Proving that a functional equation is constant

Question

A functional equation over ℝ is defined as $$ f(x+1/y) + f(x-1/y) = 2f(x)f(1/y) $$

and it's given that $f(0) = 0$.

We have to prove that $f(x)=0$ for all real x.

Substituting $y$ as $1/x$, we get that $f(2x)= 2 f^2(x)$ but I'm stuck as to any further progress.

Answer 1

Hint:

• Notice how $\frac{1}{y}$ may be changed for any $z\in \mathbb{R}^*$
• Prove that $\forall x\in \mathbb{R}, f(x)=-f(-x)$
• Prove that $\forall x\in \mathbb{R}, f(x)=f(-x)$
• Conclude from parity arguments