Positive integer solutions to $1\cdot m!\cdot(m^2)!\cdots (m^p)!=2(n^p)!$

Question

Let $m$ and $n$ be natural numbers and $p$ a positive integer such that

$$1\cdot m!\cdot(m^2)!\cdot\ldots\cdot(m^p)!=2(n^p)!$$

One solution is $(m,n,p)=(2,1,1)$. Are there any others?

Answer 1

Let $m,n,p\geq0$ be integers such that \begin{eqnarray} m!(m^2)!\ldots(m^p)!=2(n^p)!.\tag{1} \end{eqnarray} Suppose $n^p>m+m^2+\ldots+m^p$ and denote the difference by $k$. Then $$\frac{(n^p)!}{m!(m^2)!\ldots(m^p)!}=k!\cdot\frac{(n^p)!}{m!(m^2)!\ldots(m^p)!k!}=k!\cdot\binom{n^p}{m,m^2,\ldots,m^p,k},$$ where the latter is a multinomial coefficient, which is an integer. But by $(1)$ we have $$\frac{(n^p)!}{m!(m^2)!\ldots(m^p)!}=\frac{1}{2},$$ a contradiction. Hence $n^p\leq m+m^2+\ldots+m^p$, and for all $p\geq0$ we have $$(m+1)^p=\sum_{i=0}^p\binom{p}{i}m^i\geq\sum_{i=0}^pm^i>\sum_{i=1}^pm^i=m+m^2+\ldots+m^p.$$ It follows that $n\leq m$. But then $(n^p)!\leq(m^p)!$ an so from $(1)$ we get $$m!(m^2)!\ldots(m^{p-1})!\leq2,$$ which means that either $p=0$ or $m\leq2$. For $m=0$ and $m=1$ we get $$1=2(n^p)!,$$ which is of course impossible, and for $p=0$ there are clearly no solutions. For $m=2$ we get $$2!(2^2)!\ldots(2^p)!=2(n^p)!,$$ and knowing that $n\leq m$ we see that $n=0$, $n=1$ and $n=2$ are solutions for $p=1$, $p=1$ and $p=2$, respectively. So for $(m,n,p)$ we have the following solutions: $$(2,0,1)\qquad(2,1,1)\qquad(2,2,2).$$