Problem 1190

## Solution to Exercise 17

We will show that the pairs in the statement are those such that $] \alpha, \beta[\cup] \frac{1}{\beta}, \frac{1}{\alpha}[$ contains an integer.

- First, let's show that the condition is necessary. Indeed, if the condition is not satisfied, the function $f(n)=0$ for $n>1$ and $f(1)=1$ satisfies the functional equation but is not linear.
- Now, let's show that the condition is sufficient. Suppose that $] \alpha, \beta[\cup] \frac{1}{\beta}, \frac{1}{\alpha}[$ contains an integer. We first note that by replacing $(\alpha, \beta)$ with $\left(\frac{1}{\beta}, \frac{1}{\alpha}\right)$ and swapping $x$ and $y$ in the equation, we can assume without loss of generality that $] \alpha, \beta[$ contains an integer $\mathrm{N} \in \mathbb{N}^{*}$. Notably, $\alpha < 1 < \beta$.

We will show that there exist $a, b \in \mathbb{R}$ such that $\forall \mathrm{n} \geqslant M, \mathrm{f}(\mathrm{n})=\mathrm{an}+\mathrm{b}$.

Let's show that in fact $b=0$. Indeed, consider $n \geqslant M$. We have $\alpha < 1 < \beta$, so $f(n N) = a n N + b$. Since $f(n N) = N f(n)$, we get $N f(n) = a n N + b$. Therefore, $f(n) = a n + \frac{b}{N}$. Since $f(n)$ is linear for $n \geqslant M$, we have $f(n) = a n + b$.

If $N > 1$, we also have $f(n N) = a n N$, hence $f(n) = a n (N + 1) - a n N = a n$. If $N = 1$, we have $2 f(n) = 2 a n$ and $f(n) = a n$. In both cases, we obtain a contradiction.

Thus, $\mathrm{f}$ is linear and $(\alpha, \beta)$ satisfies the property in the statement.

Grader's Comment: The exercise was difficult, rarely approached, and even less successfully completed. The condition required in the statement was found in most of the attempts, which managed to conjecture that it was sufficient. The difficult part of the exercise was to find a relation of the form $f(n+k)-f(n)$ constant for $n$ large enough, almost all students who reached this point managed to conclude.