Unique solution to an Algebraic equation $$f(x_{2}-x_{1})= f(x_{2}-y)f(y-x_{1})$$ I need to find a unique solution to this algebraic equation. Any hints as to how to proceed. The Exponential function works in this case, but I explicitly need to see the details. Thanks in advance.

I'm going to suppose that $x_1,x_2\in\mathbb{R}$ are variables and not fixed real numbers, and that $f$ is differentiable.
First, if $y=x_2$ then $f(x_2-x_1)=f(0)f(x_2-x_1)$, hence $f(0)=1$ unless $f(x_2-x_1)=0$ for every $x_1,x_2$, i.e., unless $f(x)=0$ in all of $\mathbb{R}$.
Now suppose $f\neq0$ and let us differentiate the equation with respect to $y$:
$0=-f'(x_2-y)f(y-x_1)+f(x_2-y)f'(x_1-y)$,
so $$f'(x_2-y)f(y-x_1)=f(x_2-y)f'(x_1-y)$$
and substituting $y=x_1$ we get
$$f'(x_2-x_1)=f'(0)f(x_2-x_1)$$
and therefore
$$f'(x)=\alpha_f f(x) \,\,\,\, (1)$$
for some constant $\alpha_f:=f'(0)\in\mathbb{R}$.
Since the derivative of $f$ is proportional to $f$, the unique family of solutions for (1) when $f\neq0$ is $f(x)=a^x$ for $a\in(0,\infty)$.
Finally we check that $f(x)=0$ and $f(x)=a^x$, $a>1$ are solutions to the original equation, so they are the only ones.
