Problem 1. Find all pairs (a,b) of positive integers such that the numbers a!+b  and b!+a  are both powers of  5
 
Solution

Assume that $a \geq b$. We split into two cases.

Case 1: $a=b$. Then, $a!+a=5^k$ with $k \geq 1$. If $k=1$ then obiously $(a,b)=(5,5)$ works. If $k \geq 2$, then $(a-1)!+1 \equiv 1 \pmod 5$, and since $5^k=a!+a=a((a-1)!+1),$ we must have $(a-1)!+1=1,$ a contradiction.

Case 2: $a>b$. Then, $a=b+s$ with $s \geq 1$. Let $a!+b=5^k$ with $k \geq 1$. Therefore,

$b((b-1)! \cdot (b+1) \cdots (b+s)+1)=5^k$.

Thus, $b=5^\ell$ with $\ell \geq 0$, and if $b-1 \geq 5$ then $(b-1)! \cdot (b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction. Moreover, if $s \geq 5$, then $5 \mid (b+1) \cdots (b+5) \mid (b+1) \cdots (b+s),$ and so $(b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction.

Therefore, $b \leq 5$ and $s \leq 4$.

Taking all cases, we obtain $(b,s)=(1,3),$ that is $(a,b)=(4,1)$.

To sum up, we obtain the solutions $(a,b)=(5,5), (4,1)$ and $(1,4)$.