answer:# Part (a) We need to find all prime ( p ) such that ( frac{7^{p-1} - 1}{p} ) is a perfect square. 1. **Initial Setup and GCD Property**: Since (gcd(7^{frac{p-1}{2}} - 1, 7^{frac{p-1}{2}} + 1) = 2) for any odd prime ( p ), if ( frac{7^{p-1} - 1}{p} = x^2 ) for some integer ( x ), then we have two cases to consider: - Case 1: ( 7^{frac{p-1}{2}} - 1 = 2py^2 ) and ( 7^{frac{p-1}{2}} + 1 = 2z^2 ) - Case 2: ( 7^{frac{p-1}{2}} - 2 = 2y^2 ) and ( 7^{frac{p-1}{2}} + 1 = 2pz^2 ) 2. **Case 1 Analysis**: - From ( 7^{frac{p-1}{2}} - 1 = 2py^2 ), we have ( 6 mid 7^{frac{p-1}{2}} - 1 ). This implies ( 3 mid py^2 ). - If ( p = 3 ), then ( frac{7^2 - 1}{3} = 16 = 4^2 ), which is a perfect square. - If ( p geq 5 ), then ( 3 mid y^2 ) implies ( 9 mid 7^{frac{p-1}{2}} - 1 ). Thus, ( 3 mid frac{p-1}{2} ). Let ( k = frac{p-1}{6} ). Then ( 2z^2 = 7^{3k} + 1 = (7^k + 1)(7^{2k} - 7^k + 1) ). - For ( 7^{2k} - 7^k + 1 ) to be a perfect square, we need ( (7^k - 1)^2 leq 7^{2k} - 7^k + 1 leq (7^k)^2 ), which is not possible. 3. **Case 2 Analysis**: - Since 2 is a quadratic residue modulo 7, but (-1) is not, this case does not yield any valid solutions. 4. **Conclusion for Part (a)**: Therefore, the only prime ( p ) for which ( frac{7^{p-1} - 1}{p} ) is a perfect square is ( p = 3 ). # Part (b) We need to find all prime ( p ) such that ( frac{11^{p-1} - 1}{p} ) is a perfect square. 1. **Initial Setup and GCD Property**: Since (gcd(11^{frac{p-1}{2}} - 1, 11^{frac{p-1}{2}} + 1) = 2) for any odd prime ( p ), if ( frac{11^{p-1} - 1}{p} = x^2 ) for some integer ( x ), then we have: - Case 2: ( 11^{frac{p-1}{2}} - 2 = 2y^2 ) and ( 11^{frac{p-1}{2}} + 1 = 2pz^2 ) 2. **Case 2 Analysis**: - Since 2 is not a quadratic residue modulo 11, we only consider Case 2. - From ( 11^{frac{p-1}{2}} + 1 = 2pz^2 ), we have ( p mid 11^{frac{p-1}{2}} + 1 ). Thus, ( 2y^2 = 11^{frac{p-1}{2}} - 1 equiv -2 pmod{p} ), implying ( p mid y^2 + 1 ). - Therefore, ( 4 mid p-1 ). Let ( k = frac{p-1}{4} ). Then ( 11^{frac{p-1}{4}} + 1 = u^2 ) or ( 11^{frac{p-1}{4}} + 1 = 2u^2 ). - Both cases are impossible because ( u-1 ) and ( u+1 ) cannot both be powers of 11, and 2 is not a quadratic residue modulo 11. 3. **Conclusion for Part (b)**: Therefore, there is no prime ( p ) for which ( frac{11^{p-1} - 1}{p} ) is a perfect square. The final answer is ( boxed{ p = 3 } ) for part (a) and no prime ( p ) for part (b).

answer:To show that the function ( f(x) = log_{a} left( x + sqrt{b x^2 + 2 a^2} right) ) is an odd function and to find the real numbers ( (a, b) ), we will follow these steps: 1. **Understand the Property of Odd Functions**: For ( f(x) ) to be an odd function, it must satisfy ( f(-x) = -f(x) ). Additionally, evaluating ( f(0) ) should give us a clue about the relationship between ( a ) and ( b ). 2. **Calculate ( f(0) )**: Let's start by evaluating ( f(x) ) at ( x = 0 ): [ f(0) = log_{a} left( 0 + sqrt{b cdot 0^2 + 2 a^2} right) = log_{a} left( sqrt{2 a^2} right) ] Simplifying inside the logarithm: [ f(0) = log_{a} ( sqrt{2} a ) ] Given that ( f(0) = 0 ) for odd functions (assuming the domain includes 0): [ log_{a} ( sqrt{2} a ) = 0 ] This implies: [ sqrt{2} a = 1 ] Solving for ( a ): [ a = frac{1}{sqrt{2}} = frac{sqrt{2}}{2} ] 3. **Rewrite and Analyze the Function with Found ( a )**: Substitute ( a = frac{sqrt{2}}{2} ) back into the function: [ f(x) = log_{frac{sqrt{2}}{2}} left( x + sqrt{b x^2 + 1} right) ] 4. **Check if ( f(x) ) is Indeed an Odd Function**: We need ( f(x) + f(-x) = 0 ). Evaluate ( f(-x) ): [ f(-x) = log_{frac{sqrt{2}}{2}} left( -x + sqrt{b (-x)^2 + 1} right) = log_{frac{sqrt{2}}{2}} left( -x + sqrt{b x^2 + 1} right) ] Add the two expressions: [ f(x) + f(-x) = log_{frac{sqrt{2}}{2}} left( x + sqrt{b x^2 + 1} right) + log_{frac{sqrt{2}}{2}} left( -x + sqrt{b x^2 + 1} right) ] Using the property of logarithms: [ f(x) + f(-x) = log_{frac{sqrt{2}}{2}} left[ (x + sqrt{b x^2 + 1}) cdot (-x + sqrt{b x^2 + 1}) right] ] Simplify the argument: [ (x + sqrt{b x^2 + 1})(-x + sqrt{b x^2 + 1}) = (sqrt{b x^2 + 1})^2 - x^2 = b x^2 + 1 - x^2 = (b - 1) x^2 + 1 ] So, [ f(x) + f(-x) = log_{frac{sqrt{2}}{2}} left[ (b - 1) x^2 + 1 right] ] For ( f ) to be an odd function, we require: [ log_{frac{sqrt{2}}{2}} left[ (b - 1) x^2 + 1 right] = 0 ] Solving the equation: [ (b - 1) x^2 + 1 = 1 ] This implies: [ (b - 1) x^2 = 0 quad forall x ] Therefore: [ b - 1 = 0 Rightarrow b = 1 ] 5. **Conclusion**: The values of ( a ) and ( b ) that make ( f(x) ) an odd function are: [ boxed{left( frac{sqrt{2}}{2}, 1 right)} ]

answer:**Analysis** This problem examines the relationship between trigonometric functions of the same angle and the sum formula for cosine. It is a basic question. **Solution** Given sin A= frac{ sqrt{5}}{5} and sin B= frac{ sqrt{10}}{10}, and since A and B are acute angles, we have cos A=frac{2 sqrt{5}}{5} and cos B=frac{3 sqrt{10}}{10}, thus cos (A+B)=frac{2 sqrt{5}}{5} times frac{3 sqrt{10}}{10} -frac{ sqrt{5}}{5}times frac{ sqrt{10}}{10} =frac{ sqrt{2}}{2}, and since A+B is in the interval (0^{circ},180^{circ}), we conclude A+B=frac{pi}{4}. Therefore, the answer is boxed{frac{pi}{4}}.

answer:Let us examine the properties of the function f(x): 1. First, setting x = y = 0, we have: f(0) = f(0 + 0) = f(0) + f(0) This simplifies to f(0) = 2f(0), and therefore, we conclude that f(0) = 0. 2. Letting y = -x, we get: f(0) = f(x - x) = f(x) + f(-x) Since we know f(0) = 0, this gives us f(-x) = -f(x), indicating that the function is odd. 3. Now, suppose we have two real numbers x_1 and x_2 such that x_1 > x_2. Then x_1 - x_2 > 0, and using the functional equation: f(x_1) - f(x_2) = f(x_1) + f(-x_2) = f(x_1 + (-x_2)) = f(x_1 - x_2) Given that x_1 - x_2 > 0 and the condition that f(x) < 0 for x > 0, we get: f(x_1 - x_2) < 0 Thus it follows that: f(x_1) - f(x_2) < 0 This implies that f(x_1) < f(x_2) for x_1 > x_2. Therefore, f(x) is a decreasing function on mathbb{R}. [boxed{text{Decreasing}}]