Problem:
Find 3x2y2 if x and y are integers such that y2+3x2y2=30x2+517.
Solution:
Rewrite the given equation in the form
(y2−10)(3x2+1)=3⋅132
and note that, since y is an integer and 3x2+1 is a positive integer, y2−10 must be a positive integer. Consequently, y2−10=1,3,13,39,169 or 507, implying that y2=11,13,23,49,179 or 517. Since the only perfect square in the second list is 49, it follows that y2−10=39, implying that 3x2+1=13,x2=4 and 3x2y2=12⋅49=588​.
The problems on this page are the property of the MAA's American Mathematics Competitions