Fundamental theorem of arithmetic question

Let $b \in \mathbb{Z}$. Prove that if $p$ is a prime number such that $p | b^2$, then $p|b$.

A certain theorem can be used to get this proof set up. I know the general rule that this scenario is true and the concept behind it, but I am unable to find the right starting point to solve this proof. I know that if p divides bc then it dives b or c.

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Do you know the fundamental theorem of arithmetic? –  Brian Fitzpatrick Feb 10 at 3:04
every integer n except 0 and 1 is a product of primes? –  D-Man Feb 10 at 3:07
@D-Man That's the easy existence half of the theorem. You also need the other half, viz. the uniqueness of prime factorizations. What is the "certain theorem""? –  Bill Dubuque Feb 10 at 3:09
@BillDubuque that is new to me. I have only been taught that half to my knowledge. –  D-Man Feb 10 at 3:10
Do you know Euclid's Lemma, $\,\gcd(a,b)=1,\ a\mid bc\,\Rightarrow\, a\mid c,\,$ or do you know Bezout's Identity for the gcd? –  Bill Dubuque Feb 10 at 3:12

Well if you've been granted the theorem for a prime $p$, $\; p \ | \ bc \implies p \ | \ b$ or $p \ | \ c$ you're pretty much sorted. A direct application would yield $p \ | \ b^2 \implies p \ | \ b$

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Well, if you know that $p \mid ab \Rightarrow p \mid a \; \text{or} \; p \mid b$, then if $p \mid b^2$ you have $p \mid b$ or? What else, $p \mid b$. And you're done!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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You know that if $p \mid u v$ with $p$ prime then $p \mid u$ or $p \mid v$. You are given that $p \mid b^2$, that is $p \mid b b$. From the above, $p \mid b$.

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p divide b ^ 2 is

$b ^ 2 = pq, q \epsilon \mathbb{Z}$

If b is prime, it is clear that

$p = q = b$

else

$b = (q_1 ... q_n), q_1...q_n \epsilon \mathbb{Z} \Leftrightarrow b ^ 2 = (q_1 ... q_n) ^ 2$

it's clear that

$\exists q_i, 1\leq i\leq n$ that $q_i = p$

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