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Square of an integer

Let a, b be positive integers such that ab +1 divides a^2+b^2. Show that

    \[\frac{a^2+b^2}{ab+1}\]

is the square of an integer.

Solution

This is a very well known problem. It first appeared in IMO 1988 held in Canberra, Australia. It even has its own Wikipedia page.

History Background

One of the organisers had sent this exercise to all professors of Number Theory to check if the exercise is original and try to solve it within 4 hours. No one was able to do so. However, one Bulgarian student managed to solve the problem in less than 4 hours ( his solution was fallen from the sky )  and for that he was awarded a special reward beyond the medal.

Solution

 

Choose integers a,b,k such that a^2+b^2=k(ab+1) Now, for fixed k, out of all pairs (a,b) choose the one with the lowest value of \min(a,b). 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\implies c'<b' Thus, c' isn’t a positive integer (if it were, it would contradict the minimality condition). But c'=kb'-a', 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'>-1. We conclude that c'=0 so that b'^2=k.

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

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