It's impossible for the sum of the squares of two rational numbers to equal 14.
Set x=a/b and y=c/d and multiply out to get A^2 + B^2 = 14C^2, where A=(ad)^2, B=(bc)^2, C=(bd)^2.
(A,B) will be both even or both odd. In the first case, C will also be even, and after factoring 4 from each term an equation of the same form is the result.
So take (A,B) both odd. Then necessarily 7C^2 = 1 mod4, 7C^2 = 21 mod4, and C^2 = 3 mod4. But all squares are either =1 or =0 mod 4 so the conclusion follows.
Posted by xdog
on 2015-05-07 13:53:09