# Prime divisors of $n^{3} - 27$ and squarefreeness

Let $n \equiv 4 \bmod{6}$. Does there exist infinitely many $n$ such that $3 \nmid \operatorname{ord}_{p}(n^{3} - 27)$ for each prime $p \mid n^{3} - 27$?

In particular, the following (possibly harder) question is sufficient to answer the above in the affirmative: Are there infinitely many $n$ such that $n^{3} - 27$ is squarefree? Testing this out using a script I wrote, for $4 \leq n \leq 10^{6}$ there were 141927 such $n$ which had squarefree $n^{3} - 27$ and for $4 \leq n \leq 10^{7}$ there were 1419384 such $n$. I suppose something like $14\%$ of $n$ are such that $n^{3} - 27$ is squarefree?

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Why $4 \pmod 6?$ And what is wrong with cubefree? – Will Jagy Jan 30 '13 at 6:55
For the $4\bmod{6}$ part, no particular reason, I just included it for concreteness on my part. And being $n^{3} - 27$ being cubefree is also fine. The second question in the affirmative would answer the first and I was trying to work on that. However, showing that $n^{3} - 27$ is cubefree would also be fine. – user60194 Jan 30 '13 at 7:03

I don't believe it is known whether $n^3-27$ is squarefree for a positive proportion of values of $n$.