Set x=3a and y=3b giving
15a^2 - 7b^2 = 1.
Write this as 15a^2 - 6b^2 = b^2 + 1.
LHS is evenly divisible by 3 so RHS must be too.
But b = 0,1,-1 mod 3 and b^2 + 1 = 1,2 mod 3, not 0.
Consequently, there are no integer solutions.
|
Posted by xdog
on 2012-12-10 16:11:43 |