If $f(x-y)=f(x)f(y)-f(a-x)f(a+y), f(0)=1$
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a) $f(a)=0 $
b) $f(t)=f(-t)$
a) falls out easily setting $y=0, x=a$
I can get really close but I can't quite do this. I would appreciate a hint more than a full solution.
Setting $x=t, y=t$ $$f(0)=f(t)^2-f(a-t)f(a+t)$$ and $x=-t,y=-t$ $$f(0)=f(-t)^2-f(a+t)(a-t)$$
Then subtracting we find $f(t)= \pm f(-t)$. How do I eliminate the negative solution, or is there a better approach altogether?
$1) x=t, y=0$
$f(t)=f(t) f(0)-f(a-t) f(a)=f(t) f(0)$
because $ f(a)=0$
$2) x=0, y=t$
$f(-t)=f(t) f(0)-f(a-t) f(a)=f(t) f(0)$
From 1) and 2) $\Rightarrow \quad f(t)=f(-t)$