If $m$ is a positive integer, explain why each prime in the prime factorization of $m^2$ must occur an even number of times.

I did a small proof, I was wondering what's a nice way to explain it aside from my explanation below:

Proof: let $m=p_1^{e_1} p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$

Then $m^2=(p_1^{e_1})^2 (p_2^{e_2})^2(p_3^{e_3})^2\cdots (p_k^{e_k})^2 = p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}\cdots p_k^{2e_k}$

These $p_i^{e_i}$ occur an even number of times any suggestions to make a more solid answer?

  • $\begingroup$ I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good. $\endgroup$ Commented Jul 27, 2017 at 2:01
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    $\begingroup$ One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent). $\endgroup$ Commented Jul 27, 2017 at 2:02
  • $\begingroup$ Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even. $\endgroup$ Commented Jul 27, 2017 at 2:02
  • $\begingroup$ @GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either. $\endgroup$
    – OLE
    Commented Jul 27, 2017 at 2:04
  • $\begingroup$ @BillDubuque is their a nicer more stronger way of proving this? $\endgroup$
    – OLE
    Commented Jul 27, 2017 at 2:06

1 Answer 1


It's in fact an if and only if :

Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $k$ occurs an even number of times, if and only if $k$ is a perfect square.

Suppose each prime in the prime factorization of $k$ occurs an even number of times, say $k = \prod p_i^{r_i}$, where each $r_i$ is even. Then, you can see that if $b = \prod p_i^{\frac {r_i} 2}$, then $b$ is a well defined integer, and $b^2 = k$.

Conversely, note that if $k = m^2$ is a perfect square, then the prime factorization of $m = \prod p_i^{r_i}$ suggests the prime factorization for $k = \prod p_i^{2r_i}$. Since prime factorization is unique, it follows that $k$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.

Alternately, we can also go by this way : Suppose $p^r$ divides $m^2$, where $r$ is maximal. Suppose $r$ is even, then we are done. Otherwise, note that $r = 2k+1$, and we can write $p \ \mid\ \frac {m^2}{p^{2k}}$, so that $p$ divides a perfect square.

Hence, from here, using Euclid's lemma, that $p$ divides $ab$ implies $p$ divides $a$ or $p$ divides $b$, we get that taking $a=b=\frac m{p^k}$, $p^{k+1} \mid m$, or that $p^{2k+2} \mid m^2$, a contradiction by uniqueness of prime factorization.

On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.

EXTENSION : If $m$ is a perfect $k$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $k$.


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