# 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 $\mathbb{C}$.

# Prime Number Problem

Standard

Following is a problem about prime factorization of the sum of consecutive odd primes. (source: problem 80 from The Green Book of Mathematical Problems)

Prove that the sum of two consecutive odd primes is the product of at least three (possibly repeated) prime factors.

The first thing to observe is that sum of odd numbers is even, hence the sum of two consecutive odd primes will be divisible by 2. Let’s see factorization of some of the examples:

$3 + 5 = 2\times 2 \times 2$
$5 + 7 = 2 \times 2\times 3$
$7+11 = 2 \times 3\times 3$
$11+13 = 2 \times 2 \times 2 \times 3$
$13+17 = 2 \times 3 \times 5$
$17+19 = 2\times 2\times 3 \times 3$
$19+23 = 42 = 2\times 3\times 7$
$23+29 = 52 = 2\times 2 \times 13$

Now let $p_n$ and $p_{n+1}$ be the consecutive odd primes, then from above observations we can conjecture that either $p_n+p_{n+1}$ is product of at least three distinct primes or $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ such that $k+\ell \geq 3$.

To prove our conjecture, let’s assume that $p_n+p_{n+1}$ is NOT a product of three (or more) distinct primes (otherwise we are done). Now we will have to show that if $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ then $k+\ell \geq 3$.

If $\ell = 0$ then we should have $k\geq 3$. This is true since $3+5=8$.

Now let $\ell > 0$. Since $k\geq 1$ (sum of odd numbers is even), we just need to show that $k=1, \ell=1$ is not possible. On the contrary, let’s assume that $k=1,\ell = 1$. Then $p_n+p_{n+1} = 2p$. By arithmetic mean property, we have

$\displaystyle{p_n < \frac{p_n+p_{n+1}}{2}} = p

But, this contradicts the fact that $p_n,p_{n+1}$ are consecutive primes. Hence completing the proof of our conjecture.

This is a nice problem where we are equating the sum of prime numbers to product of prime numbers. Please let me know the flaws in my solution (if any) in the comments.