Number theory 4: 1988 IMO Problems

Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2$. Show that $\dfrac{a^2+b^2}{ab+1}$ is the square of an integer.

Solution

Choose integers $a,b,k$  such that $a^2+b^2=k(ab+1)$ ($k \in N*$) 

Now, for fixed $k$, out of all pairs $(a,b)$ choose the one with the lowest value of $min(ab)$. Label $b'=min(a,b)$,$a'=max(a,b)$. 

Thus,$a'^2-kb'a'+b'^2-k=0$ is a quadratic in $a'$. 

Should there be another root, c, the root would satisfy: $b'c\leq a'c=b'^2-k<b'^2\Rightarrow c<b'$ (By Viete) 

Thus, $c$  isn't a positive integer (if it were, it would contradict the minimality condition).
 But $c=kb'-a'$, (By Viete $c'+a'=kb'$) so $c$ is an integer; hence, $c\leq 0$. 

In addition, $(a'+1)(c+1)=a'c+a'+c+1=b'^2-k+b'k+1=b'^2+(b'-1)k+1 \geq 1$  so that $c \geq -1$.

 We conclude that $c=0$ so that $b'^2=k$.
$\Rightarrow k$ is the square of an integer.

This construction works whenever there exists a solution $(a,b)$ for a fixed $k$ , hence $k$  is always a perfect square.