Competition problem — Algebra — Olympiad Maths Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

Find all functions $f: \mathbb{R} \to \mathbb{R}$ , such that:

1) For all real $x, y$ the following equality holds
$f(2x) = f(x+y)f(y-x) + f(x-y)f(-x-y).$

2) $f(x) \ge 0$ for all $x$ .

Solution

Take $x = y$ , we get:
$f(2x) = f(0)f(2x) + f(0)f(-2x). \quad (1)$
Take in (1) $x = 0$ and we get $f(0) = 2f^2(0)$ . So $f(0) = 0$ or $f(0) = \frac{1}{2}$ .
If $f(0) = 0$ , (1) implies that $f(2x) = 0$ , hence $f = 0$ .

If $f(0) = \frac{1}{2}$ , (1) implies that $f(2x) = \frac{1}{2}f(2x) + \frac{1}{2}f(-2x)$ and $f(2x) = f(-2x)$ .
Taking $x = 0$ and using the fact that our function is even, we arrive at
$\frac{1}{2} = f(0) = (f(y))^2 + (f(-y))^2 = 2(f(y))^2.$
This gives us that $f = \frac{1}{2}$ .

It is easy to check that both functions satisfy all the requirements.

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