Find all natural numbers $n$ such that $5^{n}+12^{n}$ is a perfect square.
By checking the cases \( n=1,2,3 \) we get the solution \( n=2 \) and \( 13^{2}=5^{2}+12^{2} \).

If \( n=2k+1 \) is odd, we consider the equation modulo 5 and we obtain

\[
\begin{aligned}
x^{2} & \equiv 5^{2k+1} + 12^{2k+1} \pmod{5} \equiv 2^{2k} \cdot 2 \pmod{5} \\
& \equiv (-1)^{k} \cdot 2 \pmod{5} \equiv \pm 2 \pmod{5}
\end{aligned}
\]

This is not possible, because the square residue of any natural number modulo 5 is 0, 1, or 4. Therefore \( n \) is even and \( x^{2} = 5^{2k} + 12^{2k} \). Rearrange this equation in the form

\[
5^{2k} = (x - 12^k)(x + 12^k)
\]

If 5 divides both factors on the right, it must also divide their difference, that is

\[
5 \mid (x + 12^k) - (x - 12^k) = 2 \cdot 12^k
\]

which is not possible. Therefore we must have

\[
x - 12^k = 1 \text{ and } x + 12^k = 5^{2k}
\]

By adding the above equalities we get

\[
5^{2k} - 1 = 2 \cdot 12^k
\]

For \( k \geq 2 \), we have the inequality

\[
25^k - 1 > 24^k = 2^k \cdot 12^k > 2 \cdot 12^k
\]

Thus we conclude that there exists a unique solution to our problem, namely \( n=2 \).