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=\frac12`. Then


`f(x-y)=√(f(xy)+1)\implies f\left(\frac12\right)=√(f\left(\frac12\right)+1)\implies f\left(\frac12\right)=\frac{1+\sqrt5}2`


Now suppose we choose
`x,y` such that


`\begin{cases}x-y=\frac12\\\\xy=2009\end{cases}`


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


`\frac{1+\sqrt5}2=√(f(2009)+1)\implies f(2009)=\frac{1+\sqrt5}2`

Note that you can always find a solution to the system above that satisfies
`x>y>0` as long as
`x>\frac12`. What this means is that you can always find the value of
`f(x)` as a (constant) function of
`f\left(\frac12\right)`.

Answer 2

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)=√(f(xy)+1)\\\\x>0,y>0\\\\for, x=1, \text{and} y=(1)/(2)\\\\f(1-(1)/(2))=\sqrt{f(1* (1)/(2))+1}\\\\f((1)/(2))^2=f((1)/(2))+1\\\\f((1)/(2))=(1+√(5))/(2)`

Now, suppose

x=2009, y=0

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