$$\left(\sum_{n=-\infty}^{\infty}x^{n^{2}}\right)^{2}=1+4\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{x^{-\left(2n+1\right)}-1}$$ Proving: $$\left(1+2\sum_{n=1}^{\infty}x^{n^{2}}\right)^{2}=1+4\sum_{n=0}^{\infty}\left(-1\right)^{n}\frac{x^{2n+1}}{1-x^{2n+1}}$$
This is reduced to the form: $$\left(\sum_{n=1}^{\infty}x^{n^{2}}\right)^{2}+\sum_{n=1}^{\infty}x^{n^{2}}=\sum_{n=0}^{\infty}\left(-1\right)^{n}\frac{x^{2n+1}}{1-x^{2n+1}}$$ The Sum on Right can be rewritten to the form: $$\sum_{n=1}^{\infty}\frac{x^{n}}{1+x^{2n}}=\sum_{n=1}^{\infty}\frac{1}{x^{n}+x^{-n}}$$ or, $$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}x^{j^{2}+i^{2}}+\sum_{n=1}^{\infty}x^{n^{2}}=\sum_{n=1}^{\infty}\frac{1}{x^{n}+x^{-n}}$$
I am unable to manipulate it any further to prove the equality.
This is taken from the Answer made Here by User Sangchul Lee and someone made a post to ask the same question but the post seems to have been deleted and no proof is present for this so I am asking it here again.
