Solution-1

Answer: \((2,2,2)\) and \((3,4,3)\).

Clearly, \(a>1\). We consider three cases.

Case 1: We have \(a<p\). Then we either have \(a \leqslant b\) which implies \(a \mid a^{p}-b!=p\) leading to a contradiction, or \(a>b\) which is also impossible since in this case we have \(b!\leqslant a!<a^{p}-p\), where the last inequality is true for any \(p>a>1\).

Case 2: We have \(a>p\). In this case \(b!=a^{p}-p>p^{p}-p \geqslant p!\) so \(b>p\) which means that \(a^{p}=b!+p\) is divisible by \(p\). Hence, \(a\) is divisible by \(p\) and \(b!=a^{p}-p\) is not divisible by \(p^{2}\). This means that \(b<2 p\). If \(a<p^{2}\) then \(a / p<p\) divides both \(a^{p}\) and \(b!\) and hence it also divides \(p=a^{p}-b!\) which is impossible. On the other hand, the case \(a \geqslant p^{2}\) is also impossible since then \(a^{p} \geqslant\left(p^{2}\right)^{p}>(2 p-1)!+p \geqslant b!+p\).

Comment. The inequality \(p^{2 p}>(2 p-1)!+p\) can be shown e.g. by using

\[(2p-1)!=[1 \cdot(2 p-1)] \cdot[2 \cdot(2 p-2)] \cdots \cdots[(p-1)(p+1)] \cdot p < \left(\left(\frac{2 p}{2}\right)^{2}\right)^{p-1} \cdot p=p^{2 p-1}\]

where the inequality comes from applying AM-GM to each of the terms in square brackets. **Case 3:** We have \(a=p\). In this case \(b!=p^{p}-p\). One can check that the values \(p=2,3\) lead to the claimed solutions and \(p=5\) does not lead to a solution. So we now assume that \(p \geqslant 7\). We have \(b!=p^{p}-p>p!\) and so \(b \geqslant p+1\) which implies that

\[\nu_{2}((p+1)!) \leqslant \nu_{2}(b!) = \nu_{2}\left(p^{p-1}-1\right) \stackrel{LTE}{=} 2\nu_{2}(p-1) + \nu_{2}(p+1)-1 = \nu_{2}\left(\frac{p-1}{2} \cdot(p-1) \cdot(p+1)\right),\]

where in the middle we used lifting-the-exponent lemma. On the RHS we have three factors of \((p+1)!\). But, due to \(p+1 \geqslant 8\), there are at least \(4\) even numbers among \(1,2, \ldots, p+1\), so this case is not possible.