Given a prime , show has no nontrivial integer solutions.
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.
First part, to prove divisibility by the prime for and . From it's clear that and then . Dividing the equation by gives and since still divides it follows that , also . Dividing again by gives , where and are divisible by so .
and are of the forms and , respectively.Also, and are relatively prime to . Therefore, .
. If is the smallest of these, then after division we have . But then ; since and it's contradiction.
Similarly for the cases where or are the smallest powers.