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

solutions to

(EDIT) where is squarefree.