Problem 1

Find all quadruples of positive integers

(a, p, m, o)

with

o \ge 2

such that

a! + p! = m^o + 26

Step 1 of 5: Reduce to

p \le 3

Previous step Next step

o\ge2,\ p\ge4 \implies 4\mid a!,\ 4\mid p! \implies m^{o}=a!+p!-26\equiv2\pmod4

Detailed analysis

Assume without loss of generality $a\ge p$ . If $p\ge4$ then both $a!$ and $p!$ are divisible by $4$ , so $m^o=a!+p!-26\equiv-26\equiv2\pmod4$ . But for $o\ge2$ , $m^o$ is odd when $m$ is odd and divisible by $4$ when $m$ is even, so $m^o\equiv2\pmod4$ is impossible. Hence $p\in\{1,2,3\}$ .