1.2. Find all prime numbers $p$ such that $5^{p}+12^{p}$ is a perfect square.
1.2. The problem can be changed to find all integers $m$ such that $5^{m}+12^{m}$ is a perfect square. Again the only answer is $m=2$. We shall give the solution in this more general case. (The solution of the original problem is easy by considering mod 5 or mod 10.) One solution is $p=2$ and we assert that it is the only solution. If $p=2 k+1$ is odd, then $5^{2 k+1}+12^{2 k+1} \equiv 2^{2 k+1} \equiv 2 \cdot 4^{k} \equiv 2(-1)^{k} \equiv 2$ or $3(\bmod 5)$. However the square of an integer can only be 0,1 or $4(\bmod 5)$. So $5^{p}+12^{p}$ is not a square when $p$ is odd. Now suppose that $5^{2 n}+12^{2 n}=t^{2}$ with $n \geq 2$. Then
$$
5^{2 n}=t^{2}-12^{2 n}=\left(t-12^{n}\right)\left(t+12^{n}\right) .
$$

If 5 divides both factors on the right, it must also divide their difference which means it divides 12. But this is impossible. Thus $t-12^{n}=1$ and
$$
5^{2 n}=2 \cdot 12^{n}+1 \quad \text { or } \quad 2^{2 n+1} 3^{n}=\left(5^{n}-1\right)\left(5^{n}+1\right) .
$$

If $n$ is odd, then $3 \mid 5^{n}+1$ and $3 \nmid 5^{n}-1$. Thus $5^{n}+1=2 \cdot 3^{n}$ and $5^{n}-1=4^{n}$ which cannot hold for $n>1$. If $n$ is even, then $5^{n}-1=2 \cdot 3^{n}$ and $5^{n}+1=4^{n}$, which again cannot hold for $n \geq 2$. Thus there is no solution for $p=2 n, n \geq 2$.