Posted by: Alexandre Borovik | May 24, 2008

## Fermat’s Last Theorem for Polynomials

An old post by Nick Halloway at sci.math:

Someone mentioned to that an analog of FLT is true in $\mathbb{C}[x]$ as well, i.e. there are no relatively prime [non-constant] polynomials $a(x)$, $b(x)$ and $c(x)$ in $\mathbb{C}[x]$ with

$a^n + b^n = c^n$ for $n>2$.

This is true. You can show it by adapting the false proof of FLT.

It’s only necessary to show it for $n$ an odd prime and for $n = 4$. The proof is similar for $n = 4$ to $n$ an odd prime, so I will just show it for odd primes.

Proof by induction on the sum of the degrees of $a(x)$ and $b(x)$.

Suppose the sum of the degrees is $2$. Then $a^p + b^p$ factors as

$(a+b)(a+\rho b)\cdots (a+\rho^{p-1}b)$,

where $\rho$ is a primitive $p$‘th root of unity. These factors are all relatively prime polynomials of degree $1$. But $c^p$ has repeated factors. So it won’t work for this case.

Now assume the sum of the degrees is $m$. Again factor $a^p + b^p$ as

$(a+b)(a+\rho b)\cdots (a+\rho^{p-1}b).$

These factors are relatively prime. But $c^p$ has factors with multiplicity divisible by $p$. So each factor $a+\rho^ib$ must be a $p$‘th power. So we have

$a+b = d^p$,
$a+\rho b= e^p$,
$a+\rho^2 b = f^p$

So $b = (e^p - d^p)/ (\rho -1)$, and

$a = (e^p - \rho d^p) / (1-\rho)$

So $f^p = e^p (\rho+1) - \rho d^p$

Since neither $\rho +1$ nor $-\rho$ are 0, and $d(x)$ and $e(x)$ are relatively prime, this violates the induction hypothesis.

This also shows it for polynomials in $\mathbb{Z}[x]$, since polynomials that are relatively prime in $\mathbb{Z}[x]$ are also relatively prime in $\mathbb{C}[x]$.