First we choose distinct positive rational numbers $r_{1}, \ldots, r_{k+3}$ such that
$$r_{i} r_{i+1} r_{i+2} r_{i+3}=i \quad \text { for } 1 \leqslant i \leqslant k$$
Let $r_{1}=x, r_{2}=y, r_{3}=z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_{4}=\frac{1}{r_{1} r_{2} r_{3}}$ and $r_{i+4}=\frac{i+1}{i} r_{i}$. It follows that if $r_{i}$ are represented as irreducible fractions, the numerators are divisible by $x$ for $i \equiv 1(\bmod 4)$, by $y$ for $i \equiv 2(\bmod 4)$, by $z$ for $i \equiv 3(\bmod 4)$ and by none for $i \equiv 0(\bmod 4)$. Notice that $r_{i}<r_{i+4}$; thus the sequences $r_{1}<r_{5}<r_{9}<\ldots$, $r_{2}<r_{6}<r_{10}<\ldots, r_{3}<r_{7}<r_{11}<\ldots, r_{4}<r_{8}<r_{12}<\ldots$ are increasing and have no common terms, that is, all $r_{i}$ are distinct. If each $r_{i}$ is represented by an irreducible fraction $\frac{u_{i}}{v_{i}}$, choose a prime $p$ which divides neither $v_{i}, 1 \leqslant i \leqslant k+1$, nor $v_{i} v_{j}\left(r_{i}-r_{j}\right)=v_{j} u_{i}-v_{i} u_{j}$ for $i<j$, and define $a_{i}$ by the congruence $a_{i} v_{i} \equiv u_{i}(\bmod p)$. Since $r_{i} r_{i+1} r_{i+2} r_{i+3}=i$, we have
$$\begin{aligned} i v_{i} v_{i+1} v_{i+2} v_{i+3}=r_{i} v_{i} r_{i+1} v_{i+1} r_{i+2} & v_{i+2} r_{i+3} v_{i+3} \\ & =u_{i} u_{i+1} u_{i+2} u_{i+3} \equiv a_{i} v_{i} a_{i+1} v_{i+1} a_{i+2} v_{i+2} a_{i+3} v_{i+3} \quad(\bmod p) \end{aligned}$$
and therefore $a_{i} a_{i+1} a_{i+2} a_{i+3} \equiv i(\bmod p)$ for $1 \leqslant i \leqslant k$. If $a_{i} \equiv a_{j}(\bmod p)$, then $u_{i} v_{j} \equiv a_{i} v_{i} v_{j} \equiv u_{j} v_{i}(\bmod p)$, a contradiction. Comment. One can explicitly express residues $b_{i} \equiv a_{1} a_{2} \cdot \ldots \cdot a_{i}(\bmod p)$ in terms of $b_{1}, b_{2}, b_{3}$ and $b_{0}=1$ :
$$b_{i+3}=i(i-4)(i-8) \cdot \ldots \cdot(i-4 k+4) b_{r}$$
where $i+3=4 k+r, 0 \leqslant r<4$. Then the numbers $a_{i}$ are found from the congruences $b_{i-1} a_{i} \equiv b_{i}$ $(\bmod p)$, and choosing $p$ so that $a_{i}$ are not congruent modulo $p$ is done in a way very similar to the above solution.

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