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.