Show that the cube of a number is divisible by a prime p then the number is divisible by p.

Here is my attempt so far:

Call the number x. Then from the definition of divisibility, we can say $\frac{x^3}{p}=m$ where $m \in \mathbb{Z}.$

$$ \frac{x \cdot x^2}{p}= m \;\;\; \Rightarrow \;\;\; \frac{x}{p}= \frac{m}{x^2} $$

Can I say that $\frac{m}{x^2} \in \mathbb{Z}$ hence showing the thesis to be true? If so why? If not what would be the next step in such a proof.

Thank you very much!!!


3 Answers 3


You should avoid using symbols such as $\frac{m}{x^2}$ when working in the integers, as a general rule. There are no fractions in the integers.

How do you know, for instance, that $x^2\mid m$? What if $x\mid m$ but $x^2\nmid m$.

Try to generalize as follows:

If $a,b\in \mathbb{Z}$, and $p|ab$, then $p|a$ or $p|b$. This is called Euclid's Lemma and, if you prove it, it should help you prove the assertion.

Edit: Since you've proven it, I'll add some more to give you variety.

Proof 1. By assumption, $p|x^3$. By Euclid's lemma we have $p|x$ or $p|x^2$. If $p|x$, then we are done. Suppose then $p|x^2$. Again, by Euclid's lemma we have $p|x$ or $p|x$. But that is what we are trying to show. $\square$

Proof 2. Assume $p|x^3$. Since every number can be uniquely factored into a product of primes, we can write $x=p_1^{a_1}\cdots p_s^{a_s}$. Consider $x^3=(p_1^{a_1}\cdots p_s^{a_s})^3=p_1^{3a_1}\cdots p_s^{3a_s}$. Since $p|x^3$, it must be in the set $\{p_i| 1\leq i \leq s\}$. But then, $p=p_i$ for some $i$, and is thus in the prime factorization of $x$, i.e. $p|x$.$\square$

The reason why the above proof is not a good place to start learning number theory is because it is essentially claiming $p\nmid x \implies p\nmid x^2 \implies p\nmid x^3$ and does not develop tools such as the $\gcd$ or Euclid's lemma.

  • $\begingroup$ I will review Euclid's Lemma and then return if I am still having trouble. Thank you for the good hint!! $\endgroup$ Commented Feb 19, 2015 at 3:56
  • $\begingroup$ @mathamphetamines Good luck! $\endgroup$
    – Eoin
    Commented Feb 19, 2015 at 3:57
  • $\begingroup$ If fractions are known, then they can prove very useful for proving results about integers. For example, good luck trying to prove this as quickly and naturally without fractions. $\endgroup$ Commented Feb 19, 2015 at 4:23
  • $\begingroup$ @BillDubuque Sometimes dividing in the integers can lead you astray. That is the point I wanted to get across. I would argue that proof is a nicer result due to polynomial factoring then it is of fractions, however! $\endgroup$
    – Eoin
    Commented Feb 19, 2015 at 6:21
  • $\begingroup$ @Eoin I posted a solution. If you happen to have any time, would you mind taking a look at it and seeing if it is correct? $\endgroup$ Commented Feb 19, 2015 at 6:26

Hint: If $p$ is prime, and $p|(xx^2)$ then $p|x$ or $p|x^2$, by Euclid's Lemma.


Euclid's Lemma: If $p \mid ab$ then $p \mid a$ or $p \mid b$ for a prime p.

Let $ab=x^3$, that is let $a=x^2$ and $b=x.$

Next, assume that $p \nmid x^2.$ Since $p \nmid x^2,$ then $p$ and $x^2$ are relatively prime, i.e., $gcd(p,x^2)=1.$

We can write $gcd$ as a linear combination of two integers $v$ and $w.$

$$ pv + x^2w = 1 $$

Now multiplying through by x we get

$$ pxv + x^3w = x $$

Since we know that $p \mid x^3$ and that $p \mid pxv$, then $p \mid (pxv+x^3w)$, hence $p \mid x.$

  • $\begingroup$ This is a good proof. It is in fact stronger then what you needed but it is very good. It is also a proof by contradiction. Notice that you in fact showed $p|x$, $p|x^2$, and $p|x^3$. I'll add a couple of variations to my answer as well so you can get some variety. $\endgroup$
    – Eoin
    Commented Feb 19, 2015 at 6:29
  • $\begingroup$ @ Eoin Thank you for your help and patience! $\endgroup$ Commented Feb 19, 2015 at 6:30

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