Question

Is the following solvable?
(x^2 + 16) = (y^2 + 9) = (x+y)^2 + 1
where: ^ is square.
Thanks.

Answer 1

im not sure but if you expand all three you get (x+4)(x-4) = (y+3)(y-3) = (x+y-1)(x+y+1)
the only way i see to solve it is if all = 0. So x= 4, y=-3 should make it 0=0=0

Answer 2

Thansk TerrenceT;
But (x-4)(x+4) = x^2 - 4^2 is very different from mine (x^2 + 4^2).
I'm missing something? Thanks.