# proof of equivalence of Fermat’s Last Theorem to its analytic form

Consider the Taylor expansion^{} of the cosine function. We have

${\mathrm{lim}}_{s\to \mathrm{\infty}}({A}_{s})=2-\mathrm{cos}x-\mathrm{cos}y$

and

${\mathrm{lim}}_{s\to \mathrm{\infty}}({B}_{s})=1-\mathrm{cos}z$.

For $r>x,y$ the sequence^{} ${a}_{r}$ is decreasing as the denominator grows faster than the numerator.
Hence for $s>x,y$ the sequence ${A}_{s}$ is increasing as ${A}_{s+4}={A}_{s}+{a}_{s+2}-{a}_{s+4}$ and
${a}_{s+2}>{a}_{s+4}$. So if ${A}_{N}>0$ for some $N>x,y$, we have $2-\mathrm{cos}x-\mathrm{cos}y>0$.
Conversely if no such $N$ exists then ${A}_{s}\le 0$ for $s>x,y$, so its limit, $2-\mathrm{cos}x-\mathrm{cos}y$, is also less than or equal to $0$. However as this
expression cannot be negative we would have $2-\mathrm{cos}x-\mathrm{cos}y=0$.

Similarly for $r>z$ the sequence ${b}_{r}$ is decreasing, and for $s>z$ the sequence ${B}_{s}$ is increasing. So if ${B}_{M}>0$ for some $M>z$ we have $1-\mathrm{cos}z>0$. Conversely if no such $M$ exists then $1-\mathrm{cos}z\le 0$. However as this expression cannot be negative we would have $1-\mathrm{cos}z=0$.

Note that $2-\mathrm{cos}x-\mathrm{cos}y=0$ precisely when $x,y\in 2\pi \mathbb{Z}$. Also $1-\mathrm{cos}z=0$ precisely when $z\in 2\pi \mathbb{Z}$.

So the form of the theorem may be read:

If for positive reals $x,y,z$ we have ${x}^{n}+{y}^{n}={z}^{n}$ for some odd integer $n>2$, then either $x$ or $y$ not in $2\pi \mathbb{Z}$ or $z$ not in $2\pi \mathbb{Z}$.

Clearly this only fails if for positive integers $a,b,c$ and some odd $n>2$, we have

${(2\pi a)}^{n}+{(2\pi b)}^{n}={(2\pi c)}^{n}$.

Dividing through by ${(2\pi )}^{n}$ we see that ${a}^{n}+{b}^{n}={c}^{n}$.

Conversely suppose we have non-zero integers satisfying ${a}^{n}+{b}^{n}={c}^{n}$ for some $n>2$. If $n=4k$ we have ${({a}^{k})}^{4}+{({b}^{k})}^{4}={({c}^{k})}^{4}$, contradicting example of Fermat’s last theorem. Hence if $n$ is even we may replace $a,b,c$ with ${a}^{2},{b}^{2},{c}^{2}$ and $n$ with $n/2$, which will be odd and greater than 1 (and hence greater than 2 as it is odd). So without loss of generality we may assume $n$ odd.

Finally replace $a,b,c$ with their absolute values^{} and if
reorder to obtain a positive
integer solution. This would be a counterexample to the
form of the theorem as stated above.

Title | proof of equivalence of Fermat’s Last Theorem to its analytic^{} form |
---|---|

Canonical name | ProofOfEquivalenceOfFermatsLastTheoremToItsAnalyticForm |

Date of creation | 2013-03-22 16:19:04 |

Last modified on | 2013-03-22 16:19:04 |

Owner | whm22 (2009) |

Last modified by | whm22 (2009) |

Numerical id | 7 |

Author | whm22 (2009) |

Entry type | Proof |

Classification | msc 11D41 |