Find $f(x)$ from two-variable equation

Question

Let $f$ be a function which satisfies the equation \begin{equation} f(x-y+a)-f(x+y+a)=2f(x)f(y) \end{equation} where $a$ is a positive real number and for every pair of real numbers $x$ and $y$. Suppose that $f(a)\neq0$. What is the value of $f(0)$? Hence or otherwise, prove that $f$ is an odd function.

What I've done: When I saw that equation, what springs to mind is to change the subject to $f(x) $. However, it's rather complicated and messy. Alternatively, I used differentiation and integration for both sides, to see whether it'll be easier for computation. However, I figured out $f$ is not assumed to be differentiable or continuous here.

The unsuccessful attempts make me wonder, how should I approach to this question? I'm currently confused and not knowing how to proceed. Any hints, approaches or answers are appreciated.

Answer 1

Hint:

for $x=0$ and $y=0$ your equation becomes $$ f(a)-f(a)=2f(0)f(0) $$

Answer 2

Hint: Substitute $y$ with $-y$. Then use that $f$ is either identically zero (and hence odd) or you can pick $x$ such that division by $2f(x)$ is allowed. (We do not need that $f(a)\ne0$).