Gọi cặp số là \(\left(m;n\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m+1=a.n\\n+1=b.m\end{matrix}\right.\) với \(a;b\in Z^+\)
\(\Rightarrow\left\{{}\begin{matrix}m=an-1\\n+1=bm\end{matrix}\right.\) \(\Rightarrow n+1=b\left(an-1\right)\)
\(\Rightarrow\left(ab-1\right)n=b+1\Rightarrow n=\frac{b+1}{ab-1}\)
- Với \(a=1\Rightarrow n=\frac{b+1}{b-1}=1+\frac{2}{b-1}\Rightarrow b=\left\{2;3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}n=3\Rightarrow m=2\\n=2\Rightarrow m=3\end{matrix}\right.\)
- Với \(a\ge2\Rightarrow ab-1\ge2b-1\)
Mặt khác \(n\in Z^+\Rightarrow n\ge1\Rightarrow\frac{b+1}{ab-1}\ge1\Rightarrow b+1\ge ab-1\)
\(\Rightarrow b+1\ge2b-1\Rightarrow b\le2\Rightarrow b=\left\{1;2\right\}\)
Với \(b=1\Rightarrow n=\frac{2}{a-1}\Rightarrow a=\left\{2;3\right\}\Rightarrow n=\left\{1;2\right\}\)
\(\Rightarrow\left(m;n\right)=\left(1;2\right);\left(2;1\right)\)
Với \(b=2\Rightarrow n=\frac{3}{2a-1}\Rightarrow a=\left\{1;2\right\}\Rightarrow n=\left\{3;1\right\}\) (giống trên)
Vậy ta có các cặp số là \(\left(1;2\right);\left(2;3\right)\)
