# FLT for rational functions

Standard

Following is the problem 2.16 in The Math Problems Notebook:

Prove that if $n>2$, then we do not have any nontrivial solutions of the equation $F^n + G^n = H^n$ where $F,G,H$ are rational functions. Solutions of the form $F = aJ, G=bJ, H=cJ$ where $J$ is a rational function and $a,b,c$ are complex numbers satisfying $a^n + b^n = c^n$, are called trivial.

This problem is analogous to the Fermat’s Last Theorem (FLT) which states that for $n> 2$, $x^n + y^n = z^n$ has no nontrivial integer solutions.

The solution of this problem involves proof by contradiction:

Since any rational solution yields a complex polynomial solution, by clearing the denominators, it is sufficient to assume that $(f,g,h)$ is a polynomial solution such that $r=\max(\deg(f),\deg(g),\deg(h))$ is minimal among all polynomial solutions, where $r>0$.

Assume also that $f,g,h$ are relatively prime.  Hence we have $f^n+g^n = h^n$, i.e. $f^n-h^n = g^n$. Now using the simple factorization identity involving the roots of unity, we get:

$\displaystyle{\prod_{\ell = 0}^{n-1}\left(f-\zeta^\ell h\right) = g^n}$

where $\zeta = e^{\frac{2\pi i}{n}}$ with $i = \sqrt{-1}$.

Since $\gcd(f,g) = \gcd(f,h) = 1$, we have $\gcd(f-\zeta^\ell h, f-\zeta^k h)=1$ for $\ell\neq k$. Since the ring of complex polynomials has unique facotrization property, we must have $g = g_1\cdots g_{n-1}$, where $g_j$ are polynomials satisfying $\boxed{g_\ell^n = f-\zeta^\ell h}$.

Now consider the factors $f-h, f-\zeta h, f-\zeta^2 h$. Note that, since $n>2$, these elements belong to the 2-dimensional vector space generated by $f,h$ over $\mathbb{C}$.  Hence these three elements are linearly dependent, i.e. there exists a vanishing linear combination with complex coefficients (not all zero) in these three elements. Thus there exist $a_\ell\in\mathbb{C}$ so that $a_0g_0^n + a_1g_1^n = a_2g_2^n$. We then set $h_\ell = \sqrt[n]{a_\ell}g_\ell$, and observe that $\boxed{h_0^n+h_1^n = h_2^n}$.

Moreover, the polynomials $\gcd(h_\ell,h_k)=1$ for $\ell \neq k$ and $\max_{\ell}(h_\ell) = \max_{\ell} (g_\ell) < r$ since $g_\ell^n \mid f^n-h^n$. Thus contradicting the minimality of $r$, i.e. the minimal (degree) solution $f,g,h$ didn’t exist. Hence no solution exists.

The above argument fails for proving the non-existence of integer solutions since two coprime integers don’t form a 2-dimensional vector space over $\mathbb{C}$.