Problem 1719

Let $p$ be a prime number of the form $3k+2$, such that $p > \max(m, 2018)$. (Such $p$ exists because there are infinitely many prime numbers of the form $3k+2$.) We shall prove that $n = p - 2019 + i$, $1 \le i \le m$ satisfies the problem's conditions.

It suffices to prove that
$$v_p((1^3 + 2018^3)(2^3 + 2018^3)\dots(n^3 + 2018^3)) = 1,$$
as it would result in $(1^3 + 2018^3)(2^3 + 2018^3)\dots(n^3 + 2018^3)$ not being a perfect power.
$$v_p((1^3 + 2018^3)(2^3 + 2018^3)\dots(n^3 + 2018^3)) = v_p(1^3 + 2018^3) + v_p(2^3 + 2018^3) + \dots + v_p(n^3 + 2018^3).$$
However, if $i$ exists such that $i \ne p - 2018$ and $v_p(i^3 + 2018^3) > 0$,
$$\begin{align*} p|i^3 + 2018^3 &\implies i^3 \stackrel{p}{\equiv} (-2018)^3 \\ &\implies (i^3)^{\frac{p-2}{3}} \stackrel{p}{\equiv} ((-2018)^3)^{\frac{p-2}{3}} \\ &\implies (-2018)i^{p-1} \stackrel{p}{\equiv} (-2018)^{p-1}i \\ &\implies (-2018) \stackrel{p}{\equiv} i \implies p|i + 2018, \ i \ne p - 2018 \\ &\implies 2p - 2019 > p - 2019 + m \ge i \ge 2p - 2018 \end{align*}$$
which is a contradiction. Therefore,
$$v_p((1^3 + 2018^3)(2^3 + 2018^3)\dots(n^3 + 2018^3)) = v_p((p - 2018)^3 + 2018^3) = v_p(p) + v_p(3) = 1$$
Due to L.T.E., which proves the claim. ■