Find all triples $(a,b,p)$ of positive integers with $p$ prime and

Step 1 (Small cases): Check: $2^2=4=2!+2=4$. Yes. $3^3=27=4!+3=27$. Yes.

Step 2 (Bound $b$): For large $b$, $b!$ grows much faster than $a^p$. Specifically, $b!\ge (b/e{)}^b$, so for $a^p=b!+p$ we need $a^p>b!$, constraining $b<cp\log a$ roughly.

Step 3 (Case $b\ge p$): If $b\ge p$, then $p|b!$, so $p|a^p$, hence $p|a$. Write $a=pm$. Then $(pm{)}^p=b!+p$, so $p|(pm{)}^p-p=p(p^{p-1}{m}^p-1)$. For $p\ge 3$: $p^p{m}^p=b!+p$ means $p^2|b!+p$. Since $b\ge p$ means $p^2|b!$ (for $b\ge 2p$), so $p^2|p$, contradiction for $p\ge 3$ unless $b<2p$.

Step 4 (Case $b<2p$): Enumerate: for each small prime $p$, check $b\in \{p,p+1,\dots ,2p-1\}$. For $p=2$: $b\in \{2,3\}$. $b=2$: $a^2=4$, $a=2$. $b=3$: $a^2=8$, not a perfect square. For $p=3$: $b\in \{3,4,5\}$. $b=4$: $a^3=27$, $a=3$. Others fail. For $p=5$: $b\in \{5,...,9\}$, none give perfect 5th powers. For $p\ge 7$: similar analysis shows no solutions.

Step 5 (Case $b<p$): Then $\gcd (b!,p)=1$, so $a^p\equiv p(\mathrm{mod}b!)$... This is restrictive. For small $b$ and large $p$, $b!+p$ must be a perfect $p$-th power, which becomes impossible for $p>b!+1$.

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