A Proof of Fermat’s Sums of Squares Theorem using Triple Product Identity

We present a proof of Jacobi’s formula for representation number for sums of two squares due to Michael D. Hirschhorn

$\displaystyle r_{2}(n)=4\left(d_{1}(n)-d_{3}(n)\right)$

$\displaystyle d_{i}(n)=\sum_{d \mid n,~ d \equiv i(\bmod 4)} 1$

Start with the Jacobi’s Triple Product identity

$\displaystyle \prod_{n = 1}^{\infty}\left(1+z x^{2 n-1}\right)\left(1+z^{-1} x^{2 n-1}\right)\left(1-x^{2 n}\right)=\sum_{n=-\infty}^{\infty} z^{n} x^{n^{2}}$

Plugging ${-z^{2} x}$ for ${z,}$ then ${x}$ for ${x^{2},}$ multiply by ${z}$ and we obtain

$\displaystyle \left(z-z^{-1}\right) \prod_{n = 1}^{\infty}\left(1-z^{2} x^{n}\right)\left(1-z^{-2} x^{n}\right)\left(1-x^{n}\right)$

$\displaystyle =\sum_{n=-\infty}^{\infty}(-1)^{n} z^{2 n+1} x^{\left(n^{2}+n\right) / 2}$

$\displaystyle =\sum_{n=-\infty}^{\infty} z^{4 n+1} x^{2 n^{2}+n}-\sum_{n=-\infty}^{\infty} z^{4 n-1} x^{2 n^{2}-n}$

$\displaystyle =z \prod_{n=1}^{\infty}\left(1+z^{4} x^{4 n-1}\right)\left(1+z^{-4} x^{4 n-3}\right)\left(1-x^{4 n}\right) -z^{-1} \prod_{n=1}^{\infty}\left(1+z^{4} x^{4 n-3}\right)\left(1+z^{-4} x^{4 n-1}\right)\left(1-x^{4 n}\right)$

Differentiating with ${z}$ and plugging ${z=1}$ we get,

$\displaystyle \prod_{n =1}^{\infty}\left(1-x^{n}\right)^{3}= \prod_{n=1}^{\infty}\left(1+x^{4 n-3}\right)\left(1+x^{4 n-1}\right)\left(1-x^{4 n}\right) \times\left(1-4 \sum_{n= 1}^{\infty}\left(\frac{x^{4 n-3}}{1+x^{4 n-3}}-\frac{x^{4 n-1}}{1+x^{4 n-1}}\right)\right)$

Divide by

$\displaystyle \prod_{n=1}^{\infty}\left(1+x^{n}\right)^{2}\left(1-x^{n}\right)$

which equals

$\displaystyle \prod_{n =1}^{\infty}\left(1+x^{4 n-3}\right)\left(1+x^{4 n-1}\right)\left(1-x^{4 n}\right)$

to get

$\displaystyle \prod_{n=1}^{\infty}\left(\frac{1-x^{n}}{1+x^{n}}\right)^{2}=1-4 \sum_{n= 1}^{\infty}\left(\frac{x^{4 n-3}}{1+x^{4 n-3}}-\frac{x^{4 n-1}}{1+x^{4 n-1}}\right)$

We also have

$\displaystyle \begin{aligned} \prod_{n =1}^{\infty}\left(\frac{1-x^{n}}{1+x^{n}}\right) &=\prod_{n =1}^{\infty} \frac{\left(1-x^{2 n-1}\right)\left(1-x^{2 n}\right)}{\left(1+x^{n}\right)} \\ &=\prod_{n =1}^{\infty}\left(1-x^{2 n-1}\right)\left(1-x^{n}\right) \\ &=\prod_{n =1}^{\infty}\left(1-x^{2 n-1}\right)\left(1-x^{2 n-1}\right)\left(1-x^{2 n}\right) \\ &=\sum_{n=-\infty}^{\infty}(-1)^{n} x^{n^{2}} \end{aligned}$

Therefore we established

$\displaystyle \left(\sum_{n=-\infty}^{\infty}(-1)^{n} x^{n^{2}}\right)^{2}=1-4 \sum_{n =1}^{\infty}\left(\frac{x^{4 n-3}}{1+x^{4 n-3}}-\frac{x^{4 n-1}}{1+x^{4 n-1}}\right)$

Plugging in ${-x}$ for ${x}$ we get

$\displaystyle \left(\sum_{n=-\infty}^{\infty} x^{n^{2}}\right)^{2}=1+4 \sum_{n =1}^{\infty}\left(\frac{x^{4 n-3}}{1-x^{4 n-3}}-\frac{x^{4 n-1}}{1-x^{4 n-1}}\right)$