Finding the family of functions satisfying $f(x + y)=f(x)f(y) - f(x-y)$

Question

What are the functions $f$ such that $f(x + y)=f(x)f(y) - f(x-y)$?

I got $$f(0) = 2$$ $$f(x)=f(-x)$$ $$\left(f(a)^2 - 4\right)\left(f(b)^2 - 4\right) \ge 0$$ $$f'(0)=0$$ Can this be solved using the given information? Is $f(x) = a^x + \frac{1}{a^x}$ the only solution? Thanks in advance!

Attention, $2f(x) = f(0)f(x)$ does not imply $f(0) = 2$. For instance the function $f(x) = 0$ for all $x$ is solution of your problem but $f(0) = 0 \neq 2$. — Mister Da, Nov 28 '19 at 14:44
Does this answer your question? Solution of the functional equation $g(x)g(z) = g(x+z)+g(x-z)$ — Mohsen Shahriari, Dec 27 '21 at 23:31

score 1 · Accepted Answer · answered Nov 28 '19 at 16:02

$f(x)=2 \cos x$ satisfies all the conditions given here. $$f(x+y)+f(x-y)= 4 \cos x \cos y= 2 f(x) f(y) ~~~(1)$$ $$f(0)=2, f(-x)=f(x), (4 \cos^2 a-4)(4\cos^2 b-4)=16 \sin^2 a \\sin^2 b \ge 0$$ and $f'(x) =\sin x, f'(0)= 0.$ Also $f(x)=2\cos x= e^{ix}+ e^{-ix}=a^x+a^{-x} \implies a=e^{i}$.

Finding the family of functions satisfying $f(x + y)=f(x)f(y) - f(x-y)$

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