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.