Problem 5

Find all triples

(a,b,p)

of positive integers with

p

prime and

a^p=b!+p

Step 6 of 6: Rule out

p\ge5

with a Zsigmondy prime

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b!=p^p-p=p(p-1)(1+p+\cdots+p^{p-2})

Detailed analysis

For $p\ge5$ , write $b!=p(p^{p-1}-1)=p(p-1)(1+p+\cdots+p^{p-2})$ . Zsigmondy's theorem gives a primitive prime divisor $q$ of $p^{p-1}-1$ ; its multiplicative order modulo $q$ is $p-1$ , so $q\equiv1\pmod{p-1}$ . Thus $q\ge2p-1>2p-2\ge b$ , and $q\ne p$ , contradicting $q\mid b!$ . Hence no solutions exist for $p\ge5$ .