An interesting problem

Here is an interesting functional equation problem. Can you solve it?

Find all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ so that
a) $ f(2u)=f(u+v)f(v-u)+f(u-v)f(-u-v)$ for all $u,v \in \mathbb{R}$ and
b) $ f(u)\ge  0$ for all $u \in \mathbb{R}$

Here is how I solved it. Substitute u for -u and v for -v. Then our functional equation 
becomes F(-2u)= F(u+v)F(v-u)+F(u-v)F(-u-v)=F(2u). Let 2u=x, then we get F(x)=F(-x).
Now let u=0. So we get F(0)=F(v)^2+ F(v)F(-v)= 2F(v)^2. If we let v=0, we get that F(0)=2F(0)^2. Solving for F(0), we get that F(0)= 0 or F(0)= 1/2. If F(0)=0, then F(x)=0 for all x. But if F(0)= 1/2, then F(x)= 1/2 or -1/2 for all x. However, F(x)≥0, so the only solutions are F(x)=0 or F(x)= 1/2.

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