When does $\,n\mid q^3 \Rightarrow n\mid q\,$?

Question

I think this is true because of prime factorisations, i.e.

If $3$ a factor of the prime factorisation of $q^3$, then $3$ is a factor of the prime factorisation of $q$.

Therefore If $3$ divides $q^3$, then $3$ divides $q$.

Or for a general case, if $n$ divides $q^3$; then $n$ divides $q$. Is this correct?

Or, more generally, if $n$ is product of distinct primes. – Andrés E. Caicedo May 31 '14 at 03:20 — Andrés E. Caicedo, May 31 '14 at 03:20

score 1 · Accepted Answer · answered May 29 '14 at 08:39

1

The generalization is false: example: $n = 9$, and $q = 3$.

But the first claim is true.

answered May 29 '14 at 08:39

DeepSea

77,651

So, as Asal Beag Dubh said, the generalisation is true as long as $n$ is prime? – Mr Croutini May 29 '14 at 08:40
2

Yes, the primeness of $n$ is vital here. – coolpapa May 29 '14 at 08:40

Conifold · Answer 2 · 2014-05-29T09:12:18.430

1

This is an instance of Euclid's lemma: if a prime number divides a product then it divides one of the factors. Your second claim is only true if you additionally assume $n$ to be prime.

Euclid's lemma can be derived from prime factorization as you suggest, but usually it is used to prove prime factorization, an independent proof uses the Bezout identity http://en.wikipedia.org/wiki/Euclid's_lemma.

edited May 29 '14 at 09:12

answered May 29 '14 at 09:07

Conifold

11,756

Bill Dubuque · Answer 3 · 2021-06-03T00:15:06.403

$\ n\mid q^3\,\Rightarrow\,n\mid q\ $ is true $\forall q\,$ iff $\,n\,$ is squarefree. Indeed, if so, $\,n\,$ is a product of distinct primes $p_i$ so by Euclid's Lemma $\,p_i\mid q^3\Rightarrow\,p_i\mid q,\,$ thus $\,n = {\rm lcm}(p_i)\mid q.\,$

Conversely, if $\,n\,$ is not squarefree then $\,n = ab^2,\ b>1,\,$ hence $\,n\mid (ab)^3\,$ but $\,n = ab^2\nmid ab.\ $

See this answer for a handful of characterizations of "squarefree".

When does $\,n\mid q^3 \Rightarrow n\mid q\,$?

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