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For odd $n$ the Fermat equation $x^n + y^n = z^n$ factorises as $$(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n,$$

where $\zeta = e^{2 \pi i/n}$. I tried seeing this was true by multiplying the factors out by hand but got into a mess. Is there a neat way of showing this (e.g. is there a useful identity that can be used involving products of the $\zeta^i$ etc.?

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We will make use of the following theorem

Theorem Let $P(x)$ be a polynomial and $P(z)=0$, then $x-z$ divides $P(x)$.

Let $n$ be odd and write $$P(x) = x^n + y^n,$$ then $P(-\zeta^r y)=0$ for all $n$ powers of $r$.

Therefore $(x+y)(x+\zeta y)(x+\zeta^2 y)\cdots$ divides $P(x)$ but it has the same degree so its equal (up to a constant factor which is easily seen to be 1).

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Fix $x \neq 0$. Let

$$P(y)=(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) -x^n-y^n$$

Then $P(-x)=P(-\zeta x)=P(-\zeta^2 x)=..=P(-\zeta^{n-1} x)=0$ and $P(0)=x^n-x^n=0$.

Thus $P$ is a polynomial of degree at most $n$ which has at least $n+1$ roots. This proves that

$$(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) =x^n+y^n$$

for all $y$ and all $x \neq 0$.

The case $x=0$ is trivial.

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  • $\begingroup$ Rats, this is beautiful! +1 $\endgroup$ – DonAntonio May 2 '13 at 17:39
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Let $n$ be odd. The $n$ distinct roots of the polynomial $X^n-1$ are $1,\zeta,\dots,\zeta^{n-1}$ $$ X^n-1=(X-1)(X-\zeta)\cdots(X-\zeta^{n-1}) $$ Apply the transformation $X\mapsto -\frac{x}{y}$ $$ \left(-\frac{x}{y}\right)^n-1=\left(-\frac{x}{y}-1\right)\left(-\frac{x}{y}-\zeta\right)\cdots\left(-\frac{x}{y}-\zeta^{n-1}\right) $$ Finally, multiply through by $-y^n$ $$ x^n+y^n=(x+y)\cdots(x+\zeta^{n-1}y) $$

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  • $\begingroup$ You have to multiply it with $(-y)^n $ and not $-y^n $ to get the RHS, but then your LHS is $ x^n-(-y)^n $, so this can't be done unless $n$ is odd. $\endgroup$ – smiley06 May 2 '13 at 15:42
  • $\begingroup$ @smiley06 Yes, I assumed that the OP knew this identity only held for old $n$ (as stated in the question), so they are the same thing. $\endgroup$ – Warren Moore May 2 '13 at 15:52
  • $\begingroup$ I like your way very much! +3 :) $\endgroup$ – neofoxmulder Mar 1 '14 at 18:36

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