Problem 1734

a) We suppose that $\frac{s}{r} \notin \mathbb{N}$. Then, there exists $k \in \mathbb{N}$ such that $k < \frac{s}{r} < k+1 \iff kr < s < (k+1)r$. Choosing $b = a \in \mathbb{N}^*$, arbitrary, we obtain $[ar] \mid [as]$ and thus $[ar] \mid [as] - k[ar]$. (1)
From $s > kr$, we obtain that there exists $u > 0$ such that $us > ukr + 2$ and thus, for every $a > u$ we get $as > akr + 2 \implies [as] \ge [akr] + 2 > akr + 1 > k[ar]$, so $[as] > k[ar]$.
From (1) we obtain $[as] - k[ar] \ge [ar] \iff [as] \ge (k+1)[ar]$, so,
$$as > (k+1)(ar - 1) \iff k+1 > a((k+1)r - s),$$
for every $a > u$.
Thus, $a < \frac{k+1}{(k+1)r-s}$, for every $a > u$, which is a contradiction, so the assumption is false.

b) Let us show that $s$ is a positive integer.
We will show that for every $a \in \mathbb{N}$ such that $ar \ge 2$, we get $as \in \mathbb{N}$.
If $as \notin \mathbb{N}$, then there exists $n \in \mathbb{N}^*$ such that $\frac{1}{n+1} \le \{as\} < \frac{1}{n}$, so $1 \le (n+1)\{as\} < \frac{n+1}{n} \le 2$, thus $[(n+1)\{as\}] = 1$.
We obtain
$$[(n+1)as] = [(n+1)[as] + (n+1)\{as\}] = (n+1)[as] + [(n+1)\{as\}] = (n+1)[as] + 1.$$
Since $[ar] \mid [as]$ and $[ar] \mid [(n+1)as]$, we obtain that $[ar] \mid 1 \implies [ar] = 1$, which is a contradiction.
Thus $as \in \mathbb{N}$, for every $a \in \mathbb{N}$ with $ar \ge 2$, from which we get $(a+1)s \in \mathbb{N}$, so $(a+1)s - as = s \in \mathbb{N}$.

Let us prove that $r$ is a positive integer.
Let $p$ be an arbitrary prime number with $p[r] > s$ and $m = [p\{r\}]$. Since $p\{r\} < p$, we get $m < p$.
If $m \ne 0$, then $(m, p) = 1$. Since
$$[pr] \mid ps \implies [p([r] + \{r\})] \mid ps \implies p[r] + m \mid ps.$$
Since $(p[r] + m, p) = 1 \implies p[r] + m \mid s$, we obtain a contradiction as $p[r] > s$.
Thus, $m = 0 \implies p\{r\} < 1 \implies \{r\} < \frac{1}{p}$, for every prime number $p$, with $p > \frac{s}{[r]} \implies \{r\} = 0$ and thus, $r \in \mathbb{N}$.