Proof by infinite descent:
Suppose there axists a solution . Then since .
Therefore we also have .
Thus, after substitution and division by we get .
An analogous argument shows that and then and we get
a new solution .
We can proceed with this argument infinitely many times, which is of course absurd.
Note: This problem can be generalized to the statement that there are no
(EDIT) where is squarefree.