Problem 5

Find all triples $(a,b,p)$ of positives integers with $p$ prime and  $$a^p=b!+p$$

Solution

All answers are $(a,b,p) = (2,2,2), (3,4,3)$. The key claim is the following.

Claim: $a$ is not divisible by any prime $q<p$.

Proof: Assume so. Then, $q\nmid a^p-p=b!\implies b<q$. Therefore,
$$b! \leq (q-1)! < q^{p-1} < q^p-p,$$so $a < q$, a contradiction. $\blacksquare$

Now we have two cases:

  • If $p \nmid a$, then $p\nmid b!\implies b < p$, so
    $$b! \leq (p-1)! < p^{p-1} \leq p^p-p,$$or $a<p$.

This gives a contradiction to the above claim.

 

  • If $p\mid a$, then $p^2\nmid a^p-p \implies b < 2p$, so
    \begin{align*} b! &\leq (2p-1)! \\ &= (1(2p-1))(2(2p-2)) \dots ((p-1)(p+1)) p \\ &< (p^2)(p^2)\dots (p^2) p \\ &= p^{2p-1} \leq (p^2)^p-p, \\ \end{align*}

so $a < p^2$. Therefore, the claim forces $a=p$. If $p\geq 5$, then by Zsigmondy theorem, we can find a prime $q$ such that

$\operatorname{ord}_q(p) = p-1$, so that $q\geq 2(p-1)+1 = 2p+1$ and $q\mid p^p-p$. This gives a contradiction to $b<2p$.

In conclusion, we must have $a=p$ and $p\in\{2,3\}$.