Show that if $p$ is a prime number and $a$ is an integer, and if $p \mid a^2$ , then $p \mid a$.

Question

I am suppose to make use of the following lemma

If $a$, $b$ and $c$ are positive integers such that $(a, \, b) = 1$ and $a \mid bc$, then $a \mid c$

to prove that if $p$ is a prime number and $a$ is an integer, and if $p \mid a^2$ , then $p \mid a$.

I'm not sure how the lemma helps. I could write $a^2 = aa$ with $b=c=a$ in the lemma but we don't have $(a,b)=(p,a)=1$ so I don't see how it would help. I must use this lemma to prove it. I am aware that there are other methods.

@AaronMaroja An integer $> 1$ with no divisors except one and itself. — MathMajor, May 14 '15 at 21:30
Hint: If $p$ divides $a$ we are finished. Otherwise, $p$ and $a$ are relatively prime and we can apply the lemma. — André Nicolas, May 14 '15 at 21:32
You are welcome. If $d\gt 1$ divides $p$ and $a$, then in particular $d$ divides $a$, so $d=p$. — André Nicolas, May 14 '15 at 21:37
See also: A short or elegant proof for if $p \mid n^2$ then $p \mid n$ when $p$ is prime?, A quick question about: If $p$ is prime show that $p\mid a^2 \implies p\mid a$, $p$ prime, $p\mid a^k \Rightarrow p^k\mid a^k$, If $p$ is prime and $p \mid a^k$, then $p \mid a$, and hence $p^k \mid a^k$. — Martin Sleziak, Dec 28 '19 at 15:04

score 1 · Accepted Answer · edited May 14 '15 at 21:38

1

By contradiction: Given $p \mid a^2$, if $p\nmid a$ and since $p$ is prime then $(p,a)=1$ so using the mentioned result we get $p \mid a$ which is a contradiction.

edited May 14 '15 at 21:38

Joffan

39,627

answered May 14 '15 at 21:34

score 1 · Answer 2 · answered May 14 '15 at 21:34

1

Hint: Suppose $p$ does not divide $a$ then $(p,a) =1$ and now you may apply your lemma.

answered May 14 '15 at 21:34

Aaron Maroja

17,571

score 1 · Answer 3 · answered May 14 '15 at 21:36

1

Assume that $p| a^2$ but $p\not| a$. Note that $(p,a)$ divides $p$, so $(p,a)=1$ or $p$. Also, $(p,a)$ divides $a$, and $p\not| a$, so $(p,a)=1$. Then apply your lemma.

answered May 14 '15 at 21:36

Show that if $p$ is a prime number and $a$ is an integer, and if $p \mid a^2$ , then $p \mid a$.

3 Answers3