That equation has no integer solutions. Remember that

is congruent to 0 (mod 4) if

is even, and congruent t0 1 (mod 4) if

is odd.

So

unless

and

are both odd. But if

and

are both odd the

will be congruent to 2 or 3 (mod 4), so that case cannot arise.

Therefore

and hence

. But that implies that

,

and

are all even, say

,

,

. Then the equation becomes

. But this time, the right side is obviously congruent to 0 (mod 4), so

,

and

must all be even.

That argument can be continued indefinitely, so by

Fermat's method of infinite descent it follows that there are no solutions.