contest math - Prove that $f$ is additive if $f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$

Say $f:\mathbb{R}\to \mathbb{R}$ is non-constant such that $$f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$$ Prove that $f(x+y)=f(x)+f(y)$.

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If we put $a=f(0)$ and $y=0$ we get $$f(x)^2+af(x)= f(x)^2+a^2 \implies af(x) = a^2$$ If $a\ne 0$ then $f(x)=a$ is constant function which can not be, so $a=0$. Now if we put $x=y$ we get $$f(x)f(2x)=2f(x)^2$$

From here I have no more idea what to do.

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Edit after Lulu's comment: If we put also $y=-x$ we get $$2f(x)^2= f(x)f(2x) = f(x)^2+f(-x)^2\implies \boxed{f(x)^2=f(-x)^2}$$