(*) $ x = y = 1$ $ f^{2}(1) + 1 = 2f(1)$ $ f(1) = 1$ (**) $ y = 1$ $ f(x) + f\left(\frac {2008}{x}\right) = 2f(x)$ $ f\left(\frac {2008}{x}\right) = f(x)$ (***) $ y = \frac {2008}{x}$ $ f(x)f\left(\frac {2008}{x}\right) + f\left(\frac {2008}{x}\right)f(x) = 2$ $ f^{2}(x) = 1$ (****) $ f(xy) = f(x)f(y)$ $ x = y$ $ f(x^2) = f^{2}(x)\geq 0$ Answer : $ \boxed{f(x) = 1}$

This problem was proposed in the 2006 Spanish Mathematical Olympiad, except that the question stated that a $ \lambda\in\mathbb{R}$ existed such that \[ f(x)f(y) + f\left(\frac {\lambda}{x}\right)f\left(\frac {\lambda}{y}\right) = 2f(xy), \] and $ f(\lambda) = 1$.

$ \sum M_{x}=R_{m}^{|}$

http://www.mathlinks.ro/viewtopic.php?p=596307#596307

Same solution as #2, post for storage. $P(1,1)$ gives us $f(1)=1$. $P(1,y)$ gives us $f(y)=f(\frac{2008}{y})$. So we got $2f(x)f(y)=2f(xy)$, which is just $f(x)f(y)=f(xy)$. $P(x,x)$ gives us $f$ bounded, so $f(x)=1$ can be checked to be the only function that works.