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

Consider the Taylor expansion of the cosine function. We have

${\\rm lim}_{s\\to\\infty}(A_{s})=2-{\\rm cos}\\,x-{\\rm cos}\\,y$

and

${\\rm lim}_{s\\to\\infty}(B_{s})=1-{\\rm cos}\\,z$ .

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 $2-{\\rm cos}\\,x-{\\rm cos}\\,y>0$ .Conversely if no such $N$ exists then $A_{s}\\leq 0$ for $s>x,y$ , so its limit, $2-{\\rm cos}\\,x-{\\rm cos}\\,y$ , is also less than or equal to $0$ . However as thisexpression cannot be negative we would have $2-{\\rm cos}\\,x-{\\rm cos}\\,y=0$ .

Similarly for $r>z$ the sequence $b_{r}$ is decreasing, and for $s>z$ the sequence $B_{s}$ isincreasing. So if $B_{M}>0$ for some $M>z$ we have $1-{\\rm cos}\\,z>0$ .Conversely if no such $M$ exists then $1-{\\rm cos}\\,z\\leq 0$ . However as thisexpression cannot be negative we would have $1-{\\rm cos}\\,z=0$ .

Note that $2-{\\rm cos}\\,x-{\\rm cos}\\,y=0$ precisely when $x,y\\in 2\\pi\\mathbb{Z}$ . Also $1-{\\rm cos}\\,z=0$ 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 $(a^{k})^{4}+(b^{k})^{4}=(c^{k})^{4}$ , contradictingexample 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 mayassume $n$ odd.

Finally replace $a, b, c$ 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.

单词	ProofOfEquivalenceOfFermatsLastTheoremToItsAnalyticForm
释义	proof of equivalence of Fermat’s Last Theorem to its analytic form Consider the Taylor expansion of the cosine function. We have ${\\rm lim}_{s\\to\\infty}(A_{s})=2-{\\rm cos}\\,x-{\\rm cos}\\,y$ and ${\\rm lim}_{s\\to\\infty}(B_{s})=1-{\\rm cos}\\,z$ . 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 $2-{\\rm cos}\\,x-{\\rm cos}\\,y>0$ .Conversely if no such $N$ exists then $A_{s}\\leq 0$ for $s>x,y$ , so its limit, $2-{\\rm cos}\\,x-{\\rm cos}\\,y$ , is also less than or equal to $0$ . However as thisexpression cannot be negative we would have $2-{\\rm cos}\\,x-{\\rm cos}\\,y=0$ . Similarly for $r>z$ the sequence $b_{r}$ is decreasing, and for $s>z$ the sequence $B_{s}$ isincreasing. So if $B_{M}>0$ for some $M>z$ we have $1-{\\rm cos}\\,z>0$ .Conversely if no such $M$ exists then $1-{\\rm cos}\\,z\\leq 0$ . However as thisexpression cannot be negative we would have $1-{\\rm cos}\\,z=0$ . Note that $2-{\\rm cos}\\,x-{\\rm cos}\\,y=0$ precisely when $x,y\\in 2\\pi\\mathbb{Z}$ . Also $1-{\\rm cos}\\,z=0$ 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 $(a^{k})^{4}+(b^{k})^{4}=(c^{k})^{4}$ , contradictingexample 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 mayassume $n$ odd. Finally replace $a, b, c$ 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.
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