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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.

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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.

Please inform me if you think I should elaborate more.

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  • $\begingroup$ Yup I got it already. Thanks $\endgroup$ Apr 26 '17 at 0:11

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