 47 ):Equation (1): 48 = 2x² → x²=24. Not a square.( p = 53 ):Equation (1): 54 = 2x² → x²=27. Not a square.Okay, seems like after 7, primes aren't working. Maybe 7 is the only one. But how can we be sure?Wait, maybe there's a more algebraic way to approach this rather than checking primes one by one. Let me think.From equation (1): ( p = 2x² - 1 ). Since ( p ) is prime, ( 2x² - 1 ) must be prime. Similarly, equation (2): ( p² + 1 = 2y² ). So, substituting ( p ) from equation (1) into equation (2):( (2x² - 1)^2 + 1 = 2y² ).Let's expand that:( (4x⁴ - 4x² + 1) + 1 = 2y² )Simplify:( 4x⁴ - 4x² + 2 = 2y² )Divide both sides by 2:( 2x⁴ - 2x² + 1 = y² )So, we have ( y² = 2x⁴ - 2x² + 1 ). Hmm. So, we need integers ( x ) and ( y ) such that this equation holds, and ( p = 2x² - 1 ) is prime.So, perhaps if we can find all integer solutions ( (x, y) ) to this quartic equation, then we can check which of them give prime ( p ).Alternatively, maybe this equation can be transformed into a Pell-type equation or something similar. Let's see.Let me rearrange the equation:( y² = 2x⁴ - 2x² + 1 )Hmm. Let me try to write it as:( y² = 2x⁴ - 2x² + 1 )Let me see if we can factor this or complete the square or something. Let's consider substituting ( z = x² ), so the equation becomes:( y² = 2z² - 2z + 1 )So, ( y² = 2z² - 2z + 1 ). Let's rearrange:( y² = 2(z² - z) + 1 )Perhaps complete the square for the quadratic in z:( z² - z = z² - z + 1/4 - 1/4 = (z - 1/2)^2 - 1/4 )So,( y² = 2[(z - 1/2)^2 - 1/4] + 1 = 2(z - 1/2)^2 - 1/2 + 1 = 2(z - 1/2)^2 + 1/2 )Hmm, not sure if that helps. Maybe another approach.Let me consider specific values of x. Since we already saw that for x=2, we get p=7, which works, and for x=3, p=17, which doesn't work. Let's check x=1:x=1: p=2(1)^2 -1 =1. But 1 is not prime. So invalid.x=0: p= -1. Not prime. Negative.x=-1: same as x=1. p=1. Not prime.x=2: p=7, which works.x=3: p=17, which we saw didn't work for equation (2).x=4: p=2(16) -1=31, which we checked and it didn't work.x=5: p=2(25)-1=49. Wait, 49 is not prime. So, p=49 is invalid.x=6: p=2(36)-1=71. Let's check equation (2):p² +1= 5041 +1=5042=2y² → y²=2521. 2521 is between 50²=2500 and 51²=2601. Not a square. So nope.x=7: p=2(49)-1=97. Check equation (2):97² +1=9409 +1=9410=2y² → y²=4705. Not a square. 68²=4624, 69²=4761. So, no.x=8: p=2(64)-1=127. Equation (2):127² +1=16129 +1=16130=2y² → y²=8065. Not a square. 89²=7921, 90²=8100. Close, but nope.x=9: p=2(81)-1=161. 161 is not prime (7*23). So invalid.x=10: p=2(100)-1=199. Check equation (2):199² +1=39601 +1=39602=2y² → y²=19801. 19801 is 140.71² approximately. Wait, 140²=19600, 141²=19881. So, 19801 is between them. Not a square.Hmm. So, up to x=10, only x=2 gives a valid prime p=7. Let's check x=11:p=2(121)-1=241. Prime. Equation (2):241² +1=58081 +1=58082=2y² → y²=29041. Let's see: 170²=28900, 171²=29241. So, 29041 is between them. Not a square.x=12: p=2(144)-1=287. 287 is not prime (7*41). So invalid.x=13: p=2(169)-1=337. Prime. Equation (2):337² +1=113569 +1=113570=2y² → y²=56785. 238²=56644, 239²=57121. Not a square.x=14: p=2(196)-1=391. Not prime (17*23). So invalid.x=15: p=2(225)-1=449. Prime. Equation (2):449² +1=201601 +1=201602=2y² → y²=100801. Let's see, 317²=100489, 318²=101124. So, nope.This seems like it's not working. Maybe there's a higher x that works, but it's getting too big. Maybe we need a different approach.Looking back at the equation ( y² = 2x⁴ - 2x² + 1 ). Let me see if I can write this in another way. Maybe factor it.Alternatively, perhaps rearrange terms:( 2x⁴ - 2x² + 1 - y² = 0 )Not sure. Let's consider writing the equation as ( y² = 2x⁴ - 2x² + 1 ). Maybe we can consider this as an elliptic curve? Hmm, quartic equations can sometimes be difficult. Alternatively, maybe see if this equation has solutions beyond x=2.Alternatively, let's note that both equations must hold, so perhaps we can relate them.From equation (1), ( p = 2x² - 1 ). Then equation (2) becomes:( (2x² -1)^2 + 1 = 2y² )Which simplifies to:( 4x⁴ -4x² +1 +1 = 2y² )( 4x⁴ -4x² +2 = 2y² )Divide both sides by 2:( 2x⁴ -2x² +1 = y² )So, we have ( y² = 2x⁴ -2x² +1 ). Let me denote ( t = x² ), so:( y² = 2t² - 2t +1 )So, this is a quadratic in t: ( 2t² -2t +1 - y² =0 ). Perhaps solving for t:Using quadratic formula:( t = [2 ± sqrt(4 - 8(1 - y²)}]/4 )Wait, let's compute discriminant:Discriminant D = (-2)^2 - 4*2*(1 - y²) = 4 - 8(1 - y²) = 4 -8 +8y²= -4 +8y²So, for t to be integer, discriminant must be a perfect square:So, -4 +8y² = k² for some integer k.So, ( k² =8y² -4 )Rearranged as:( k² +4 =8y² )Which implies:( k² ≡ -4 mod 8 )But squares modulo 8 are 0,1,4. Let's see:If ( k² ≡ -4 mod 8 ), then ( k² ≡4 mod 8 ), since -4 mod8 is 4. So, possible. Because 2²=4 mod8, 6²=36≡4 mod8, etc. So, k must be even. Let k=2m:Then, equation becomes:( (2m)² +4 =8y² )→ (4m² +4=8y²)→ Divide by 4: (m² +1=2y²)So, now we have:( m² +1 =2y² )Wait, this is a Pell-type equation. The equation ( y² - (m² +1)/2=0 ), but not exactly. Wait, let's write it as:( 2y² -m² =1 )Which is similar to the Pell equation ( x² - 2y² =1 ), but here it's ( 2y² -m² =1 ). Let's rearrange:( m² = 2y² -1 )So, m² ≡ -1 mod 2. But squares mod2 are 0 or1. So, -1 mod2 is 1, so m²≡1 mod2 → m must be odd.But this seems like another diophantine equation. Let me check small values of y.For y=1: m²=2(1)-1=1 → m=±1. So, possible.For y=2: m²=8-1=7. Not a square.y=3: m²=18-1=17. Not a square.y=5: m²=50-1=49. So, m=±7.Wait, y=5 gives m=±7. Let's check that. Then, recalling back:We had k=2m. So, if m=7, then k=14. Then discriminant D=k²=196. Then t=(2 ±k)/4=(2 ±14)/4. So t=(16)/4=4 or t=(-12)/4=-3. Since t=x² must be non-negative, t=4. So x²=4 → x=±2. Which is the solution we found earlier.Similarly, for y=1, m=±1. Then k=±2. Then t=(2 ±2)/4 → t=(4)/4=1 or t=0/4=0. t=1 gives x²=1 →x=±1. Then p=2x² -1=1, which isn't prime. t=0 gives x=0, p=-1, invalid.But y=5 gives a valid solution. What about higher y?y=7: m²=2(49)-1=97. Not a square.y=10: m²=200 -1=199. Not a square.y=12: m²=288 -1=287. Not a square.Wait, but Pell equations usually have infinite solutions. Let's see. The equation ( m² =2y² -1 ) is similar to the negative Pell equation ( m² - 2y² = -1 ). The solutions to this equation can be generated using continued fractions or recurrence relations. The minimal solution is (m,y)=(1,1), then (7,5), (41,29), etc., following the recurrence relations.So, for each solution (m,y) of ( m² -2y² = -1 ), we can get possible x values.Wait, if ( m² =2y² -1 ), then substituting back into our previous steps:From earlier, we had:From the discriminant, we arrived at ( m² =2y² -1 ). Then, in the quadratic equation for t:t=(2 ±k)/4, but k=2m. So, t=(2 ±2m)/4=(1 ±m)/2.But t=x² must be a non-negative integer, so (1 ±m)/2 must be integer and non-negative.Given that m is positive (since we can take m positive without loss of generality), let's see:For the minimal solution (m,y)=(1,1):t=(1 +1)/2=1 or t=(1 -1)/2=0. So t=1 or 0. As before, x²=1 or 0. But p=2x² -1=1 or -1, which are invalid.Next solution (m,y)=(7,5):t=(1 +7)/2=4, t=(1-7)/2=-3. Discarding negative, t=4. Then x²=4 →x=±2. Which gives p=7. That's our solution.Next solution (m,y)=(41,29):t=(1 +41)/2=21. So x²=21. Not a perfect square. So invalid. Hmm.Wait, but x²=21 is not a square. So, even though (41,29) solves the Pell equation, it doesn't lead to a valid x here.Similarly, next solution (m,y)=(239, 169):t=(1 +239)/2=120. x²=120. Not a square. So invalid.So, seems like only the first non-trivial solution (m=7, y=5) gives a valid x=2. Then, p=7. The other solutions of the Pell equation don't give square x². Therefore, perhaps p=7 is the only prime that satisfies both equations.But how do we know there aren't other solutions?Well, looking at the structure of the equations, since the Pell equation ( m² -2y² = -1 ) has solutions generated by the fundamental solution (1,1) and then using recurrence, but each subsequent solution leads to t=(1 +m)/2 being non-square for x². Since x² must be a square, unless t is a square, which only happens when m +1 is twice a square. Wait, because t=(1 +m)/2, so if t is a square, then (1 +m)/2 =x², so m=2x² -1. But from equation (1), p=2x² -1. So, m=p. Wait, but m is defined as part of solving the discriminant, which came from the original equations. So, substituting back, we have that m=p. Wait, let's check.From equation (1): p=2x² -1. From the Pell equation solution, we have m=2x² -1? Wait, no. Let me retrace.Wait, the Pell equation solution gives m=7 when y=5. Then, t=(1 +7)/2=4=x², which gives x=2, then p=2x² -1=7. So, indeed, m=p in this case. So, in general, for each solution of the Pell equation, m=2x² -1=p. But p must be prime. So, in the Pell equation solutions, m must be prime. However, the subsequent solutions give m=41, which is prime, but t=(1 +41)/2=21, which is not a perfect square, so x would be sqrt(21), which isn't integer. So, even though m=41 is prime, since x isn't integer, that solution doesn't contribute to a valid p.Similarly, m=239 is prime, but t=120 isn't a square, so p=239 would require x²=120, which isn't integer. Hence, only when t is a perfect square does the solution give a valid prime p. Since in the Pell equation solutions, t=(1 +m)/2 must be a square, and m itself must be prime. But for m=7, t=4 is square, and m=7 is prime. For m=41, t=21 isn't square; m=239, t=120 isn't square, etc. Therefore, the only solution where both m is prime and t is a square is m=7, t=4. Hence, p=7 is the only prime satisfying both equations.Alternatively, we can formalize this reasoning.Suppose that (m,y) is a solution to the Pell equation ( m² -2y² = -1 ). Then, we have:From the earlier steps, t=(1 +m)/2, which must be a perfect square, say t=k². Therefore:( k² = (1 +m)/2 implies m=2k² -1 )But m is also given by the Pell equation solution. So substituting m=2k² -1 into the Pell equation:( (2k² -1)² -2y² =-1 )Expanding:( 4k⁴ -4k² +1 -2y² = -1 implies 4k⁴ -4k² +2 =2y² implies 2k⁴ -2k² +1 =y² )But this is exactly the equation we had earlier for y² in terms of x. Wait, here k is x. Because earlier, we set t=x², and t=(1 +m)/2= k² (here). So, this implies that k=x. Therefore, the equation is the same as the original equation. Hence, the only solutions where m=2x² -1 and t=x²=(1 +m)/2 are such that x must satisfy the original equation. Therefore, the only time when both the Pell equation is satisfied and t is a square is when x is such that the Pell equation solution m equals 2x² -1. But from the Pell equation solutions, the minimal solution is m=1, y=1, which gives x²=1, but p=1, invalid. Next solution is m=7, y=5, which gives x=2, p=7, valid. The next solutions for m in the Pell equation do not satisfy m=2x² -1 with x integer. Hence, p=7 is the only prime solution.Therefore, after thorough exploration, the only prime number satisfying both equations is 7.Final AnswerThe prime number ( p ) that satisfies the system of equations is boxed{7}.

✅:To find the prime numbers ( p ) that satisfy the system of equations:[p + 1 = 2x^2][p^2 + 1 = 2y^2]where ( x ) and ( y ) are integers, we start by substituting ( p = 2x^2 - 1 ) from the first equation into the second equation. This gives:[(2x^2 - 1)^2 + 1 = 2y^2]Expanding and simplifying this equation, we get:[4x^4 - 4x^2 + 2 = 2y^2]Dividing by 2:[2x^4 - 2x^2 + 1 = y^2]We need to find integer solutions ( (x, y) ) such that ( p = 2x^2 - 1 ) is a prime number. By checking small primes and corresponding values of ( x ):- For ( p = 7 ), we find ( x = 2 ) (since ( 7 + 1 = 2 cdot 2^2 )) and ( y = 5 ) (since ( 7^2 + 1 = 2 cdot 5^2 )). This satisfies both equations.Next, we consider the equation ( y^2 = 2x^4 - 2x^2 + 1 ) and transform it into a Pell-type equation. By solving the Pell equation ( m^2 - 2y^2 = -1 ), we find that the minimal solution leading to a prime ( p ) is when ( m = 7 ) and ( y = 5 ), corresponding to ( x = 2 ) and ( p = 7 ).Checking higher solutions of the Pell equation, we find that they do not yield integer values for ( x ) that result in prime ( p ). Thus, the only prime number ( p ) that satisfies both equations is:[boxed{7}]

🔑:Given the system of equations:[ p + 1 = 2x^2 ][ p^2 + 1 = 2y^2 ]We need to find the prime numbers ( p ) for which the system has integer solutions.1. Analyzing Constraints on ( x ) and ( y ): If ((x, y)) is an integer solution, then ((-x, y)), ((-x, -y)) and ((x, -y)) are also solutions. Without loss of generality, we can assume ( x > 0 ) and ( y > 0 ). From the given equations: [ p + 1 = 2x^2 implies p = 2x^2 - 1 ] [ p^2 + 1 = 2y^2 implies 2y^2 = (2x^2 - 1)^2 + 1 ] Clearly, ( y > x ) and ( p > y ), so ( p > x ), which implies ( x + y < 2p ).2. Combining the Equations: Subtract the first equation from the second: [ p^2 + 1 - (p + 1) = 2y^2 - 2x^2 ] Simplify: [ p^2 - p = 2y^2 - 2x^2 ] Factor out common terms: [ p(p - 1) = 2(y^2 - x^2) ] Which factors further to: [ p cdot frac{p - 1}{2} = (y + x)(y - x) ] Since ( p ) is a prime number and ( p neq 2 ), hence ( frac{p - 1}{2} ) is an integer.3. Analyzing Factors: Prime ( p ) does not divide ( (y - x) ) since ( 0 < y - x < y < p ). Therefore, ( p ) must divide ( y + x ). Let ( y + x = kp ) for some integer ( k geq 1 ): [ x + y = kp ] From the earlier observation ( x + y < 2p ), ( k ) can be only 1, so: [ x + y = p ]4. Solving for ( y - x ): [ y - x = frac{p - 1}{2} ]5. Solve for ( y ) and Substitute: Adding and solving ( x + y = p ) and ( y - x = frac{p - 1}{2} ): [ 2y = p + frac{p - 1}{2} = frac{3p - 1}{2} ] [ y = frac{3p - 1}{4} ] Substituting ( y = frac{3p - 1}{4} ) into the second equation [ p^2 + 1 = 2y^2 ]: [ p^2 + 1 = 2 left( frac{(3p - 1)^2}{16} right) ] Simplifies to: [ 16p^2 + 16 = 2(9p^2 - 6p + 1) ] [ 16p^2 + 16 = 18p^2 - 12p + 2 ] Rearrange: [ 16p^2 - 18p^2 + 12p + 16 - 2 = 0 ] Which simplifies to: [ -2p^2 + 12p + 14 = 0 ] Dividing by -2: [ p^2 - 6p - 7 = 0 ]6. Solving the Quadratic Equation: [ p^2 - 6p - 7 = 0 ] Using the quadratic formula ( p = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) with ( a = 1, b = -6, c = -7 ): [ p = frac{6 pm sqrt{36 + 28}}{2} ] [ p = frac{6 pm sqrt{64}}{2} ] [ p = frac{6 pm 8}{2} ] [ p_1 = 7, quad p_2 = -1 ]7. Validating the Prime Solution: The valid prime solution is: [ p = 7 ]8. Conclusion:Since ( p = 7 ) is a prime number, we conclude that the system has integer solutions for ( p = 7 ).Thus, the solution to the problem is:[boxed{7}]

🧠:Okay, so I need to prove that there's no convex polyhedron with 2 square faces and all other faces being regular pentagons. Hmm, let's start by recalling some basics about polyhedrons. Maybe Euler's formula will come into play here. Euler's formula states that for any convex polyhedron, the number of vertices minus the number of edges plus the number of faces equals 2. So, V - E + F = 2. That might be useful.First, let's denote the number of square faces as F4 and the number of pentagonal faces as F5. According to the problem, there are 2 square faces, so F4 = 2. All the remaining faces are regular pentagons. So the total number of faces F = F4 + F5 = 2 + F5.Now, let's think about the edges. Each square has 4 edges, and each pentagon has 5 edges. But in a polyhedron, each edge is shared by exactly two faces. So the total number of edges E can be calculated as (4*F4 + 5*F5)/2. Substituting F4 = 2, we get E = (4*2 + 5*F5)/2 = (8 + 5F5)/2.Similarly, let's consider the vertices. Each vertex in a convex polyhedron must be where at least three faces meet, right? Because if only two faces meet at a vertex, it would be an edge, not a vertex. So the degree of each vertex (the number of edges meeting there) is at least 3. However, since all faces are regular polygons (squares and pentagons), the polyhedron would be vertex-transitive if it's a regular polyhedron, but here it's not regular because there are two different types of faces. Wait, but maybe the arrangement of squares and pentagons around each vertex has to be the same? Not necessarily. But since it's convex and the faces are regular, the angles at each vertex must add up to less than 360 degrees. Also, each vertex is part of some number of faces.Let me think. Each square has a corner angle of 90 degrees, and each regular pentagon has a corner angle of 108 degrees. In a convex polyhedron, the sum of the angles around each vertex must be less than 360 degrees. So if a vertex is where, say, a square and two pentagons meet, the total angle would be 90 + 108 + 108 = 306 degrees, which is less than 360. Alternatively, maybe two squares and a pentagon? 90 + 90 + 108 = 288, which is still okay. But maybe the arrangement of faces around each vertex is more uniform.Wait, but in order to have a regular structure, maybe each vertex is part of the same arrangement. For example, in a soccer ball pattern (truncated icosahedron), each vertex is two hexagons and a pentagon. But here, since there are two squares, maybe each square is surrounded by pentagons? But I need to figure out the possible configurations.Alternatively, let's try to use Euler's formula. Let's denote V, E, F as usual. We already have F = 2 + F5. And E = (8 + 5F5)/2. Now, we need to find V. Each vertex is where at least three edges meet, so if we count the total number of face corners (which is 4*2 + 5*F5) and divide by the number of faces meeting at each vertex. But since each vertex is shared by multiple faces, the total number of vertices would be (sum of all face corners)/number of faces per vertex. However, the problem is that we don't know how many faces meet at each vertex. So perhaps we need to consider the average degree of each vertex.Alternatively, using the formula for the number of edges. Since each edge is shared by two faces, we have E = (8 + 5F5)/2. Also, in terms of vertices, each edge is shared by two vertices. So the total number of edges is also related to the number of vertices. If we let V be the number of vertices and assume that each vertex has degree d (number of edges meeting at the vertex), then the total number of edges E is (d*V)/2. But the problem is that the degree d can vary per vertex. So unless all vertices have the same degree, this might complicate things.But perhaps in this hypothetical polyhedron, the vertices are all of the same type, given the regularity of the faces. Let's suppose that each vertex is adjacent to one square and two pentagons. Let's check if that works. Each square has four vertices, each pentagon has five. Let’s see, the total number of face-vertex incidences is 4*2 + 5*F5. Since each vertex is part of three faces (assuming all vertices are 3-valent), then V = (8 + 5F5)/3. But also, from Euler's formula, V - E + F = 2. Let's substitute E and F in terms of F5.So, E = (8 + 5F5)/2F = 2 + F5V = (8 + 5F5)/3Then, substituting into Euler's formula:(8 + 5F5)/3 - (8 + 5F5)/2 + (2 + F5) = 2Let me compute that. First, find a common denominator for the fractions. Let's use 6.Multiply all terms by 6 to eliminate denominators:6*(8 + 5F5)/3 - 6*(8 + 5F5)/2 + 6*(2 + F5) = 6*2Simplify:2*(8 + 5F5) - 3*(8 + 5F5) + 6*(2 + F5) = 12Compute each term:2*(8 + 5F5) = 16 + 10F5-3*(8 + 5F5) = -24 -15F56*(2 + F5) = 12 + 6F5Adding them together:(16 + 10F5) + (-24 -15F5) + (12 + 6F5) = 12Combine like terms:16 -24 +12 = 410F5 -15F5 +6F5 = 1F5So total: 4 + F5 = 12Hence, F5 = 8So, according to this, F5 = 8. Then, let's check if that works.Compute V = (8 +5*8)/3 = (8 +40)/3 = 48/3 = 16E = (8 +5*8)/2 = 48/2 =24F = 2 +8 =10Then Euler's formula: 16 -24 +10 = 2, which holds.So Euler's formula is satisfied here. But wait, so according to this, there exists a polyhedron with 2 squares, 8 pentagons, 16 vertices, 24 edges. But the problem states that such a polyhedron does not exist. Therefore, perhaps there's another constraint that isn't satisfied here.Wait, but maybe the angles at each vertex don't add up properly. Let's check the angles. If each vertex is part of one square and two pentagons, then the angles at each vertex would be 90 (from the square) + 108 +108 (from the pentagons) = 306 degrees. Since it's a convex polyhedron, this is okay because 306 < 360. So that seems okay.But wait, but regular pentagons and squares in a convex polyhedron would require that the dihedral angles also match up. Wait, dihedral angles are the angles between adjacent faces along an edge. For regular faces, the dihedral angles can be calculated.The dihedral angle of a square (in a cube) is 90 degrees. The dihedral angle of a regular pentagon (in a dodecahedron) is 116.565 degrees. If we have edges where a square meets a pentagon, the dihedral angle would need to be compatible. Hmm, but in reality, the dihedral angles must be consistent around each edge.Wait, maybe the problem is that the square and pentagons can't be arranged in such a way that the dihedral angles work out? Let me think. If two squares are adjacent, their dihedral angle is 90 degrees. If a square is adjacent to a pentagon, then the dihedral angle would have to be some combination. But in a regular polyhedron with squares and pentagons, is that possible?Alternatively, maybe the issue is that such a polyhedron would require all edges to be the same length, since the faces are regular. But squares have edges of one length and pentagons another? Wait, no. If the polyhedron is convex with regular faces, the edges must all be congruent because the edges are shared between faces. So a square and a pentagon sharing an edge must have edges of the same length. However, regular squares and regular pentagons have different edge lengths if they are to form a convex polyhedron. Wait, but in a regular polyhedron like the rhombicuboctahedron, which has squares and triangles, all edges are the same length. So maybe it's possible here. Wait, but squares and pentagons: can they have the same edge length and still have compatible angles?Wait, in a regular square, the edge length is arbitrary, but the internal angles are fixed. Similarly for pentagons. But when you build a polyhedron, the edges must be the same length for adjacent faces. So if a square and a pentagon share an edge, their edges must be the same length. But the problem is that the dihedral angles between a square and a pentagon may not match up. Let me check.The dihedral angle of a square (as in a cube) is 90 degrees. The dihedral angle of a regular pentagon (as in a dodecahedron) is approximately 116.565 degrees. If we have an edge where a square meets a pentagon, the dihedral angle would have to be something that allows both faces to meet along that edge. However, in reality, the dihedral angle is determined by the way the faces are arranged around the edge. For two regular polygons meeting along an edge, the dihedral angle can be calculated based on their angles. Wait, perhaps it's more complicated.Alternatively, maybe the problem is that the regular pentagons and squares cannot be arranged in the way required by the vertex configuration. Let's suppose each vertex is part of one square and two pentagons, as we previously considered. Then, the angles at each vertex are 90 + 108 +108 = 306 degrees. But for the geometry of the polyhedron, the way the faces meet must create a closed solid. However, perhaps the local geometry works, but the global topology does not. For example, in the case of the hypothetical polyhedron with 2 squares and 8 pentagons, maybe such a structure can't close up properly.Wait, but we have satisfied Euler's formula here, which is a necessary condition. However, Euler's formula is not sufficient. So even though the numbers add up, the actual geometric constraints might prevent such a polyhedron from existing.Another approach: maybe using the concept of dual polyhedrons or other properties. Alternatively, think about known polyhedrons with squares and pentagons. For example, the Johnson solids. There are 92 Johnson solids, which are all convex polyhedrons with regular faces. Let me check if any of them have 2 squares and the rest pentagons. But the Johnson solids are usually named by their face types. For example, the square pyramid has a square base and four triangular sides. There's also a pentagonal pyramid, which has a pentagon base and five triangular sides. But the problem here requires two squares and the rest being pentagons. If such a Johnson solid existed, it would probably be listed. However, I don't recall any Johnson solid with two squares and the rest pentagons.Another thought: the regular dodecahedron has 12 regular pentagons. If we modify it by replacing two adjacent pentagons with squares, maybe? But when you replace a pentagon with a square, you change the number of edges and vertices. However, such a modification might not preserve regularity or convexity. Moreover, the angles might not fit.Alternatively, let's check the number of edges. If F5 =8, then total edges E =24. Each square has four edges, but each edge is shared. So each square is surrounded by some number of pentagons. Let's imagine constructing the polyhedron. Each square must be adjacent to four pentagons. But each pentagon has five edges. If there are two squares, each adjacent to four pentagons, but some pentagons might be adjacent to both squares. Wait, but each edge of the square is shared with a pentagon. So each square has four edges, each shared with a pentagon. So there are four pentagons adjacent to each square. However, there are two squares, so that's 4*2=8 edges where squares meet pentagons. But each such edge is shared between a square and a pentagon, so those edges are already counted. The remaining edges of the pentagons must be shared among themselves.Each pentagon has five edges. So for each pentagon, one edge is shared with a square (if it's adjacent to a square), and the other four edges are shared with other pentagons. Wait, but there are 8 pentagons. Let's see: Each of the two squares is adjacent to four pentagons, so that's 8 pentagons. Wait, but if there are only 8 pentagons, each pentagon is adjacent to exactly one square. Because each pentagon can be adjacent to at most one square (since if a pentagon is adjacent to two squares, it would have two edges connected to squares, but then those squares would be adjacent to each other via that pentagon, which isn't possible unless the squares are adjacent, which would require another edge shared between them, but in this case, the squares are separate). So each pentagon is adjacent to one square. Therefore, each pentagon has one edge connected to a square and four edges connected to other pentagons.So the total number of edges from the pentagons is 5*8=40. But subtract the edges shared with squares: 8 edges (since each square has four edges, two squares have eight edges shared with pentagons). So the remaining edges are 40 -8=32 edges, which are shared among the pentagons themselves. Therefore, the number of edges between pentagons is 32/2=16 (since each edge is shared by two pentagons). Therefore, total edges: 8 (square-pentagon) +16 (pentagon-pentagon) =24, which matches our earlier calculation.So, edges check out. Now, let's think about the vertices. Each square has four vertices, each pentagon has five. Total face-vertex incidences: 2*4 +8*5=8 +40=48. Since each vertex is part of three faces (as we assumed earlier), then the number of vertices is 48/3=16, which matches.So Euler's formula is satisfied, and the edge counts check out. So why can't such a polyhedron exist? Maybe the problem is with the angles or the actual geometric construction.Let's consider the angles at each vertex. As calculated earlier, each vertex is part of one square and two pentagons. The angles contributed by each face are 90 degrees (square) and two times 108 degrees (pentagons), totaling 90 + 108 +108=306 degrees. Since the polyhedron is convex, the sum of the face angles at each vertex must be less than 360 degrees, which is satisfied here.But in reality, when you have regular polygons in a polyhedron, the angles not only have to sum to less than 360 degrees but also have to form a three-dimensional structure where the dihedral angles (angles between adjacent faces) are consistent around each edge.Let me recall that in a regular polyhedron (Platonic solid), all vertices are identical, and all faces are regular and identical. In an Archimedean solid, the vertices are identical, but the faces can be different regular polygons. For example, the rhombicuboctahedron has triangles and squares. So maybe this hypothetical polyhedron is an Archimedean solid. However, the Archimedean solids are well-known, and none of them have two squares and eight pentagons.Wait, the Archimedean solids include the truncated icosahedron (soccer ball), which has pentagons and hexagons, the truncated dodecahedron, the snub dodecahedron, etc. None have squares and pentagons. The rhombicuboctahedron has triangles and squares. The truncated octahedron has squares and hexagons. So maybe such a combination of squares and pentagons isn't possible in an Archimedean solid.But even if it's not an Archimedean solid, the problem is about any convex polyhedron with regular faces. So maybe it's a Johnson solid. But as I thought earlier, there's no Johnson solid with two squares and eight pentagons.Alternatively, perhaps the problem is that when trying to assemble the polyhedron, the arrangement of pentagons around the squares leads to a conflict. Let's try to visualize.Imagine two square faces. Each square is adjacent to four pentagons. So each square is like a "band" of pentagons around it. But since there are two squares, maybe these two squares are connected via the pentagons. However, with eight pentagons, each adjacent to a square, each pentagon is adjacent to one square. So there are eight pentagons, each connected to one square. But two squares, so four pentagons per square. That would mean each square is surrounded by four pentagons, forming a sort of cube-like structure, but with pentagons instead of squares on the sides. However, a regular pentagon can't be connected to a square in such a way that the angles and edges match up properly.Wait, maybe the dihedral angles are incompatible. Let's calculate the dihedral angles between a square and a pentagon. For a square, the dihedral angle (angle between two adjacent faces) in a cube is 90 degrees. For a regular pentagon in a dodecahedron, the dihedral angle is about 116.565 degrees. If a square and a pentagon meet along an edge, what would the dihedral angle be?Wait, the dihedral angle between a square and a pentagon isn't a fixed number like in the Platonic solids. It depends on how they're arranged. However, in a regular-faced polyhedron, the dihedral angles must be such that the faces can be folded up into three dimensions without overlapping.Alternatively, perhaps we can use the concept of angular defects and the Descartes theorem. Descartes' theorem states that the sum of the angular defects at all vertices is 4π (720 degrees). The angular defect at a vertex is 360 degrees minus the sum of the face angles at that vertex.In our case, each vertex has an angular defect of 360 - 306 = 54 degrees. There are 16 vertices, so total angular defect is 16*54=864 degrees. But 4π radians is 720 degrees. So 864≠720, which contradicts Descartes' theorem. Therefore, such a polyhedron cannot exist.Ah, this seems promising. Wait, so if each vertex has a defect of 54 degrees, then the total defect would be 16*54=864, which is more than 720, violating Descartes' theorem. Therefore, even though Euler's formula is satisfied, the angular defects don't add up to 4π, so such a polyhedron is impossible.Therefore, the key issue is that while Euler's formula is satisfied, the total angular defect does not match the required 720 degrees. Hence, the polyhedron cannot exist.Let me verify this calculation. The angular defect at each vertex is 360 - sum of face angles. In our case, each vertex has one square (90 degrees) and two pentagons (108 each), so total angle sum 90 + 108 +108=306. Therefore, angular defect is 360 -306=54 degrees per vertex. Number of vertices is 16, so total defect 54*16=864. But Descartes' theorem says total defect must be 720 degrees. Therefore, this is a contradiction, proving that such a polyhedron cannot exist.Therefore, the conclusion is that even though the combinatorial counts (vertices, edges, faces) satisfy Euler's formula, the geometric constraint of angular defects required by Descartes' theorem is violated, making such a polyhedron impossible.

✅:To prove that no convex polyhedron with 2 square faces and all other faces being regular pentagons exists, we analyze both combinatorial and geometric constraints:1. Euler's Formula Check: - Let ( F_4 = 2 ) (squares) and ( F_5 ) (pentagons). Total faces: ( F = 2 + F_5 ). - Total edges: ( E = frac{4 cdot 2 + 5 cdot F_5}{2} = frac{8 + 5F_5}{2} ). - Total vertices: Assume each vertex joins one square and two pentagons. Face-vertex incidences: ( 4 cdot 2 + 5 cdot F_5 = 8 + 5F_5 ). Thus, ( V = frac{8 + 5F_5}{3} ). - Substituting into Euler's formula ( V - E + F = 2 ): [ frac{8 + 5F_5}{3} - frac{8 + 5F_5}{2} + (2 + F_5) = 2 ] Solving yields ( F_5 = 8 ), ( V = 16 ), ( E = 24 ), and ( F = 10 ), satisfying Euler's formula.2. Angular Defect Analysis (Descartes' Theorem): - Each vertex has angles ( 90^circ ) (square) and ( 2 times 108^circ ) (pentagons), summing to ( 306^circ ). - Angular defect per vertex: ( 360^circ - 306^circ = 54^circ ). - Total angular defect: ( 16 times 54^circ = 864^circ ). - Descartes' Theorem requires total defect ( = 720^circ ). The excess ( 864^circ neq 720^circ ) violates this theorem.Conclusion: While Euler's formula is satisfied, the geometric constraint imposed by Descartes' Theorem on angular defects is violated. Hence, no such convex polyhedron exists.

🔑:1. Step 1: Introduction We need to prove that there is no convex polyhedron with 2 faces being squares and the rest being regular pentagons.2. Step 2: Initial Consideration Consider a convex polyhedron model with 2 square faces and an unknown number of regular pentagon faces. Let's denote the number of pentagonal faces as ( x ).3. Step 3: Count of Vertices and Edges Let's calculate the total number of vertices (( V )) and edges (( E )) in terms of ( x ): - For ( x ) regular pentagons, each pentagon has 5 edges. - For 2 squares, each square has 4 edges. - Therefore, the total number of edges can be calculated as: [ E = frac{5x + 2 cdot 4}{2} = frac{5x + 8}{2} ] Since each edge is shared by 2 faces. Every vertex of a convex polyhedron must have the internal angles sum to less than ( 360^circ ). Each internal angle of a regular pentagon is ( 108^circ ). Since 4 such angles would exceed ( 360^circ ), no vertex can be formed by more than 3 faces. Thus, each vertex is formed by 3 faces. Therefore, the total number of vertices ( V ) is: [ V = frac{5x + 8}{3} ]4. Step 4: Euler's Formula Use Euler's polyhedron formula, which states for any convex polyhedron that ( V - E + F = 2 ). Here, ( F ) represents the number of faces, which is ( x + 2 ): [ frac{5x + 8}{3} - frac{5x + 8}{2} + (x + 2) = 2 ] Simplifying the above equation step-by-step: [ frac{5x + 8}{3} - frac{5x + 8}{2} + x + 2 = 2 ] Find a common denominator and combine terms: [ frac{2(5x + 8)}{6} - frac{3(5x + 8)}{6} + x + 2 = 2 ] which simplifies to: [ frac{2(5x + 8) - 3(5x + 8)}{6} + x + 2 = 2 ] [ frac{- (5x + 8)}{6} + x + 2 = 2 ] [ - frac{5x + 8}{6} + x + 2 = 2 ]5. Step 5: Solve the Equation Multiply through by 6 to clear denominators: [ - (5x + 8) + 6x + 12 = 12 ] Simplify and solve for ( x ): [ -5x - 8 + 6x + 12 = 12 ] [ x + 4 = 12 ] [ x = 8 ] So, there would be 8 pentagonal faces if such a polyhedron existed.6. Step 6: Verification with Vertex Count For ( x = 8 ), substitute back to find the number of vertices: [ V = frac{5 cdot 8 + 8}{3} = frac{40 + 8}{3} = 16 ] We established earlier total number of faces ( F = 10 ), and edges ( E = 24 ) since: [ E = frac{5 cdot 8 + 8}{2} = 24 ] Euler's formula ( V - E + F = 2 ) holds: [ 16 - 24 + 10 = 2 ]7. Step 7: Geometric Consideration Consider the geometry of the polyhedron, starting with vertices where three pentagonal faces meet. Let’s call such vertices ( alpha )-type vertices. Each meeting point of the pentagonal faces will create more ( alpha )-type vertices adjacent to each vertex. This rapidly reaches an inconsistency, as all vertices would need to be of this ( alpha )-type, leaving no room for the square faces.8. Step 8: Potential ( beta )-type or ( gamma )-type vertices Realizing that inserting square faces disrupts the continuous arrangement of pentagons, while trying to adhere to the ( alpha ) arrangement geometry further demonstrates the model's inconsistency logically and spatially, proving infeasibility.Conclusion:Given these derived geometrical and topological contradictions and inconsistencies, we conclude that a convex polyhedron with exactly 2 square faces and the remainder being pentagonal faces does not exist.[boxed{text{Not Possible}}]

🧠:Okay, let's try to figure out how to find the volume of a hemisphere inscribed in a cone with height 4 cm and slant height 5 cm. The hemisphere's base is on the base of the cone. Hmm, first, I need to visualize this. A cone with height 4 cm, and the slant height is 5 cm. So, the radius of the cone's base can be found using the Pythagorean theorem, right? Because in a right triangle formed by the height, radius, and slant height, so ( r^2 + h^2 = l^2 ). Here, h is 4, l is 5, so ( r^2 + 16 = 25 ), which means ( r^2 = 9 ), so radius r is 3 cm. Got that.Now, there's a hemisphere inscribed in the cone, with its base on the cone's base. So the hemisphere is sitting inside the cone, touching the cone's sides. The hemisphere's flat face is on the cone's base, and the curved part touches the cone's lateral surface. To find the volume of the hemisphere, I need its radius. Once I have the radius, the volume is ( frac{2}{3}pi r^3 ). So the key is to find the radius of the hemisphere.How do I find the radius of the hemisphere? Let me think. The hemisphere is tangent to the cone's surface. So, the center of the hemisphere's flat face is at the center of the cone's base. The hemisphere extends upward into the cone. The highest point of the hemisphere will be at a height equal to its radius from the base. Wait, but the hemisphere is inside the cone, so the top of the hemisphere must touch the cone's lateral surface. So, the hemisphere's radius is such that from the center of the base, moving up by radius r, the point (0,0,r) in some coordinate system, lies on the cone's surface.Wait, maybe I should set up coordinates. Let me place the vertex of the cone at the origin (0,0,0), but wait, no. Actually, the cone has its base on, say, the xy-plane, and the vertex at (0,0,4). The base is a circle with radius 3 cm in the plane z=0. The hemisphere is sitting on the base, so its center is at (0,0,0), and it extends up to z=r. The hemisphere equation would be ( x^2 + y^2 + (z)^2 = r^2 ), with z ≥ 0. But the cone's equation is... Let's see.The cone has height 4, base radius 3, so its equation can be written as ( sqrt{x^2 + y^2} = frac{3}{4}z ). Because at height z, the radius is (3/4)z. So, for any point on the cone, the distance from the z-axis is (3/4)z. The hemisphere is tangent to the cone, so there must be a point where both equations are satisfied and the tangent planes are the same.Alternatively, since the hemisphere is inside the cone, the hemisphere's surface must touch the cone's surface at exactly one point. Let's parametrize the hemisphere. Any point on the hemisphere satisfies ( x^2 + y^2 + z^2 = r^2 ), with z ≥ 0. The cone is ( sqrt{x^2 + y^2} = (3/4)z ). To find where they touch, substitute the cone equation into the hemisphere equation.Let me substitute ( sqrt{x^2 + y^2} = (3/4)z ) into the hemisphere equation. Squaring both sides, ( x^2 + y^2 = (9/16)z^2 ). Plugging into the hemisphere equation: ( (9/16)z^2 + z^2 = r^2 ). Combine terms: ( (25/16)z^2 = r^2 ), so ( z^2 = (16/25)r^2 ), so z = (4/5)r. Therefore, the point of tangency is at z = (4/5)r. Then, substituting back into the cone equation: ( sqrt{x^2 + y^2} = (3/4)(4/5)r = (3/5)r ). So, at the point of tangency, the distance from the z-axis is (3/5)r.But the hemisphere's equation at that z is ( x^2 + y^2 + z^2 = r^2 ). We already have x² + y² = (9/25)r² from the cone, and z = (4/5)r, so indeed, substituting those into the hemisphere equation: (9/25)r² + (16/25)r² = (25/25)r² = r². So that checks out.But how does this help us find r? Wait, maybe we need another condition. Since the hemisphere is entirely inside the cone, the slope of the hemisphere's surface at the point of tangency must match the slope of the cone's surface. That is, their normals must be in the same direction, or their tangent planes are the same. Maybe we need to compute the gradient vectors of both surfaces at the point of tangency and set them to be scalar multiples.Let's compute the normal vector to the cone at the point (x, y, z). The cone's equation is ( F(x, y, z) = sqrt{x^2 + y^2} - (3/4)z = 0 ). The gradient of F is ( (x)/sqrt(x² + y²), (y)/sqrt(x² + y²), -3/4 ). At the point of tangency, which is ( (3/5)r, 0, (4/5)r ) if we take a point along the x-axis for simplicity due to symmetry. So, let's take x = (3/5)r, y = 0, z = (4/5)r. Then, sqrt(x² + y²) = (3/5)r. So, the gradient of F at this point is ( (3/5)r / (3/5)r, 0 / (3/5)r, -3/4 ) = (1, 0, -3/4).Now, the hemisphere's equation is ( G(x, y, z) = x² + y² + z² - r² = 0 ), with z ≥ 0. The gradient of G is (2x, 2y, 2z). At the same point ( (3/5)r, 0, (4/5)r ), the gradient is ( 2*(3/5)r, 0, 2*(4/5)r ) = ( (6/5)r, 0, (8/5)r ).Since the surfaces are tangent at that point, their gradient vectors must be parallel. Therefore, the gradient of F is a scalar multiple of the gradient of G. So, there exists a scalar λ such that:(1, 0, -3/4) = λ*( (6/5)r, 0, (8/5)r )So, component-wise:1 = λ*(6/5)r0 = λ*0 (which is okay)-3/4 = λ*(8/5)rFrom the first equation: λ = 1 / ( (6/5)r ) = 5/(6r)From the third equation: -3/4 = λ*(8/5)rSubstituting λ from the first equation: -3/4 = (5/(6r))*(8/5)r = (8/6) = 4/3Wait, that gives -3/4 = 4/3, which is not possible. That can't be right. That suggests a contradiction, which must mean an error in my reasoning.Hmm, maybe the gradients aren't parallel? But if the surfaces are tangent, their normals should be parallel. So perhaps I made a mistake in setting up the equations. Let me check my calculations.Gradient of F for the cone: Let me re-derive it. The cone equation is ( sqrt{x^2 + y^2} = (3/4)z ). Let me write this as ( F(x, y, z) = sqrt{x^2 + y^2} - (3/4)z = 0 ). The partial derivatives:dF/dx = (x)/sqrt(x² + y²)dF/dy = (y)/sqrt(x² + y²)dF/dz = -3/4So, at the point ( (3/5)r, 0, (4/5)r ), as above, sqrt(x² + y²) = (3/5)r. So, dF/dx = (3/5)r / (3/5)r = 1dF/dy = 0dF/dz = -3/4So, gradient vector (1, 0, -3/4). Correct.Gradient of G for the hemisphere: ( G(x, y, z) = x² + y² + z² - r² ). So, gradient is (2x, 2y, 2z). At the point ( (3/5)r, 0, (4/5)r ), that's (6/5 r, 0, 8/5 r ). Correct.So, if they are parallel, then (1, 0, -3/4) = λ*(6/5 r, 0, 8/5 r )Thus, 1 = λ*(6/5) rand -3/4 = λ*(8/5) rSo, solving for λ from the first equation: λ = 5/(6r)Substitute into the second equation:-3/4 = (5/(6r)) * (8/5) r = (8/6) = 4/3But -3/4 is not equal to 4/3. This is a contradiction, which means that my assumption that the gradients are parallel is wrong, or my setup is wrong. But if the surfaces are tangent, their normals should be parallel. So, maybe the point of tangency isn't along the x-axis? Wait, no, due to symmetry, the point of tangency should lie along any radial direction, so it's sufficient to check along the x-axis.Alternatively, perhaps the hemisphere is not tangent at that point. Maybe there's another condition. Wait, perhaps the hemisphere touches the cone along a circle, but that doesn't make sense because a hemisphere and a cone would intersect in a circle only if the hemisphere is larger, but in this case, it's inscribed. Wait, maybe the hemisphere touches the cone at only one point? But in 3D, the intersection of a cone and a sphere (hemisphere) is generally a circle. Hmm, but if it's tangent, then the intersection is a single point. So maybe we need the hemisphere and cone to intersect at exactly one point. But how?Alternatively, perhaps the hemisphere's "top" point is at the center axis, but since it's a hemisphere, the top point is at height r. But the cone's height is 4 cm, so the hemisphere's radius can't be more than 4 cm. But the cone's radius at the base is 3 cm, so maybe the hemisphere's radius is constrained by the cone's slope.Wait, another approach. Imagine a cross-sectional view of the cone and hemisphere. The cross-section through the axis of the cone is a triangle, and the hemisphere would appear as a semicircle tangent to the sides of the triangle.So, let's do that. Take a cross-sectional view: the cone becomes an isoceles triangle with height 4 cm, base 6 cm (since radius is 3 cm). The hemisphere inscribed in this triangle, with its base on the base of the triangle. So, the semicircle must be tangent to the two sides of the triangle.This might be easier to solve in 2D first, then translate back to 3D.In the cross-sectional view, the triangle has vertices at (-3, 0), (3, 0), and (0, 4). The semicircle is sitting on the base from (-r, 0) to (r, 0), centered at (0,0), with radius r. The semicircle is the upper half: ( x^2 + y^2 = r^2 ), y ≥ 0.The sides of the triangle are the lines from (0,4) to (3,0) and (-3,0). Let's find the equation of the right side: from (0,4) to (3,0). The slope is (0-4)/(3-0) = -4/3. So the equation is y = -4/3 x + 4.The semicircle must be tangent to this line. So, the distance from the center of the semicircle (0,0) to the line y = -4/3 x + 4 must be equal to the radius r.The distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a² + b²). Let's write the line in standard form: 4/3 x + y - 4 = 0. Multiply both sides by 3 to eliminate fractions: 4x + 3y - 12 = 0.Wait, original line is y = -4/3 x + 4. So rearranged: 4x + 3y - 12 = 0. So a = 4, b = 3, c = -12.The distance from (0,0) to this line is |4*0 + 3*0 -12| / sqrt(16 + 9) = | -12 | / 5 = 12/5 = 2.4 cm. But this distance must equal the radius r. So, r = 12/5 = 2.4 cm.Wait, that seems straightforward. So in the cross-sectional view, the radius of the semicircle is 12/5 cm. Therefore, in 3D, the hemisphere has radius 12/5 cm. Hence, the volume is ( frac{2}{3}pi (12/5)^3 ).But wait, let me confirm. If in the cross-section the semicircle has radius 12/5 cm, then in 3D, the hemisphere has the same radius. Because in the cross-section, the semicircle is part of the hemisphere. So yes, the radius is 12/5 cm.So, converting that to a volume: ( frac{2}{3}pi left( frac{12}{5} right)^3 ). Let's compute that. 12/5 is 2.4, so cubed is 13.824, multiplied by 2/3 gives 9.216, times π is approximately 9.216π. But let me do the exact calculation.( left( frac{12}{5} right)^3 = frac{1728}{125} )So, volume is ( frac{2}{3} times frac{1728}{125} times pi = frac{3456}{375} pi = frac{1152}{125} pi ). Simplifying, 1152 ÷ 125 = 9.216. So, 1152/125 π cm³.Therefore, the volume is ( frac{1152}{125}pi ) cm³.But let me verify why the previous approach with gradients gave a contradiction. Maybe because I considered the hemisphere as a full sphere equation, but it's only a hemisphere. Wait, no, the hemisphere is just the upper half, so maybe the tangency condition is different. Wait, but in the cross-sectional view, solving it as a 2D problem gives the correct radius, and that seems valid because of symmetry. The 3D problem reduces to the 2D cross-section because everything is rotationally symmetric.So, probably the first approach with gradients was incorrect because of a miscalculation or wrong setup, but the cross-sectional method is accurate. Therefore, the radius is indeed 12/5 cm, and the volume is 1152/125 π cm³.But let's check with another method to ensure.Alternative approach: Using similar triangles. In the cross-sectional view, the cone is a triangle with base 6 cm and height 4 cm. The hemisphere inscribed in it has radius r. The center of the hemisphere is at the base, so at (0,0). The top of the hemisphere is at (0, r). The point of tangency on the right side of the triangle is somewhere on the line y = -4/3 x + 4. The distance from the center (0,0) to this line is r, which we already found as 12/5. Alternatively, using similar triangles.Consider the triangle formed by the cone. The hemisphere touches the cone's side at a point. The line from the apex of the cone (0,4) to the point of tangency (x, y) must be tangent to the hemisphere. The tangent line to the hemisphere at that point is perpendicular to the radius. So, the radius from (0,0) to (x, y) is perpendicular to the tangent line.But the tangent line is also the side of the cone, which has slope -4/3. Therefore, the radius vector (x, y) must have a slope that is the negative reciprocal of -4/3, which is 3/4. So, the slope of the radius is y/x = 3/4. Therefore, y = (3/4)x.But the point (x, y) is also on the cone's side: y = -4/3 x + 4. So, substituting y = (3/4)x into that:(3/4)x = -4/3 x + 4Multiply both sides by 12 to eliminate denominators:9x = -16x + 489x + 16x = 4825x = 48x = 48/25Then, y = (3/4)(48/25) = 36/25So, the point of tangency is (48/25, 36/25). The distance from (0,0) to this point is sqrt( (48/25)^2 + (36/25)^2 ) = sqrt( (2304 + 1296)/625 ) = sqrt(3600/625) = 60/25 = 12/5. So, the radius is 12/5 cm, which confirms the previous result.Therefore, the radius of the hemisphere is 12/5 cm, and the volume is ( frac{2}{3}pi (12/5)^3 = frac{2}{3}pi times frac{1728}{125} = frac{3456}{375}pi = frac{1152}{125}pi ).Simplifying ( frac{1152}{125} ), which is 9.216. So, 9.216π cm³. But since the question asks for the volume, we can leave it as a fraction times π. So, the exact volume is ( frac{1152}{125}pi ) cm³.Therefore, after going through the cross-sectional analysis, similar triangles, and verifying with coordinate geometry, I'm confident that the radius is 12/5 cm and the volume is 1152/125 π cm³.Final AnswerThe volume of the hemisphere is boxed{dfrac{1152}{125}pi} cubic centimeters.

✅:To find the volume of a hemisphere inscribed in a cone with height 4 cm and slant height 5 cm, we start by determining the radius of the cone's base using the Pythagorean theorem. The radius ( r ) of the cone is found as follows:[r^2 + 4^2 = 5^2 implies r^2 + 16 = 25 implies r^2 = 9 implies r = 3 text{ cm}]Next, we consider the cross-sectional view of the cone and hemisphere. The cone is an isosceles triangle with height 4 cm and base 6 cm (radius 3 cm). The hemisphere inscribed in this triangle has its base on the base of the triangle. The equation of the cone's side in the cross-sectional view is ( y = -frac{4}{3}x + 4 ).The distance from the center of the hemisphere (0,0) to the line ( 4x + 3y - 12 = 0 ) (which is the standard form of the cone's side) must equal the radius ( r ) of the hemisphere. The distance formula gives:[text{Distance} = frac{|4 cdot 0 + 3 cdot 0 - 12|}{sqrt{4^2 + 3^2}} = frac{12}{5} = 2.4 text{ cm}]Thus, the radius ( r ) of the hemisphere is ( frac{12}{5} ) cm. The volume ( V ) of the hemisphere is calculated as:[V = frac{2}{3} pi r^3 = frac{2}{3} pi left( frac{12}{5} right)^3 = frac{2}{3} pi left( frac{1728}{125} right) = frac{3456}{375} pi = frac{1152}{125} pi]Therefore, the volume of the hemisphere is:[boxed{dfrac{1152}{125}pi}] cubic centimeters.

🔑:1. Identify the given values and relationships: The height ((h)) of the cone is 4 cm and the slant height ((l)) is 5 cm. Let's sketch a right triangle formed by the height, radius of the base ((R)), and slant height of the cone.2. Calculate the radius of the cone's base: From the Pythagorean theorem applied to (triangle SOA), we have:[ l^2 = h^2 + R^2 ]Substituting (l = 5) cm and (h = 4) cm:[ 5^2 = 4^2 + R^2 ][ 25 = 16 + R^2 ][ R^2 = 25 - 16 ][ R^2 = 9 ][ R = sqrt{9} ][ R = 3 , text{cm} ]3. Volume formula for the inscribed hemisphere: The volume ((V)) of a hemisphere is given by:[ V = frac{2}{3} pi r^3 ]4. Establish relationships for the hemisphere radius ((r)): Since the hemisphere is inscribed in the cone, with its base on the cone's base, let the radius of the hemisphere be (r) and the distance from the cone's vertex to the top of the hemisphere be (x).5. Using Pythagorean theorem in triangles: From (triangle SBO) (where (S) is the cone's vertex and (O) is the center of the base):[ r^2 + (5-x)^2 = 16 ]From (triangle ABO) (where (A) is the point where height meets the base and (B) is directly below the top of the hemisphere):[ x^2 + r^2 = 9 ]6. Solve the system of equations: Solve the equations[ begin{cases}r^2 + (5 - x)^2 = 16 r^2 + x^2 = 9 end{cases}]Subtract the second from the first:[ r^2 + 25 - 10x + x^2 - r^2 = 16 ][ 25 - 10x + x^2 = 16 ][ 9 = 10x - x^2 ]This can be rearranged as:[ x^2 - 10x + 9 = 0 ]Solve for (x) using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}):[ x = frac{10 pm sqrt{100 - 36}}{2} ][ x = frac{10 pm 8}{2} ]Hence, we get two values for (x):[ x = frac{10 + 8}{2} = 9 , text{(not valid as } x < 5) ][ x = frac{10 - 8}{2} = 1 , text{(valid value)} ]Thus, (x = frac{9}{5}).7. Calculate (r): [ r^2 = 9 - x^2 ][ r^2 = 9 - left(frac{9}{5}right)^2 ][ r^2 = 9 - frac{81}{25} ][ r^2 = frac{225}{25} - frac{81}{25} ][ r^2 = frac{144}{25} ][ r = sqrt{frac{144}{25}} ][ r = frac{12}{5} , text{cm} ]8. Calculate the volume of the hemisphere:[ V = frac{2}{3} pi r^3 ][ r = frac{12}{5} ][ V = frac{2}{3} pi left(frac{12}{5}right)^3 ][ = frac{2}{3} pi cdot frac{1728}{125} ][ = frac{2 cdot 1728}{3 cdot 125} pi ][ = frac{1152}{125} pi , text{cm}^3 ] Conclusion:[boxed{frac{1152}{125} pi , text{cm}^3}]

🧠:Okay, so I need to find all integer values that the expression (pq + p^p + q^q)/(p + q) can take, where p and q are prime numbers. Hmm, primes can be 2, 3, 5, 7, etc. Let me start by trying small primes for p and q and see what results I get. Maybe patterns will emerge.First, let me consider the case where both p and q are 2. Then the expression becomes:(2*2 + 2^2 + 2^2)/(2 + 2) = (4 + 4 + 4)/4 = 12/4 = 3. So 3 is a possible value.Next, try p=2 and q=3. Let's compute:Numerator: 2*3 + 2^2 + 3^3 = 6 + 4 + 27 = 37Denominator: 2 + 3 = 537 divided by 5 is 7.4, which is not an integer. So this combination doesn't work. Hmm. Maybe p=2 and q=5? Let me check that.Numerator: 2*5 + 2^2 + 5^5 = 10 + 4 + 3125 = 3139Denominator: 2 + 5 = 73139 divided by 7. Let me do the division: 7*448 = 3136, so 3139 - 3136 = 3. So 3139/7 = 448 + 3/7, not integer. Not valid.How about p=3 and q=2? Let's swap them. Numerator: 3*2 + 3^3 + 2^2 = 6 + 27 + 4 = 37. Denominator: 5. Same as before, 37/5 = 7.4, still not integer.Wait, maybe both primes are odd. Let's try p=3 and q=3.Numerator: 3*3 + 3^3 + 3^3 = 9 + 27 + 27 = 63Denominator: 3 + 3 = 663/6 = 10.5, not integer. Hmm. What about p=3 and q=5?Numerator: 3*5 + 3^3 + 5^5 = 15 + 27 + 3125 = 3167Denominator: 83167 divided by 8: 8*395 = 3160, so 3167-3160=7. 3167/8=395.875. Not integer.Wait, maybe primes where one is 2 and the other is odd? Let's see. We tried p=2, q=3 and p=3, q=2. Both gave 37/5. Let me try p=2, q=7.Numerator: 2*7 + 2^2 +7^7. 2*7=14, 2^2=4, 7^7 is a big number. 7^2=49, 7^3=343, 7^4=2401, 7^5=16807, 7^6=117649, 7^7=823543. So total numerator: 14 +4 +823543=823561Denominator: 2 +7=9823561 divided by 9. Let's check divisibility: 9*91,506=823,554. 823,561 -823,554=7. So 823,561/9≈91,506.777… Not integer.This seems difficult. Maybe trying even larger primes is not the way. Let me try p=2 and q=2 first, which worked. Then maybe check p=2 and q=5, but that didn't work. How about p=5 and q=5?Numerator: 5*5 +5^5 +5^5=25 +3125 +3125=6275Denominator:106275/10=627.5. Not integer.Hmm. Maybe there's a pattern when one prime is 2 and the other is an odd prime. Let me check p=2 and q=7 again, but that was not integer. What about p=2 and q=13?Numerator:2*13 +2^2 +13^13. 13^13 is a gigantic number. Probably not helpful. Let me think differently.Wait, maybe the expression can be rewritten. Let's see:(pq + p^p + q^q)/(p + q). Maybe factor something out? Not sure. Let's try to consider modulo p + q. For the numerator to be divisible by p + q, then (pq + p^p + q^q) ≡ 0 mod (p + q). Let's explore this.Let me consider modulus (p + q). So, if we take modulo p + q, since p ≡ -q mod (p + q). Therefore, p ≡ -q mod (p + q). Similarly, q ≡ -p mod (p + q). Let's substitute these into the numerator.First term pq: since p ≡ -q, then pq ≡ -q^2 mod (p + q)Second term p^p: since p ≡ -q, p^p ≡ (-q)^p mod (p + q)Third term q^q: q^q ≡ (-p)^q mod (p + q)So the numerator becomes:- q^2 + (-q)^p + (-p)^q ≡ 0 mod (p + q)Hmm, this seems complicated, but maybe for specific primes, this congruence holds.Let me consider cases where one of the primes is 2. Let's say p=2 and q is an odd prime.Then, p + q = 2 + q.So, in the numerator:pq = 2qp^p = 2^2 =4q^q = q^qSo the numerator is 2q +4 + q^q.We need (2q +4 + q^q) divisible by (2 + q). Let's denote q as an odd prime.So let's write (q^q + 2q +4)/(q + 2). Let's see if this is an integer.Note that q + 2 divides q^q + 2q +4. Let's check for q=3:(3^3 +6 +4)/5=(27 +6 +4)/5=37/5=7.4, not integer.q=5:(5^5 +10 +4)/7=(3125 +14)/7=3139/7=448.428..., not integer.q=7:(7^7 +14 +4)/9=(823543 +18)/9=823561/9≈91506.777… Not integer.q=11:11^11 is huge, but let's compute (11^11 +22 +4)/13. Not sure. But seems like when p=2 and q is odd, it's not working. Except maybe q=2? Wait, q=2 is even prime, so p=2 and q=2 gives (4 +4 +4)/4=12/4=3. Which works.What if p and q are both odd primes? Then p and q are odd, so p + q is even. The numerator: pq is odd, p^p is odd (since p is odd), q^q is odd. So numerator is odd + odd + odd = odd. Denominator is even. So odd divided by even cannot be integer. Therefore, if both primes are odd, the expression cannot be integer. Therefore, the only possibility is when at least one of the primes is 2. But since we saw that when one is 2 and the other is odd, the result is not integer except maybe in some case. Wait, when p=2 and q=2, which are both even primes, the expression is integer. So all possible integer values must come from cases where at least one prime is 2. But if both primes are 2, we get 3. If one is 2 and the other is 2, same thing. If one is 2 and the other is an odd prime, maybe we can find another case where the division results in an integer.Wait, let's check p=2 and q=7 again. The numerator was 2*7 + 2^2 +7^7=14 +4 +823543=823561. The denominator is 9. 823561 divided by 9. 9*91,506=823,554. Remainder 7. So not divisible.Wait, what about p=2 and q=5? That was 3139/7. 3139 divided by 7: 7*448=3136. Remainder 3. Not divisible. Hmm. Maybe p=2 and q=3? 37/5=7.4. Not integer.Wait, perhaps there's another prime where p=2 and q is such that (2q +4 + q^q) is divisible by (q +2). Let's consider q=2:(2*2 +4 +2^2)/(2 +2)= (4 +4 +4)/4=12/4=3. Which works. What about q=2 and p=3? Wait, no, p and q are both primes, but p=3 and q=2 would be similar to p=2 and q=3. We already checked that.So maybe the only case where the expression is integer is when both primes are 2. But let's check another possibility. Wait, suppose p=2 and q=7. Let me check modulo (q + 2)=9. As before, the numerator modulo 9:2q +4 + q^q mod9. For q=7:2*7=14≡5 mod9, 4≡4 mod9, q^q=7^7. Compute 7 mod9=7. 7^1=7, 7^2=49≡4, 7^3=7*4=28≡1, 7^4=7*1=7 mod9. The cycle repeats every 3. 7^7: exponent 7 divided by 3 is 2 with remainder 1. So 7^7≡7^1≡7 mod9. Therefore numerator≡5 +4 +7=16≡7 mod9. 7≠0 mod9. So not divisible.Alternatively, maybe p=3 and q=3. Wait, but both are odd, so denominator is 6, numerator is 63. 63/6=10.5. Not integer.Wait, let me try p=2 and q=13. Maybe q=13. Compute (2*13 +4 +13^13)/(15). 13^13 is huge, but maybe check modulo 15.13 ≡ -2 mod15. So 13^13 ≡ (-2)^13 mod15. (-2)^13= -2^13. 2^4=16≡1 mod15. So 2^13=2^(4*3 +1)= (1)^3 *2=2 mod15. Therefore, (-2)^13≡-2 mod15. So 13^13≡-2 mod15.Numerator: 2*13=26≡11 mod15, 4≡4 mod15, 13^13≡-2 mod15. Total numerator≡11 +4 -2=13 mod15. 13≡13 mod15. Denominator is 15. So 13 mod15≠0. Therefore, not divisible.Alternatively, maybe p=2 and q=7. Wait, we did that already.Alternatively, try p=2 and q=5. Let's compute numerator modulo (2 +5)=7. 2*5=10≡3 mod7, 2^2=4≡4 mod7, 5^5. 5 mod7=5. 5^2=25≡4 mod7, 5^4≡4^2=16≡2 mod7, 5^5=5^4 *5≡2*5=10≡3 mod7. So numerator≡3 +4 +3=10≡3 mod7. 3≠0 mod7. Not divisible.Hmm. So maybe the only case where it's divisible is when both primes are 2. Let me check p=2 and q=2 again. Yep, gives 3. What if one prime is 2 and the other is 2. Same thing. What about other primes?Wait, wait. Let me check p=2 and q=2, which we already did. How about p=2 and q=7? We saw that the numerator is 823561, denominator is 9. 823561 divided by 9 is not integer. How about p=2 and q=3? 37/5=7.4. Not integer.Wait, maybe there's another case where p and q are different but give an integer. Let me think of primes where p + q divides pq + p^p + q^q.Alternatively, suppose p=3 and q=2. Let's check modulus 5 (since 3 +2=5). The numerator is 3*2 +3^3 +2^2=6 +27 +4=37. 37 mod5=2. Not divisible.Wait, maybe p=5 and q=2. Then denominator is 7. Numerator is 5*2 +5^5 +2^2=10 +3125 +4=3139. 3139 mod7. Let's compute 3139 divided by7. 7*448=3136. 3139-3136=3. So 3139≡3 mod7. Not divisible.Wait, maybe trying primes in reverse. If p=7 and q=2. Numerator:7*2 +7^7 +2^2=14 +823543 +4=823561. Denominator:9. Same as before, not divisible.Hmm. So far, the only valid case is p=q=2 giving 3. Wait, but maybe there's another case where p and q are different primes but the expression is integer. Let me try p=2 and q=7 again. Wait, no. Let's think differently.Perhaps the expression can be written as:[pq + p^p + q^q]/(p + q) = [p^p + q^q + pq]/(p + q)If we can write this as some combination. For example, if we consider p^p + q^q, maybe there's a factorization. But p^p and q^q are both powers. Not sure.Alternatively, maybe use the fact that p + q divides the numerator. So let's suppose that p + q divides pq + p^p + q^q. As we did before, use modular arithmetic. Let me think again for p=2 and q=2, which gives 3. What about p=2 and q= any other prime. Let's see:If p=2 and q is prime, then:Numerator: 2q +4 + q^qDenominator: q + 2We need (q^q +2q +4) divisible by (q +2). Let's denote q +2 as a divisor. Let's use polynomial division or factor theorem. For (q +2) to divide the numerator, then substituting q = -2 into the numerator should give 0. But q is a prime number, so q cannot be -2. Wait, but in algebra, if a polynomial f(x) is divisible by (x - a), then f(a)=0. But here, we have a similar concept. If we consider the numerator as a function of q, then if (q +2) divides the numerator, then substituting q = -2 into the numerator would yield zero. Let's compute the numerator at q=-2:(-2)^(-2) + 2*(-2) +4. Wait, this is problematic since (-2)^(-2) is 1/((-2)^2)=1/4. So 1/4 + (-4) +4=1/4. Which is not zero. Therefore, (q +2) is not a factor of the numerator in general.But perhaps for specific primes q, even if q ≠ -2, the numerator is divisible by q +2. For example, let's check q=2:Numerator:2^2 +2*2 +4=4 +4 +4=12. 12 divided by (2 +2)=4 gives 3. So works.q=3:3^3 +2*3 +4=27 +6 +4=37. 37 divided by5=7.4. Not divisible.q=5:5^5 +2*5 +4=3125 +10 +4=3139. 3139/7=448.428… Not integer.q=7:7^7 +2*7 +4=823543 +14 +4=823561. 823561/9=91506.777… Not integer.q=11:11^11 +2*11 +4. Let's compute 11^11 mod (11 +2)=13. 11 ≡ -2 mod13. So 11^11 ≡ (-2)^11 mod13. (-2)^11= (-2)^10 * (-2)= (1024)*(-2). But mod13:(-2)^1= -2(-2)^2=4(-2)^3=-8≡5(-2)^4= -10≡3(-2)^5= -6≡7(-2)^6= -14≡-1≡12(-2)^7= -12≡1(-2)^8= -2... This is getting too complicated. Let's use Fermat's little theorem: since 13 is prime, (-2)^12 ≡1 mod13. So (-2)^11= (-2)^(12-1)= (-2)^-1 mod13. The inverse of -2 mod13 is 6, since (-2)*6=-12≡1 mod13. Therefore, (-2)^11≡6 mod13.So numerator:11^11 +2*11 +4 ≡6 +22 +4 mod13. 22 mod13=9, 4 mod13=4. Total:6 +9 +4=19≡6 mod13. 6≠0 mod13. Therefore, not divisible.So it seems like for q=2, we get divisibility, but for other primes q, even when p=2, we don't. So maybe the only case is when both primes are 2.Wait, let me check p=3 and q=3 again. The expression was 63/6=10.5. Not integer. p=5 and q=5:6275/10=627.5. Not integer.Wait, but what if p and q are different primes but one of them is 2. Wait, we tried p=2 and q=3, 5,7,11, etc., none worked. So perhaps the only integer value is 3 when both primes are 2.But wait, let me check if there's another possible pair. Maybe p=2 and q= a higher prime where the numerator is divisible by q +2. Let's think about q=2: works. q=3:37/5=7.4. q=5:3139/7=448.428. q=7:823561/9≈91506.777. q=11: huge number divided by13, which we saw modulo13 gives 6, so not divisible. q=13: similarly, numerator modulo15 was 13, which isn't 0. q=17: even larger, likely not.Alternatively, maybe p=3 and q=2. We saw it gives 37/5=7.4. Not integer.Wait, another approach: suppose p=2 and q is a prime such that q +2 divides the numerator 2q +4 + q^q. Let's denote d = q +2. Then, d divides q^q +2q +4. Let's express everything in terms of d:Since d = q +2, then q = d -2. Substitute into the numerator:q^q +2q +4 = (d -2)^{d -2} + 2(d -2) +4.We need d divides this. Since q is prime and d = q +2, then d = prime +2. For example, q=2 gives d=4, which is composite. q=3 gives d=5, which is prime. q=5 gives d=7, prime. q=7 gives d=9, composite. q=11 gives d=13, prime. Etc.So maybe check d=5 (q=3):Numerator:3^3 +6 +4=27 +6 +4=37. 37 divided by5=7.4. Not integer.d=7 (q=5):5^5 +10 +4=3125 +14=3139. 3139/7=448.428… Not integer.d=13 (q=11):11^11 +22 +4. As before, numerator mod13=6≠0. Not divisible.d=9 (q=7): composite.d=4 (q=2): works, gives 3.So even when d is prime, except for d=4, which corresponds to q=2, the numerator is not divisible by d.Therefore, the only possible case where the expression is integer is when p=q=2, resulting in 3. Hence, the only integer value the expression can take is 3.But wait, before finalizing, let me check if there's another combination where both primes are same but not 2. For example, p=q=3:(3*3 +3^3 +3^3)/(3 +3)= (9 +27 +27)/6=63/6=10.5. Not integer.p=q=5:(25 +3125 +3125)/10=6275/10=627.5. Not integer.p=q=7:(49 +7^7 +7^7)/14. 7^7=823543. So numerator=49 +2*823543=49 +1647086=1647135. Divided by14:1647135/14=117652.5. Not integer.Hence, only when both primes are 2, the expression is integer. Therefore, the only integer value is 3.

✅:The expression (frac{p q + p^{p} + q^{q}}{p + q}) can only take the integer value 3 when both primes (p) and (q) are 2. Here's the systematic breakdown:1. Case (p = q = 2): [ frac{2 cdot 2 + 2^2 + 2^2}{2 + 2} = frac{4 + 4 + 4}{4} = frac{12}{4} = 3. ] This yields an integer result of 3.2. Cases with one prime as 2 and the other odd: - For (p = 2) and (q = 3, 5, 7, dots), calculations show the numerator modulo (p + q) does not vanish, making the division non-integer. For example: - (p = 2, q = 3): Numerator = 37, Denominator = 5 → (37/5) is not integer. - (p = 2, q = 5): Numerator = 3139, Denominator = 7 → (3139/7) is not integer.3. Cases with both primes odd: - The numerator (pq + p^p + q^q) is odd (sum of three odd terms), while the denominator (p + q) is even. An odd number divided by an even number cannot be an integer.4. Verification for other primes: - Testing (p = q = 3, 5, 7,) etc., results in non-integer divisions (e.g., (63/6 = 10.5) for (p = q = 3)).Conclusion: The only integer value attainable is (boxed{3}).

🔑:1. Let's begin by considering the given expression: [ frac{p q + p^{p} + q^{q}}{p + q} ] where ( p ) and ( q ) are prime numbers.2. We will investigate different combinations of ( p ) and ( q ) to determine when this expression yields integer values.3. Case 1: Both ( p ) and ( q ) are odd primes. - Since both ( p ) and ( q ) are odd, their product ( pq ) is odd. - ( p^p ) and ( q^q ) are also odd, since the power of an odd number remains odd. Hence, the numerator ( pq + p^p + q^q ) is odd. - The denominator ( p + q ) is the sum of two odd numbers, which is even. - An odd number divided by an even number cannot be an integer, hence the expression is not an integer in this case.4. Case 2: One of ( p ) or ( q ) is 2 (since 2 is the only even prime). Without loss of generality, let ( q = 2 ). We substitute ( q ) with 2: [ frac{2p + p^p + 2^2}{p + 2} = frac{2p + p^p + 4}{p + 2} ]5. Simplify the expression: [ frac{2p + p^p + 4}{p + 2} = frac{2(p + 2) + p^p}{p + 2} = 2 + frac{p^p}{p + 2} ]6. For our expression ( 2 + frac{p^p}{p + 2} ) to be an integer, the fraction (frac{p^p}{p + 2}) must also be an integer. This means ( p + 2 ) must divide ( p^p ).7. Divisibility condition analysis: - Note that for any prime number ( p ) other than 2, the positive divisors of ( p^p ) are ( 1, p, p^2, ldots, p^{p-1}, p^p ). - We must have ( p + 2 ) as one of these divisors to make ( frac{p^p}{p + 2} ) an integer.8. Evaluate the bounds between divisors: [ p < p + 2 < p + p = 2p < p^2 ] This inequality shows that ( p + 2 ) is strictly between ( p ) and ( p^2 ), hence ( p + 2 ) cannot be one of the positive divisors of ( p^p ). - Therefore, (frac{p^p}{p+2}) is not an integer for ( p > 2 ) and primes greater than 2.9. The only remaining possibility is ( p = 2 ). Substituting ( p = 2 ): [ frac{2 times 2 + 2^2 + 2^2}{2 + 2} = frac{4 + 4 + 4}{4} = frac{12}{4} = 3 ] This shows the expression simplifies correctly to integer 3. Conclusion:The only integer value the expression can take is:[boxed{3}]