Problem A. 432. (September 2007)
A. 432. Find all integer a for which there exists distinct positive integers x and y such that (axy+1) divides (ax^{2}+1)^{2}.
(5 pont)
Deadline expired on October 15, 2007.
Solution. We show that the desired positive integers x,y exist if and only if a1.
For a=1, a suitable pair is x=1, y=2 since (ax^{2}+1)^{2}=0.
For a0, the pair x=1, y=a+2 is suitable since x<y and (ax^{2}+1)^{2}=axy+1=(a+1)^{2}.
Now consider the case a<1. Let b=a2. Call a pair (x,y) of positive integers bad, if xy and axy+1=(bxy1) divides (ax^{2}+1)^{2}=(bx^{2}1)^{2}. To the contrary, suppose that there exist at least one bad pair and consider that pair where x is minimal; if there are more than one such pair, then take the smallest possible y for that value of x.
Case 1: x>y. Since
and
(by^{2}1)^{2} is also divisible by (bxy1). So the pair (y,x) is bad. But y<x which contradicts that we chose the smallest possible x.
Case 2: x<y. Consider the number . Both the numerator and the denominator are positive because b2 and x,y1; therefore c is a positive integer. Considering modulo bx,
hence c=bxz1 with some positive integer z; then
(bx^{2}1)^{2}=(bxy1)(bxz1).
From x<y we obtain z<x. Then the pair (x,z) is bad. Since z<x<y, this contradicts the choice of y.
Both cases lead to contradiction; bad pairs do not exist for b2.
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