Disclaimer: I don't know what the source of this problem is, but I would guess it's from a (non-Putnam) 2009 contest.
Let $f(x)$ be a function defined for all positive real numbers satisfying the conditions,
- $f(x)>0$ for all $x>0$, and
- $f(x-y)=\sqrt{f(xy)+1}$ for all $x>y>0$.
Find the value of $f(2009)$.
If I set $x=1$ and $y=\frac12$, then I find
$$f\left(\frac12\right)=\sqrt{f\left(\frac12\right)+1}\implies f\left(\frac12\right)=\frac{1+\sqrt5}2$$
Then I figured that, if I can find a solution to
$$\begin{cases}x-y=\frac12\\[1ex]xy=n&(n>0)\end{cases},$$
then I can always find the value of $f(n)$ in terms of $f\left(\frac12\right)$. In particular,
$$f(2009)=f\left(\frac12\right)^2-1=\frac{1+\sqrt5}2,$$
and moreover, it would seem $f(x)=\frac{1+\sqrt5}2$ for all values of $x>\frac12$. (This condition comes from the solution set for $x-y=\frac12,x>y>0$). Is this correct?
Yes, this is correct. There is always a solution to $$ \begin{align*} x - y & = \frac{1}{2} \\ xy & = n. \end{align*} $$
It is given by $$ x = \frac{1}{4} + \sqrt{n + \frac{1}{16}} \quad \text{and} \quad y = -\frac{1}{4} + \sqrt{n + \frac{1}{16}}. $$
This was found by noting that $x$ and $-y$ are the roots of the polynomial $$ \lambda^2 - \frac{1}{2} \lambda - n = 0 $$ and using the quadratic formula.