# IMO 1993 b2 proof

"Let $$\mathbb{N}=\{ 1,2,3,\ldots \}$$. Determine if there exists a strictly increasing function $$f:\mathbb{N}\to\mathbb{N}$$ such that

1) $$f(1)=2$$

2) $$f(f(n))=f(n)+n$$ for all $$n$$."

Of the solutions I've seen online, one such $$f(n)$$ that should work is $$f(n)=\left[\phi n + \frac{1}{2}\right]$$ where $$\phi\approx 1.618$$ is the golden ratio and the square brackets denote the integral part. It is easy to see that $$f(1)=2$$ and that $$f(n)$$ is strictly increasing - however the proof that this satisfies condition 2 seems to be incorrect or unclear in the sources I have found. A clear proof that this $$f(n)$$ satisfies 2) would be appreciated.

• I think if you use condition 2 repeatedly, you will find that for the n'th fibonacci number $F_n$, our $f$ must give $f(F_n)=F_{n+1}$. – Merk Zockerborg Apr 28 at 15:41
• Can you show the proofs that you claim are incorrect or unclear? Perhaps they can be clarified. – kccu Apr 28 at 15:49

Claim: $$f(n)=\left[\phi n + \frac{1}{2}\right]$$ satisfies $$f(f(n))=f(n)+n$$.
Now $$f(f(n))-f(n)-n=\left[\phi\left[\phi n+\frac{1}{2}\right] +\frac{1}{2}\right]-\left[\phi n +\frac{1}{2}\right]-n$$, so it suffices to show $$X(n)=\phi\left[\phi n + \frac{1}{2}\right]+\frac{1}{2}-\left[\phi n +\frac{1}{2}\right]-n=(\phi-1)\left[\phi n +\frac{1}{2}\right]+\frac{1}{2}-n$$ satisfies $$0\leq X(n) < 1$$.
For the lower bound, $$\left[\phi n +\frac{1}{2}\right]\geq \phi n -\frac{1}{2}$$, giving $$X(n)\geq \phi^2n-\frac{\phi}{2}-\phi n+1-n.$$ But $$\phi=\frac{1+\sqrt{5}}{2}$$, so $$n(\phi^2-\phi-1)=0$$. Hence $$X(n)\geq 1-\frac{\phi}{2}>0.$$ For the upper bound, $$\left[\phi n +\frac{1}{2}\right] \leq \phi n + \frac{1}{2}$$, giving $$X(n)\leq \phi^2 n+\frac{\phi}{2}-\phi n-n=\frac{\phi}{2}<1$$ as required.