Let $f(n)=\left(1^3+2018^3\right)\left(2^3+2018^3\right)\cdots \left(n^3+2018^3\right)$. Let $a=p-2018$, where $p$ is a prime number large enough, which satisfies $p\equiv 11 \pmod{12}$, then we have that $v_p(f(a))=v_p(f(a+1))=\cdots = v_p(f(2a+2017))=1$, thus none of the numbers $f(a), f(a+1), \cdots f(2a+2017)$ can be a perfect power.

rmtf1111 wrote: Let $f(n)=\left(1^3+2018^3\right)\left(2^3+2018^3\right)\cdots \left(n^3+2018^3\right)$. Let $a=p-2018$, where $p$ is a prime number large enough, which satisfies $p\equiv 11 \pmod{12}$, then we have that $v_p(f(a))=v_p(f(a+1))=\cdots = v_p(f(2a+2017))=1$, thus none of the numbers $f(a), f(a+1), \cdots f(2a+2017)$ can be a perfect power. Maybe U haven’t shown that the statement is true for any interger m>0

bz0103 wrote: rmtf1111 wrote: Let $f(n)=\left(1^3+2018^3\right)\left(2^3+2018^3\right)\cdots \left(n^3+2018^3\right)$. Let $a=p-2018$, where $p$ is a prime number large enough, which satisfies $p\equiv 11 \pmod{12}$, then we have that $v_p(f(a))=v_p(f(a+1))=\cdots = v_p(f(2a+2017))=1$, thus none of the numbers $f(a), f(a+1), \cdots f(2a+2017)$ can be a perfect power. Maybe U haven’t shown that the statement is true for any interger m>0 If the statement true for some M then it is true for all m<M so solution works

I have a solution without Dirichlet.I will use the following well-known lemma: Lemma:If $p$ is a prime with $p \equiv 2 \pmod 3$ then $p|a^2+b^2-ab$ implies $p|a,p|b$. Now write $n^3+2018^3=(n+2018)(n^2-(-2018)(n) +(-2018)^2)$ this motivates to use $p \equiv 2 \pmod 3$ so that $p>2018$ then $v_p(n^3+2018^3)=v_p(n+2018)$ so we can take the numbers to be $p-2018,p-2017, \dots , 2p-1$.It is well-known that there are infinity many primes with $ p \equiv 2 \pmod 3$ so we are done.

Note that $p \mid a^3+b^3 \implies O_p(\frac ab) \mid 2 \text{gcd}(p-1,3)$ but $\nmid \text{gcd}(p-1,3)$, hence for $p \equiv 2 \pmod{3}$ $p \mid a^3+b^3$ iff $p \mid a+b$ (for $\text{gcd}(p,ab)=1$, hence choose $p \nmid 2018$ for this problem) Thus take $n=p-2018$ and run it to $2p-2019$, forcing $v_p$ of expression to be 1, hence satisfying our condition and finishing our problem.

ubermensch wrote: Note that $p \mid a^3+b^3 \implies O_p(\frac ab) \mid 2 \text{gcd}(p-1,3)$ but $\nmid \text{gcd}(p-1,3)$, hence for $p \equiv 2 \pmod{3}$ $p \mid a^3+b^3$ iff $p \mid a+b$ (for $\text{gcd}(p,ab)=1$, hence choose $p \nmid 2018$ for this problem) Thus take $n=p-2018$ and run it to $2p-2019$, forcing $v_p$ of expression to be 1, hence satisfying our condition and finishing our problem. another way to find a prime that satisfies this statement $p \mid a^3+b^3$ iff $p \mid a+b$ is as follow if $p|a^3+b^3 \implies p|(a+b)(a^2-ab+b^2) $ suppose $p|a^2 -ab +b^2 \implies p|4(a^2 -ab +b^2) \implies p|(2a-b)^2 +3 b^2 \implies (\frac{3}{p})=1 $ so choose $p$ such that $(\frac{3}{p})=-1$

Too easy for a TST #5 Pick a huge prime $p \equiv 2\pmod 3 $ such that $p > \operatorname{max}(m,2019)$ We claim that $n=p-2019+i$ with $1 \leq i \leq m$ works . We show that : $$\nu_p \left (\prod_{i=1}^{n}(i^3+2018^3) \right)= 1 $$This obviously finishes. Note that by a trivial application of orders we have that since $p \equiv 2\pmod 3 \implies p \mid i^3+2018^3 \implies p \mid i+2018$ Next suppose FTSOC that there exists $j \neq p-2018$ such that $p \mid j+2018$ . Then we have the dumb estimates $:=$ $$ i+2018 \geq 2p \implies 2p-2019 >p+m-2019 \geq i \geq 2p-2018$$Absurd . $\blacksquare$