Problem 1

Find all quadruples of positive integers

(a, p, m, o)

with

o \ge 2

such that

a! + p! = m^o + 26

Step 3 of 5: Eliminate

p=2

Previous step Next step

p=2:\ a!=m^{o}+24,\ a=5\Rightarrow m^{o}=96\text{ (impossible)};\ a\ge6\Rightarrow m^{o}\equiv3\pmod9\Rightarrow o=1

Detailed analysis

If $p=2$ , then $a\ge2$ . Values $a\le4$ would give $a!-26\le-2$ , impossible because $m^o>0$ , so $a\ge5$ . If $a=5$ , then $m^o=96=2^5\cdot3$ , which is not a perfect $o$ -th power for any $o\ge2$ . If $a\ge6$ , then $9\mid a!$ , so $m^o\equiv-24\equiv3\pmod9$ ; but an $o$ -th power with $o\ge2$ is either $0\pmod9$ when $3\mid m$ or a unit modulo $9$ when $3\nmid m$ , and neither gives $3\pmod9$ . Thus $p=2$ gives no solution.