Exercise 2. Find all integers $m \geqslant 1$ and $n \geqslant 1$ such that $\frac{5^{m}+2^{n+1}}{5^{m}-2^{n+1}}$ is the square of an integer.
Solution to Exercise 2 The following proof is valid for \( m, n \in \mathbb{N} \).

First, \( 5^m - 2^{n+1} \) must divide \( 5^m + 2^{n+1} \), so it divides \( 5^m + 2^{n+1} - (5^m - 2^{n+1}) = 2^{n+2} \), therefore it is a power of 2. However, \( 5^m - 2^{n+1} \) is odd, so \( 5^m - 2^{n+1} = 1 \).

Write \( 5^m + 2^{n+1} = a^2 \). Then we have \((a-1)(a+1) = a^2 - 1 = 5^m + 2^{n+1} - 5^m + 2^{n+1} = 2^{n+2} \), so \( a-1 \) and \( a+1 \) are powers of 2.

Write \( a-1 = 2^c \) and \( a+1 = 2^d \) with \( c+d = n+2 \). Then \( c < d \) so \( a-1 = 2^c \) divides \( 2^d - 2^c = (a+1) - (a-1) = 2 \), thus \( a-1 = 1 \) or \( a-1 = 2 \).

If \( a = 2 \) then \( 2^d = a+1 = 3 \), which is impossible.

We deduce that \( a = 3, c = 1, d = 2 \) and \( n+2 = 3 \), which gives \( n = 1 \) and \( 5^m = 1 + 2^{n+1} = 5 \), then \( m = 1 \).

Finally, \( m = n = 1 \).