How can $p^{q+1}+q^{p+1}$ be a perfect square? How can one find all primes $(p,q)$  such that $p^{q+1}+q^{p+1}$ is a perfect square 
I considered it $\mod 2$ and found a trival solution .
Im curious about an eventual answer Diophantine equations are extremely hard.
This seems harder than IMO Q2 of this year .
edit1:
I think one should consider it $\mod 4$ . 

First of all, note that $n^2\equiv 0,1\pmod 3$ for every $n\in\mathbb Z$.
Let us separate it into cases : 


*

*For $p\equiv q\equiv 0\pmod 3$, having $p=q=3$ leads $3^4+3^4=162$ which is not a perfect square.

*For $p\equiv 0,q\not\equiv 0\pmod 3$, having $3^{q+1}+q^{3+1}=m^2$ leads $3^{q+1}=(m-q^2)(m+q^2)$. Thus we have $m-q^2=3^s,m+q^2=3^t$ where $0\le s\le t$, so adding these gives $2m=3^s+3^t$. Here suppose that $1\le s$, then $m\equiv 0\pmod 3$, so $q\equiv 0\pmod 3$ which contradicts $q\not\equiv 0\pmod 3$. Hence, we have $s=0,t=q+1$. Thus, we have $3^{q+1}-2q^2=1$ which has no positive integer solution. 

*For $p\equiv 1,q\not\equiv 0\pmod 3$, $p^{q+1}+q^{p+1}\equiv 2$, so this cannot be a perfect square.

*For $p\equiv q\equiv 2\pmod 3$, if $p=q=2$, then $2^3+2^3=4^2$. For odd $p,q$, there exist $s,t$ such that $p=6s-1,q=6t-1$. Then, $p^{q+1}+q^{p+1}\equiv (-1)^{6t}+(-1)^{6s}\equiv 2\pmod 3$ which cannot be a perfect square.
Hence, $p=q=2$ is the only solution.
