JBMO 2023 | Problem 1 - Solution - 25 June 2023

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 $(1,4)$ .