⎪mn=pnp=mmp=n⇒mn.np.mp=p.m.n⇔m2.n2.p2=m.n.p

⇒m.n.p.(m.n.p−1)=0

⇔[m.n.p=0m.n.p=1

With m.n.p=0⇒p2=0⇒p=0⇒m=n=0

With m.n.p=1⇒p2=1⇒p=±1

  If p=1

 then {mn=1m=n⇒[m=n=1m=n=−1

  If  p=−1

 then {mn=−1m=−n⇒[m=1,n=−1m=−1,n=1

 So have 5 ordered triples of integers satisfied

\(\left\{{}\begin{matrix}mn=p\\np=m\\mp=n\end{matrix}\right.\)\(\Rightarrow mn.np.mp=p.m.n\Leftrightarrow m^2.n^2.p^2=m.n.p\)

\(\Rightarrow m.n.p.\left(m.n.p-1\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}m.n.p=0\\m.n.p=1\end{matrix}\right.\)

With \(m.n.p=0\Rightarrow p^2=0\Rightarrow p=0\Rightarrow m=n=0\)

With \(m.n.p=1\Rightarrow p^2=1\Rightarrow p=\pm1\)

  If \(p=1\) then \(\left\{{}\begin{matrix}mn=1\\m=n\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}m=n=1\\m=n=-1\end{matrix}\right.\)

  If  \(p=-1\) then \(\left\{{}\begin{matrix}mn=-1\\m=-n\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}m=1,n=-1\\m=-1,n=1\end{matrix}\right.\)

 So have 5 ordered triples of integers satisfied