1988 IMO ProblemsProblem 6 Problem Let a and b be positive integers such that ab 1 divides a2 b2 Show that frac a2 b2ab 1 is the square of an integer
Solution Choose integers abk such that a2b2=k(ab1) Now for fixed k out of all pairs (ab) choose the one with the lowest value of min(ab) Label b'=min(ab) a'=max(ab) 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-kb'2implies 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-kb'k1=b'2(b'-1)k1geq 1 so that c'-1 We conclude that c'=0 so that b'2=k
This construction works whenever there exists a solution (ab) for a fixed k hence k is always a perfect square
To modify or delete round reports, edit the associated round.