Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $m$ cubefree and $k$ integer $k^{6}|27m^{2}\Rightarrow k=1$ or $3$.

The "1" makes sense from $k^{6}|27m^{2}\Rightarrow ak^{3}=3m$ but not the 3.

So by assuming $9|m$

Any hints?

share|cite|improve this question
up vote 2 down vote accepted

Hint: Note that both $k^6$ and $27$ are perfect cubes; if $p$ is a prime divisor of $k$ that is not $3$, try concluding that $p^3 | m$.

share|cite|improve this answer
@TKM Right: $k$ is a multiple of $3$ (or is $1$). If $k$ is divisible by $9$, then $k^6$ is divisible by an awfully large power of $3$, so $m$.... – user61527 Feb 27 '14 at 5:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.