Saudi Arabian Mathematical Competitions

By assumption, there is an integer $t$ such that $n^{2}-1 = t\left(m^{2}+1-n^{2}\right)$. Put $k = t+1$. It is clear that $k \neq 0$. We have $k\left(n^{2}-1\right) = t m^{2}$. Note that $(k, t) = 1$ we get there is an integer $l$ such that $n^{2}-1 = l t$ (which gives $l = m^{2}+1-n^{2}$) and $m^{2} = l k = k\left(m^{2}+1-n^{2}\right) = (t+1)\left(m^{2}+1-n^{2}\right)$. Therefore, it suffices to prove that $k$ is a perfect square.
Let $S$ be the set of ordered pairs $(x, y)$ of integers such that
$$(x+y)^{2} = k(1+4 x y) .$$
Then, the pair $\left(\frac{m+n}{2}, \frac{m-n}{2}\right) \in S$, and this shows that $S$ is non-empty. Put
$$a = \min \{|x| ; \exists(x, y) \in S\} .$$
We show that $a=0$ whence $k = y^{2}$.
If $(x, y) \in S \Rightarrow (-x, -y) \in S$, then $(a, y) \in S$. We have $(a+y)^{2} = k(1+4 a y)$. The two roots $b_{1}, b_{2}$ of the equation are integers with $\left|b_{1}\right|, \left|b_{2}\right| \geqslant a$. (Since $\left(a, b_{i}\right) \in S \Rightarrow \left(b_{i}, a\right) \in S$ ). On the other hand,
$$b_{1}+b_{2} = 4 a k - 2 a, \quad b_{1} b_{2} = a^{2} - k .$$
We get $\left(a+b_{1}\right)\left(a+b_{2}\right) = \left(4 a^{2}-1\right) k$.
If $k<0$ then $b_{1} b_{2}>0, b_{1}+b_{2} \leqslant 0 \Rightarrow b_{1}, b_{2}<0$. Since $\left|b_{i}\right|>a$, it follows that $a+b_{i}<0 \Rightarrow \left(a+b_{1}\right)\left(a+b_{2}\right)>0$. This means that $\left(4 a^{2}-1\right) k>0$. Since $k<0$, we get $a=0$. Now, assume that $k>0$ and $a \neq 0$ ($a>0$).
In this case, $b_{1}+b_{2}>0, \left(a+b_{1}\right)\left(a+b_{2}\right)>0 \Rightarrow b_{1}, b_{2}>0$. By the choice of $a$, we get $a^{2} \leqslant b_{1} b_{2}$, which contradicts to the equality $b_{1} b_{2} = a^{2} - k$. So $a=0$.