1988 IMO Problems/Problem 6
Problem
Let and be positive integers such that divides . Show that is the square of an integer.
Solution 1
Choose integers such that Now, for fixed , out of all pairs choose the one with the lowest value of . Label . Thus, is a quadratic in . Should there be another root, , the root would satisfy: Thus, isn't a positive integer (if it were, it would contradict the minimality condition). But , so is an integer; hence, . In addition, so that . We conclude that so that .
This construction works whenever there exists a solution for a fixed , hence is always a perfect square.
Solution 2 (Sort of Root Jumping)
We proceed by way of contradiction.
WLOG, let and fix to be the nonsquare positive integer such that such that or Choose a pair out of all valid pairs such that is minimized. Expanding and rearranging, This quadratic has two roots, and , such that WLOG, let . By Vieta's, and From , is an integer, because both and are integers.
From is nonzero since is not square, from our assumption.
We can plug in for in the original expression, because yielding . If then and and because is a positive integer.
We construct the following inequalities: since is positive. Adding , contradicting the minimality of
-Benedict T (countmath1)
Video Solution
https://www.youtube.com/watch?v=usEQRx4J_ew ~KevinChen_Yay
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Solution 3
Given that divides , we have for some integer .
Expanding the right side, we get . Rearranging terms, we have .
Consider this as a quadratic equation in . By the quadratic formula, we have
For to be an integer, the discriminant must be a perfect square. Let for some integer .
Rearranging terms, we get . Factoring the right side, we have .
Thus, and or . In either case, we have .
Substitute back into , we get .
Simplifying, we have . Therefore, , which is the square of an integer. By M. Nazaryan.