$p^{q+1}+ q^{p+1}$ is a perfect square. Find all the prime numbers $p$ and $q$ such that $p^{q+1} + q^{p+1}$ is a perfect square.  I tried this but I can't imagine where should I start fromCan some one give me a hint please.

Lemma: $a^2\equiv 0\text{ or }1 \mod 4$ for all $a\in\Bbb{Z}$. 
In fact, 


*

*if $a$ is even, then $a^2=(2k)^2=4k^2\equiv 0 \mod 4$

*if $a$ is odd, then $a^2=(2k+1)^2=4k^2+4k+1\equiv 1 \mod 4$.


Case 1: $p,q$ both odds.
Then $p+1$ and $q+1$ are even and $p^{q+1}\equiv q^{p+1}\equiv 1\mod 4$. Their sum can not be a square because is $2 \mod 4$.
Case 2: $p=q=2$.
Then $2^3+2^3=16=4^2$ is a square. Thus, $(p,q)=(2,2)$ is solution.
Case 3: wlog $p=2$ and $q$ is odd.
We have $2^{q+1}+q^3=t^2$. Because $q+1$ is even,
$q^3=t^2-2^{q+1}=\underset{A}{(t-2^{(q+1)/2})}\underset{B}{(t+2^{(q+1)/2})}$
Now, we have 2 subcases: 
a) $q^3=B$ and $1=A$.
b) $q^2=B$ and $q=A$.
Can you finish?
