Monthly Archives: July 2018

FLT for rational functions


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}.