2. Determine the functions $f, g: \mathbb{N} \rightarrow \mathbb{N}$ knowing that:
a) $f(0)=0$ and $f(n) \neq 0$ for any non-zero natural number $n$;
b) $n^{2} f(n)+m^{2} f(m)=\left(f^{2}(n)+f^{2}(m)-f(n) \cdot f(m)\right) \cdot g(m+n)$, for any $m, n \in \mathbb{N}$.

## Cristian Amorăriţei, Suceava
Solution. In the relation from point b) we take $m=0$ and obtain:

$$
n^{2} f(n)=f^{2}(n) g(n) \Leftrightarrow f(n)\left(n^{2}-f(n) g(n)\right)=0, \forall n \in \mathbb{N} \Rightarrow g(n)=\frac{n^{2}}{f(n)}, \forall n \in \mathbb{N}^{*}(1)
$$

$g(1) f(1)=1, g(1), f(1) \in \mathbb{N} \Rightarrow g(1)=f(1)=1$.

In the relation from point b) we take $m=1$ and obtain: $n^{2} f(n)+1=\left(f^{2}(n)+1-f(n)\right) \cdot g(n+1), \forall n \in \mathbb{N}$. Using (1) we deduce: $n^{2} f(n)+1=\left(f^{2}(n)+1-f(n)\right) \cdot \frac{(n+1)^{2}}{f(n+1)}, \forall n \in \mathbb{N} \Leftrightarrow$ $f(n+1)=\left(f^{2}(n)+1-f(n)\right) \cdot \frac{(n+1)^{2}}{n^{2} f(n)+1}, \forall n \in \mathbb{N}$. It is demonstrated using mathematical induction that $f(n)=n, \forall n \in \mathbb{N}$. In the end, we get $g(n)=\left\{\begin{array}{l}a, n=0 \\ n, n \geq 1\end{array}, a \in \mathbb{N}\right.$.

Grading.

| ![](https://cdn.mathpix.com/cropped/2024_06_07_cb3b0e17698baf178e07g-1.jpg?height=126&width=1289&top_left_y=2225&top_left_x=315) | 2p |
| :---: | :---: |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_cb3b0e17698baf178e07g-1.jpg?height=56&width=1289&top_left_y=2347&top_left_x=315) | $1 \mathrm{p}$ |
| Prove $f(n+1)=\left(f^{2}(n)+1-f(n)\right) \cdot \frac{(n+1)^{2}}{n^{2} f(n)+1}, \forall n \in \mathbb{N} \ldots \ldots \ldots \ldots \ldots \ldots$. | 2p |
| Determine the function $f \ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . .$. | $1 \mathrm{p}$ |

![](https://cdn.mathpix.com/cropped/2024_06_07_cb3b0e17698baf178e07g-2.jpg?height=60&width=1583&top_left_y=111&top_left_x=301)