Determine all one to one functions $f:\mathbb{N}^* \rightarrow \mathbb{N}^*$ having the following property:

Solution 1:

We have $f(1)=1$ and $f(n) \ge 2, \forall n \ge 2$.

Let $ n ≥ 2$ be un integer. Using Egyptian fractions theorem, we can write: $$ 1 − \frac {1}n = \sum_{s \in S} \frac{1}{s} $$ where $S$ is a set of integers greater than $n(n + 1)$. Therefore: $$ 1=\frac {1}n + \sum_{s \in S} \frac{1}{s} =\frac 1{n+1} + \frac 1{n(n+1)} + \sum_{s \in S} \frac{1}{s} $$ From f property, we have: $$ \frac {1}{f(n)} + \sum_{s \in S} \frac{1}{f(s)} \in \mathbb{N} $$ and $$ \frac 1{f(n+1)} + \frac 1{f(n(n+1))} + \sum_{s \in S} \frac{1}{f(s)} \in \mathbb{N} $$ therefore $$ \frac 1{f(n+1)} + \frac 1{f(n(n+1))} - \frac {1}{f(n)} \in \mathbb{Z} $$ But: $$ \frac {-1}2 \le - \frac {1}{f(n)} \lt \frac 1{f(n+1)} + \frac 1{f(n(n+1))} - \frac {1}{f(n)} \lt \frac 1{f(n+1)} + \frac 1{f(n(n+1))} \le \frac1{2} + \frac1{2} $$ so $$ \frac 1{f(n+1)} + \frac 1{f(n(n+1))} = \frac {1}{f(n)} \tag 1 $$ It follows that f is increasing and $f(n) \ge n$. To conclude, it's easy to show, using induction, that $f(n)=n, \forall n$.

Disclaimer

This prove has been sent to me, in a hand written form, by a friend who allowed me to post it here.

Update

I was requested to continue the prove (the induction part). First, because f is increasing and f injective, we have: $f(n) \ge n, \forall n$.

Now suppose $f(k) = k$ and $f(k + 1) > k + 1$ for some $k$. From (1) we have: $$ \frac 1{f(k+1)} + \frac 1{f(k(k+1))} = \frac {1}{k} \tag 2 $$ and, because $f(n) \ge n, \forall n$: $$ \frac 1{k+1} + \frac 1{k(k+1)} \gt \frac {1}{k} \tag 3 $$ From (3): $$ \frac {1}{k} \gt \frac {1}{k} \tag 4 $$ Therefore $f(k+1) = k+1$ if $f(k) = k$.