Croatia_2018

Let $n$ be a positive integer such that $$p^{q-1}+q^{p-1}=n^2.$$ We will divide our solution into cases, depending on the parity of numbers $p$ and $q$.

If both are even, the only possibility is that both are equal to $2$. This leads to a solution $(p,q)=(2,2)$ for $n=2$.

If both $p$ and $q$ are odd, we conclude that there are no solutions by looking at the remainder of division by $4$. If $p$ and $q$ are odd, then $p-1$ and $q-1$ are even, so that $p^{q-1}$ and $q^{p-1}$ are squares of odd numbers. This means that they always give remainder $1$ when divided by $4$. This shows that the left-hand side gives remainder $2$ when divided by $4$, whereas $n^2$ must give remainder $1$ or $0$ when divided by $4$.

The only remaining case is when one of the numbers $p$ and $q$ is even, while the other is odd. The equation being symmetric, we can, without loss of generality, assume that $p$ is odd, i.e. $p=2k+1$ for some positive integer $k$, and that $q$ equals $2$.

Now we have $$p+2^{p-1}=n^2,\text{ i.e. }p=n^2-2^{2k}=(n-2^k)(n+2^k).$$ Since $p$ is prime, the only way in which we can represent it as a product $(n-2^k)(n+2^k)$ is if one of the factors equals $\pm 1$, while the other is $\pm p$. Notice that $n+2^k>0$, and also that $n+2^k>n-2^k$, which yields $$n-2^k=1\text{and}n+2^k=p.$$ Subtracting these two equalities, we get $$2^{k+1}+1=p=2k+1.$$ Let us prove that this equation has no solution. We know that $2^{k+1}+1$ gives remainder $1$ when divided by $4$, so the same must hold for $2k+1$. This implies that $k$ is even, i.e. $k=2l$ for some positive integer $l$. In that case, $$p=2^{k+1}+1=2^{2l+1}+1=2\cdot 4^l+1$$ is divisible by $3$ (because $4^l$ gives remainder $1$ when divided by $3$). We conclude that $p$ is divisible by $3$, so $p=3$. Plugging this into the initial equation, we see that $p=3$ and $q=2$ is not a solution.

This means that the only solution is $(p,q)=(2,2)$.