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 $\mathbb{C}[x]$.

## Responses

1. The remark at the end is false. For example, consider the constant polynomials f=2 and g=3. These are relatively prime in Z[x], but are multiples of each other in C[x].

2. Two elements f, g in a UFD are relatively prime if there is no irreducible or prime element which divides them both. Since 2 and 3 are units in C[x], neither has a prime factor, so they are relatively prime in this ring.

3. Yes, you are right (I wasn’t thinking clearly), but then the result is false. We have that 2, 3, and (2^3+3^3)^(1/3) are all units in the ring, and clearly these yield a solution of a^n+b^n=c^n, with n=3.

In point of fact, now that I look at it, the proof assumes that the polynomials are nonconstant, but that isn’t stated in the statement of the theorem.