\(\dfrac{p}{m-1}=\dfrac{m+n}{p}\)
\(\Rightarrow p^2=\left(m-1\right).\left(m+n\right)\)
\(\Rightarrow p^2⋮m-1\)
\(\Rightarrow p⋮m-1\Leftrightarrow\left\{{}\begin{matrix}m-1=1\\m-1=p\end{matrix}\right.\)
Nếu \(m-1=p\Rightarrow m+n=p\)
\(\Rightarrow m-1=m+n\)
\(\Rightarrow n=-1\) ( loại )
Nếu \(m-1=1\Rightarrow m=2\left(TM\right)\)
Khi đó: \(p^2=\left(2-1\right).\left(2+n\right)\)
\(\Rightarrow p^2=2+n\)