# Need hints for solving 2008 A6: $f\left(x+\frac{1}{y}\right)=f\left(y+\frac{1}{x}\right)$

Let $$f(x)$$ be function that satisfies $$f\left(x+\frac{1}{y}\right)=f\left(y+\frac{1}{x}\right) \Big| f:\mathbb{R} \to \mathbb{N}$$. Prove that there exists a positive integer that is not in the range of the function.

I'm aware that I can easily access the solution on aops, but I wish to try and solve it on my own and would like some hints in doing so. Some of the possibly nontrivial stuff I've managed to come up with:

1. For any open interval of 2, there exists infinitely many $$x$$ (cardinality continuum) such that $$f(x) = n$$ For all integers $$n$$.

2. Assuming axiom of choice, if we choose any interval of real numbers there exists an integer $$n$$ such that there are infinitely (cardinality continuum) many $$x$$ such that $$f(x) = n$$.

3. It is possible to segregate (aka disjoint sets) real numbers such that there exists no $$f(x) = f\left(x+\frac{1}{n}\right) \big| n\in \mathbb{N}$$

Would appreciate if noone spoils the solution for me, thanks.

Here's a hint: you can show that in fact the function is constant on $\mathbb R \backslash \{ 0 \}$!
To show that, substitute $\frac 1 y$ for $y$ in the original equation to get $f ( x + y ) = f \left( \frac { x + y } { x y } \right)$ (for nonzero $x$ and $y$, of course). Now note that you can choose different $x$'s and $y$'s with equal sum but different product.