nicetry007 wrote: $ gcd(nm-1,\; nm) = 1 \;\Rightarrow\; gcd(nm-1,\; m) = 1 \;\Rightarrow \;gcd(nm-1,\; m^{3}) = 1$ $ \frac{n^{3}\;+\;1}{mn \;-\;1}\in Z \;\Leftrightarrow \;\frac{(mn)^{3}\;+\;m^{3}}{mn \;-\;1}\in Z\; \Leftrightarrow \;\frac{(mn)^{3}\;-\;1\;+\;m^{3}\;+\;1}{mn \;-\;1}\in Z \;\Leftrightarrow\; \frac{m^{3}\;+\;1}{mn \;-\;1}\in Z$ This implies that if $ (n,m)$ is a solution, then $ (m,n)$ is also a solution. Suppose $ n = m$. $ \frac{n^{3}\;+\;1}{n^{2}\;-\;1}\;=\; n \;+\; \frac{n \;+\;1}{n^{2}\;-\;1}\;=\; n \;+\; \frac{1}{n \;-\;1}\in Z \;\Leftrightarrow \;n \; = \; 2$ Suppose $ n < m$. [Note: Since $ (n,m)$ and $ (m,n)$ are both solutions, it is enough to consider $ n < m$] $ \frac{n^{3}\;+\;1}{mn \;-\;1}\in Z \;\Leftrightarrow\; \frac{mn^{3}\;+\;m}{mn \;-\;1}\in Z \;\Leftrightarrow\; \frac{mn^{3}\;-\;n^{2}\;+\;n^{2}\;+\;m}{mn \;-\;1}\in Z\; \Leftrightarrow \;\frac{n^{2}\;+\;m}{mn \;-\;1}\in Z$ Since $ \frac{n^{2}\;+\;m}{mn \;-\;1}\in Z$, $ n^{2}\;+\;m \;\geq \;mn \;-\;1 \;\Leftrightarrow\;\frac{n^{2}\;+\;1}{n \;-\;1}\;\geq\; m$ (i.e.) $ \;m \;\leq\; n\;+\;1 \;+\;\frac{2}{n\;-\;1}\;\Rightarrow\; m \;\leq\; n\;+\;3$. We need to consider 3 cases $ (i)\; m = n+1, \;(ii)\; m = n+2$ and $ (iii)\; m = n+3$. Plugging $ m = n+1, \; n+2$ and $ n+3$ in $ \frac{n^{2}\;+\;m}{mn \;-\;1}$, we get $ (n,m)$ that satisfy the relation. The pairs $ (n,m)$ that satisfy the relation are $ (2,2),\;(1,2),\;(2,1),\;(1,3), \;(3,1),\;(3,5),\;(5,3), \;(2,5),\;(5,2)$.

$ mn-1|n^{3}+1$ $ \implies$ $ mn-1|n^{2}+m$ $ \implies$ $ k(mn-1)=n^{2}+m$ $ \implies$ $ n^{2}-kmn+m+k=0$ $ \Delta=k^{2}m^{2}-4(k+m)$ $ \implies$ $ (km-1)^{2}<k^{2}m^{2}-4(k+m)<(km)^{2}$ for $ 2km>4(k+m)+1$ if $ 2km\leq 4(k+m)+1$

cnyd wrote: $ mn - 1|n^{3} + 1$ $ \implies$ $ mn - 1|n^{2} + m$ $ \implies$ $ k(mn - 1) = n^{2} + m$ $ \implies$ $ n^{2} - kmn + m + k = 0$ $ \Delta = k^{2}m^{2} - 4(k + m)$ $ \implies$ $ (km - 1)^{2} < k^{2}m^{2} - 4(k + m) < (km)^{2}$ for $ 2km > 4(k + m) + 1$ if $ 2km\leq 4(k + m) + 1$ Actually since $ (km - 1)^2 = k^2m^2 - 4(k + m) \Rightarrow - 2km + 1 = - 4(k + m) \Rightarrow 1 = 0 \bmod 2$ we just have to check the cases where $ (km - 2)^2 \ge k^2m^2 - 4(k + m) \iff (m - 1)(k - 1) \le 2$, giving $ k = 1$ or $ m \in \{1,2,3\}$. $ k=1$: $ (m-2)^2-8$ is a square and hence $ m=5$ giving $ (m,n)=(5,2);(5,3)$. $ mn-1 \mid n^3 + 1 \iff mn-1 \mid m^3+1$ so: $ m=1$: $ n-1 \mid 2 \iff n=1,2$ $ m=2$: $ 2n-1 \mid 9 \iff n=1,2,5$ $ m=3$: $ 3n-1 = 28 \iff n=1,5$