1. ## Function

Find all functions $\displaystyle f: \mathbb{N^{*}} \rightarrow \mathbb{N^{*}}$ which satisfy the following conditions
$\displaystyle i)$ $\displaystyle f$ is strictly increasing;
$\displaystyle ii)$ $\displaystyle f(f(n))= 4n+9$ for all $\displaystyle n \in \mathbb{N^{*}};$
$\displaystyle iii)$ $\displaystyle f(f(n)-n)=2n+9$ for all $\displaystyle n \in \mathbb{N^{*}}$.

2. I found one, but only by trial and errors.

$\displaystyle f(n)=2n+3$

$\displaystyle f(2n+3)=2(2n+3)+3=4n+9$

$\displaystyle f(2n+3-n)=2(n+3)+3=2n+9$

3. Originally Posted by unlimited
Find all functions $\displaystyle f: \mathbb{N^{*}} \rightarrow \mathbb{N^{*}}$ which satisfy the following conditions
$\displaystyle i)$ $\displaystyle f$ is strictly increasing;
$\displaystyle ii)$ $\displaystyle f(f(n))= 4n+9$ for all $\displaystyle n \in \mathbb{N^{*}};$
$\displaystyle iii)$ $\displaystyle f(f(n)-n)=2n+9$ for all $\displaystyle n \in \mathbb{N^{*}}$.
Originally Posted by Sorombo
I found one, but only by trial and errors.

$\displaystyle f(n)=2n+3$

$\displaystyle f(2n+3)=2(2n+3)+3=4n+9$

$\displaystyle f(2n+3-n)=2(n+3)+3=2n+9$
If you assume the function is linear then it's easy. Let f(n) = ax + b and you'll get two equations in a and b. They're pretty easy to solve. The only possible result is 2n + 3. Now, why can't we do this with a quadratic?

-Dan