Mathematical Olympiad Rioplatense

All functions of the form $f(x) = cx$ with $c \in \mathbb{R}$ are solutions; they are the only ones. More generally let $b > a > 0$, and let the open interval $\Delta = (a, b)$ contain an integer (in our case $a = 10^6 - 10^{-6}$, $b = 10^6 + 10^{-6}$). Consider any function $f: \mathbb{N} \to \mathbb{R}$ such that $f(x + y) = f(x) + f(y)$ holds whenever $\frac{x}{y} \in \Delta$. We prove that $f(n) = cn$ for all $n \in \mathbb{N}$ with a real constant $c$.

To begin with let us show that $f(n+1) - f(n) = f(n) - f(n-1)$ for all sufficiently large $n$. The reason is that for each sufficiently large $n \in \mathbb{N}$ there is a $z \in \mathbb{N}$ such that
$$f(n + 1) - f(n) = f(z + 1) - f(z) = f(n) - f(n - 1).$$
To ensure the first equality it is enough to take a $z$ so that $\frac{z}{n+1} \in \Delta$ and $\frac{z+1}{n} \in \Delta$. Then by hypothesis $f(x + y) = f(x) + f(y)$ will hold with $x = z$, $y = n + 1$ and also with $x = z + 1$, $y = n$. Hence $f(z) + f(n+1) = f(n+z+1) = f(z+1) + f(n)$, as desired. Likewise the second equality will hold provided that $\frac{z}{n} \in \Delta$ and $\frac{z+1}{n-1} \in \Delta$. Since $\frac{z}{n+1} < \frac{z}{n} < \frac{z+1}{n} < \frac{z+1}{n-1}$, it suffices to find an integer $z$ so that $a < \frac{z}{n+1}$ and $\frac{z+1}{n-1} < b$, i.e. $a(n + 1) < z < b(n - 1) - 1$. Such an integer does exist for $n$ large enough. Indeed $b(n - 1) - 1$ and $a(n + 1)$ differ by $(b - a)n - (a + b + 1)$ which is greater than 1 for $n > \frac{a+b+2}{b-a}$.

In summary there exists a $k \in \mathbb{N}$ such that $f(n+1) - f(n)$ has the same value for all $n \ge k$. Then by standard induction
$$(*) \quad f(n) = (n-k)[f(k+1)-f(k)] + f(k) \quad \text{for all } n \ge k.$$
There are $x, y \in \mathbb{N}$ such that $x, y \ge k$ and $\frac{x}{y} \in \Delta$. For instance choose a rational $\frac{r}{s} \in \Delta$ ($r, s \in \mathbb{N}$) and set $x = rk, y = sk$. Take one such pair $x, y$ and compute $f(x), f(y), f(x+y)$ by the formula $(*)$; this can be done because $x, y, x+y \ge k$. Replace the obtained values in $f(x+y) = f(x)+f(y)$, which holds because $\frac{x}{y} \in \Delta$. Simplification leads to $kf(k+1) = (k+1)f(k)$. Hence $\frac{f(k)}{k} = \frac{f(k+1)}{k+1} = c \in \mathbb{R}$; equivalently $f(k) = ck$ and $f(k+1) = c(k+1)$. Then $(*)$ takes the form $f(n) = cn$ for all $n \ge k$. It remains to show that $f(n) = cn$ for all $n \in \mathbb{N}$.

Only finitely many $n \in \mathbb{N}$ may possibly disobey $f(n) = cn$. Suppose that such values exist, and let $q$ be the greatest one of them. Choose an integer $w \in \Delta$ and set $x = wq, y = q$. Then $x/w \in \Delta$, hence $f((w+1)q) = f(wq) + f(q)$. If $w > 1$ then $(w+1)q > wq > q$, so by the choice of $q$ we have $f((w+1)q) = c(w+1)q$, $f(wq) = cwq$. However then $f(q) = c(w+1)q - cwq = cq$, contrary to the assumption that $q$ violates $f(n) = cn$. And if $w = 1$ then $(w+1)q = 2q > q$, so $f(2q) = 2cq$. On the other hand $f((w+1)q) = f(wq) + f(q)$ takes the form $f(2q) = 2f(q)$ which leads to the impossible $f(q) = cq$ again. This completes the proof.

Remark. The main assumption is $f(x + y) = f(x) + f(y)$ whenever $\frac{x}{y} \in \Delta$, where $\Delta$ is an arbitrary open interval with positive endpoints. It ensures $f(n) = cn$ for all sufficiently large values of $n$. However the additional assumption that $\Delta$ contains an integer is essential to infer that $f(n) = cn$ for all $n \in \mathbb{N}$. Consider for instance the function $f: \mathbb{N} \to \mathbb{R}$ defined by $f(n) = n$ for $n \ge 5$ and $f(n) = 2010$ for $n \in \{1, 2, 3, 4\}$ (in fact $f(1), f(2), f(3), f(4)$ can be arbitrary). The interval $\Delta = (3/2, 5/3)$ does not contain fractions with denominators $1, 2, 3, 4$. So $\frac{x}{y} \in \Delta$ implies $x \ge y \ge 5$; the equation $f(x + y) = f(x) + f(y)$ is satisfied for such values. Thus $f$ is a function that satisfies the main assumption and $f(n) = n$ for $n \ge 5$ but not for all $n$.