Diophantine equation : x^{2n} + y^{2m} = z^2

Conjecture: If (x, y, z) are nonzero co prime integers, and (n, m) are positive integers > 1, then this Diophantine equation : x^{2n} + y^{2m} = z^2 Doesn't have any integer solution x,y,z are coprime Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd. If all three are odd, there can be no integer solution to x^{2n} + y^{2m} = z^2 because x^{2n} + y^{2m} is always even z^2 is always odd and the sqaure root of an odd number cannot be even. If one of x,y,z is even, the other two are odd: z = { x^{2n} + y^{2m} }^(1/2) But, again, x,y,z are coprime. z = { x^{2n} + y^{2m} }^(1/2) directly implies For some value v v1 + v2 ... + vv = v1 + v2 ... + vv Where z^2 = v1 + v2 ... + vv { x^{2n} + y^{2m} } = v1 + v2 ... + vv But, since x,y,z are coprime, no such value v exists. Q.E.D.