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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.

Sebastian Valencia
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D-Man
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  • Do you know the fundamental theorem of arithmetic? – Brian Fitzpatrick Feb 10 '14 at 03:04
  • every integer n except 0 and 1 is a product of primes? – D-Man Feb 10 '14 at 03:07
  • @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 '14 at 03:09
  • @BillDubuque that is new to me. I have only been taught that half to my knowledge. – D-Man Feb 10 '14 at 03:10
  • 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 '14 at 03:12
  • yes I know Euclid's Lemma, just didnt have it written down as that. – D-Man Feb 10 '14 at 03:14
  • Which version of Euclid's Lemma do you know? – Bill Dubuque Feb 10 '14 at 03:15
  • the only thing that was taught was the proof of if p divides bc then p divides c or d. and that thm. was proved using the fact that (p,b) = 1 and p divides bc so p divides c, then was used to show p was not 1 and not ppositive or negative p to prove the theorem. Im not sure if that answers your question. – D-Man Feb 10 '14 at 03:19
  • @D-Man That version of Euclid's Lemma implies that if a prime divides a product then it divides one of the factors of the product. So what can you conclude from this lemma when you are given that the prime $,p,$ divides the product $,b^2 = b\cdot b,?\ \ $ – Bill Dubuque Feb 11 '14 at 16:47

4 Answers4

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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 $p \ | \ b^2 \implies p \ | \ b$

Ishfaaq
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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!!!

Robert Lewis
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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, $p \mid b$.

vonbrand
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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 $q_i = p$

botika
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