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 (ax2+1)2.
Deadline expired on October 15, 2007.
Solution. We show that the desired positive integers x,y exist if and only if a-1.
For a=-1, a suitable pair is x=1, y=2 since (ax2+1)2=0.
For a0, the pair x=1, y=a+2 is suitable since x<y and (ax2+1)2=axy+1=(a+1)2.
Now consider the case a<-1. Let b=|a|2. Call a pair (x,y) of positive integers bad, if xy and |axy+1|=(bxy-1) divides (ax2+1)2=(bx2-1)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
(by2-1)2 is also divisible by (bxy-1). 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=bxz-1 with some positive integer z; then
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.