proof of equivalence of Fermat’s Last Theorem to its analytic form
Consider the Taylor expansion of the cosine function. We have
and
.
For the sequence is decreasing as the denominator grows faster than the numerator.Hence for the sequence is increasing as and. So if for some , we have .Conversely if no such exists then for , so its limit, , is also less than or equal to . However as thisexpression cannot be negative we would have .
Similarly for the sequence is decreasing, and for the sequence isincreasing. So if for some we have .Conversely if no such exists then . However as thisexpression cannot be negative we would have .
Note that precisely when . Also precisely when .
So the form of the theorem may be read:
If for positive reals we have for some odd integer , then either or not in or not in .
Clearly this only fails if for positive integers and some odd , we have
.
Dividing through by we see that .
Conversely suppose we have non-zero integers satisfying for some . If we have , contradictingexample of Fermat’s last theorem. Hence if is even we may replace with and with , which will be odd and greater than 1 (and hence greater than 2 as it is odd). So without loss of generality we mayassume odd.
Finally replace with their absolute values and if reorder to obtain a positiveinteger solution. This would be a counterexample to the form of the theorem as stated above.