I have been struggling with this equation for some time and I do not seem to find any conclusive answer (it's from my research, not a homework).

It has to do with the **real** solutions $x$ to the following equation

$$ x + x f(x) = 1 + f(1),$$ where $$ f(x) = 2\sum_{n=1}^\infty \mathrm{e}^{(-ax^2-b) n^2} $$ with $a$ and $b$ strictly positive.

I know that $x=1$ solves the equation trivially; from simulations, I cannot find a contradiction to the fact that it should be the only solution. However, I cannot prove nor disprove that $x=1$ is the **only** solution.

I have tried using the upper bound $$ f(x) \leq \frac{ax^2+b+1}{ax^2+b}\mathrm{e}^{-ax^2 -b},$$ and i have tried relating $f(x)$ with the Elliptic theta function $$ f(x) = -1 + \theta_3(0,\mathrm{e}^{-ax^2 -b});$$ I have also tried to prove that $x = -x f(x) + 1 +f(1)$ is a contractive mapping; however, I have only found (quite restrictive) sufficient conditions on $a$ and $b$ for it to be true.

If someone manages to solve it or help me find a counterexample, I will gladly acknowledge their contribution in the paper i am writing.

EDIT: $x = -x f(x) +1 +f(1)$ is **not** a contractive mapping in general. I have found counterexamples where it is not (but the equation still only has one solution $x=1$).