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 $k\ne 0$. 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 $n^2-1=lt$ (which gives $l=m^2+1-n^2$) and $m^2=lk=k\left(m^2+1-n^2\right)=(t+1)\left(m^2+1-n^2\right)$. 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 $$(x+y{)}^2=k(1+4xy).$$ 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 $(a+y{)}^2=k(1+4ay)$. The two roots $b_1,b_2$ of the equation are integers with $\left|b_1\right|,\left|b_2\right|⩾a$. (Since $\left(a,b_i\right)\in S\Rightarrow \left(b_i,a\right)\in S$ ). On the other hand, $$b_1+b_2=4ak-2a,b_1{b}_2=a^2-k.$$ We get $\left(a+b_1\right)\left(a+b_2\right)=\left(4a^2-1\right)k$. If $k<0$ then $b_1{b}_2>0,b_1+b_2⩽0\Rightarrow b_1,b_2<0$. 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 $\left(4a^2-1\right)k>0$. Since $k<0$, we get $a=0$. Now, assume that $k>0$ and $a\ne 0$ ($a>0$). 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 $a^2⩽b_1{b}_2$, which contradicts to the equality $b_1{b}_2=a^2-k$. So $a=0$.