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How to prove $p | a^2 \implies p | a$? Given that $a,p \in \mathbb{Z}$.

Bill Dubuque
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jessegerritsen
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    It will be hard. After all, $4\mid6^2$, but $4\nmid6$. – José Carlos Santos Jul 09 '22 at 16:11
  • true if $p$ is squarefree, otherwise false, take $p=4, a=2$ – Will Jagy Jul 09 '22 at 16:12
  • As José notes, this statement is not true. However, assuming $p$ is prime, the statement is true by Euclid's lemma. – Joe Jul 09 '22 at 16:12
  • @Joe how about if $p$ is squarefree as mentioned above? Does this automatically imply that $p$ is a prime? – jessegerritsen Jul 09 '22 at 16:14
  • @jessegerritsen: No, every prime number is squarefree, but not every squarefree number is prime. Assuming you are familiar with the definitions of "squarefree" and "prime", it should be easy to come up with the smallest example of a squarefree number that is not prime. – Joe Jul 09 '22 at 16:16
  • @Joe could it also be a relative prime to $a$ or does it have to be an actual prime number? – jessegerritsen Jul 09 '22 at 16:16
  • @jessegerritsen: If $n|a^2$ and $n$ is coprime to $a$, then $n|a$. However, this theorem is not very interesting because the hypotheses "$n|a^2$" and "$n$ is coprime to $a$" can only be satisfied when $n=1$. – Joe Jul 09 '22 at 20:32
  • @jessegerritsen: Let $S$ equal the set of numbers $p$ such that, for all $a$, if $p|a^2$ then $p|a$. It can be proven that $S$ equals the set of squarefree integers. In other words, the set of numbers $p$ for which the implication always holds is precisely the set of squarefree integers. – Joe Jul 09 '22 at 21:03

1 Answers1

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Since $p$ is prime, $p$ and $a$ have no common divisors other than $1$ if $p \nmid a$. In this case you can use Bezout's lemma to find integers $m$ and $n$ with $am + pn = 1$. This leads to $a^2m + apn = a$, and since $p|a^2$ there exists $b$ with $bpm + apn = a$. That is, $p (bm + an) = a$, implying $p|a$ after all.

The same argument will work if $p$ is not prime but $\gcd(a,p) = 1$.

Viktor Vaughn
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Umberto P.
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