Question

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 $x > y > 0$. Determine $f(2009)$.

Answer 1

Suppose we choose $x=1$ and $y=\dfrac12$. Then

$f(x-y)=\sqrt{f(xy)+1}\implies f\left(\dfrac12\right)=\sqrt{f\left(\dfrac12\right)+1}\implies f\left(\dfrac12\right)=\dfrac{1+\sqrt5}2$


Now suppose we choose $x,y$ such that

$\begin{cases}x-y=\dfrac12\\\\xy=2009\end{cases}$


where we pick the solution for this system such that $x>y>0$. Then we find

$\dfrac{1+\sqrt5}2=\sqrt{f(2009)+1}\implies f(2009)=\dfrac{1+\sqrt5}2$

Note that you can always find a solution to the system above that satisfies $x>y>0$ as long as $x>\dfrac12$. What this means is that you can always find the value of $f(x)$ as a (constant) function of $f\left(\dfrac12\right)$.

Answer 2

Solution:

 It is given that, f(x) is a function such that, defined for all positive real numbers satisfying the conditions ,f(x) > 0 ,for all x > 0 , and also

    $f(x-y)=\sqrt{f(xy)+1}\\\\x>0,y>0\\\\for, x=1, \text{and} y=\frac{1}{2}\\\\f(1-\frac{1}{2})=\sqrt{f(1\times \frac{1}{2})+1}\\\\f(\frac{1}{2})^2=f(\frac{1}{2})+1\\\\f(\frac{1}{2})=\frac{1+\sqrt{5}}{2}$

Now, suppose

x=2009, y=0

$f(2009-0)=\sqrt{f(2009*0)+1}\\\\f(2009)=\sqrt{f(0)+1}\\\\f(2009)=\sqrt{f(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})+1}\\\\f(2009)=\sqrt{\sqrt{f(\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}})+1}+1}\\\\f(2009)=\sqrt{\sqrt{f(\frac{1}{2})+1}+1}\\\\f(2009)=\sqrt{\sqrt{\frac{1+\sqrt{5}}{2}+1}+1}\\\\f(2009)=\sqrt\sqrt{\frac{3+\sqrt{5}}{2}}+1}\\\\f(2009)=\sqrt \sqrt{{5.236}{2}}+1}\\\\=\sqrt{3.6180}\\\\=1.9021$