$p^m = (n+3)(n^2-3n+9)$, and $((n+3),(n^2-3n+9)) = (n+3, 27)$, so $p=3$. $n+3$ is a power of $3$, so it can only be $3, 9$, since $n+3< n^2 - 3n + 9$, so $n+3$ can only be a power of $3$ that is less than $27$. So $n=0, n=6$. So the only solutions are $n=0$, and $p=3=m$; and $n=6, p=3, m=5$.

Alternate beginning $p^m=x(x^2-9x+27)$, where $x=n+3$ Now, $x\neq 1$, so $p|x$ and $p|(x^2-9x+27)$. Thus, $p|27$. So $p=3$ Now proceed as solution by WizardMath

Math-wiz wrote: Alternate beginning $p^m=x(x^2-9x+27)$, where $x=n+3$ Now, $x\neq 1$, so $p|x$ and $p|(x^2-9x+27)$. Thus, $p|27$. So $p=3$ Now proceed as solution by WizardMath I think u missed the solution (n,p,m)=(-2,19,1),

Hmm. I think they both missed that. It says all integers but they assumed n+3 was not 1. Also, a bit of an explanation from $n+3< n^2 - 3n + 9$ to n+3<27: If n is not 0 then n+3 is at least 9, or n is at least 6. Then $n+3<n^2-3n+9=(n-3/2)^2+27/4\leq(6-3/2)^2+27/4=27$, so testing values now works. BTW, the reason why you missed -2 is because n+3 could be 1, which means n=-2. So our sols in (p, m, n) are (3, 3, 0), (3, 5, 6), and (19, 1, -2).