Problem 5

Find all triples

(a,b,p)

of positive integers with

p

prime and

a^p=b!+p

Step 1 of 6: Bound

b

using

p

-adic valuation and Wilson's theorem

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In plain words

If $b!$ already contains two factors of $p$ , then $b!+p$ has exactly one factor of $p$ , which a perfect $p$ -th power can never have.

b\le2p-2

Detailed analysis

If $b\ge2p$ , then $p^2\mid b!$ , so $v_p(b!+p)=1$ , which is impossible for the perfect $p$ -th power $a^p$ . The case $b=2p-1$ is ruled out separately: $(2p-1)!+p=p\big[(p-1)!(p+1)(p+2)\cdots(2p-1)+1\big]$ , and Wilson's theorem shows the bracket is $\equiv2\pmod p$ for $p>2$ , contradicting divisibility by $p$ again; for $p=2$ , $3!+2=8$ is not a perfect square. Hence $b\le2p-2$ .