Problem:
Solution:
The two trivial solutions for both odd and even powers are
$ x = 0 $, $ y = 1 $
$ x = 1 $, $ y = 0 $
The two more trivial solutions for only even powers are
$ x = 0 $, $ y = -1 $
$ x = -1 $, $ y = 0 $
Suppose we have a non-trivial solution in $ \mathbf{Q} $, suppose the answer is $ x = \frac{p}{q} $, $ y = \frac{r}{s} $, then we have
$ (\frac{p}{q})^n + (\frac{r}{s})^n = 1 $
Multiply both side by $ q^ns^n $, we get a non-trivial solution for Fermat's last theorem.
$ (ps)^n + (rq)^n = (qs)^n $.
Solution:
The two trivial solutions for both odd and even powers are
$ x = 0 $, $ y = 1 $
$ x = 1 $, $ y = 0 $
The two more trivial solutions for only even powers are
$ x = 0 $, $ y = -1 $
$ x = -1 $, $ y = 0 $
Suppose we have a non-trivial solution in $ \mathbf{Q} $, suppose the answer is $ x = \frac{p}{q} $, $ y = \frac{r}{s} $, then we have
$ (\frac{p}{q})^n + (\frac{r}{s})^n = 1 $
Multiply both side by $ q^ns^n $, we get a non-trivial solution for Fermat's last theorem.
$ (ps)^n + (rq)^n = (qs)^n $.
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