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Im having trouble proving the converse of Euclid's lemma, and was wondering if anyone could point me in the right direction.

Euclid's Lemma: Let $p$ be a prime number and $a$ and $b$ be natural numbers greater than 1, then if $p|ab$ we know $p|a$ or $p|b$

Converse: For some number $p$, if for all natural numbers $a$ and $b$ greater than 1, $p|ab$ implies $p|a$ or $p|b$, then $p$ is prime.

I tried showing that $p$ had only factors 1 and $p$, but I didn't really manage to make any progress, I'm just really not sure where to start, I'd really appreciate any help, thanks.

CoffeeCrow
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  • $$p^aq^b$$ divides $p^{a+1}q^{b+1}r^{c+1}\cdots$ $$p^aq^b$$ divides $$p^{a+1}q^{b+1}$$ and not $$r^{c+1}\cdots$$ where $p,q,r$ are co-primes – lab bhattacharjee Jul 30 '16 at 07:32

3 Answers3

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If $p$ is not prime, then there are $a,b \neq p$ with $p = ab$. Now $p \mid ab$ but $p \not \mid a, p \not \mid b$.

longer version:

We already have the implication

($p$ prime) $\implies$ (if $p \mid a \cdot b$, then $p \mid a$ or $p \mid b$)

We want to show the implication

(if $p \mid a \cdot b$, then $p \mid a$ or $p \mid b$) $\implies$ ($p$ prime)

We do this my contraposition:

($p$ not prime) $\implies$ (there exist $a,b$ with $p \mid a \cdot b$ and $p \not \mid a, p \not \mid b$)

Now, if $p$ is not prime, then there is a factorization $p = a \cdot b$ with $a \neq p \neq b$. Hence we have $p \mid a \cdot b$, but $p \not \mid a, p \not \mid b$, which shows the above implication.

  • $p=ab$? – – – – – Kenny Lau Jul 30 '16 at 07:41
  • @KennyLau reformulated, should be understandable now. –  Jul 30 '16 at 07:46
  • Im sorry, I'm still pretty new to proof writing in general, so I'm not 100% sure if I follow and would like to check my understanding. My understanding is this is a proof by contradiction. You assume p isn't prime, recognize it must then be composite, then obviously p=1*ab so p|ab, but p then can't divide a or b, so this is a contradiction, thus p is prime, is this right? Thanks for your help by the way! – CoffeeCrow Jul 30 '16 at 08:17
  • @CoffeeCrow added a longer version. –  Jul 30 '16 at 09:24
  • Can you explain why $ p \not \mid a, p \not \mid b$ . Is it because by substitution $a \cdot b \not \mid a $ and $a \cdot b \not \mid b $ ? Though for this to hold we would have to know that $a, b \neq 1 $ (we would have to show that in $p = a\cdot b $ , it is the case that $a,b \neq 1 $ ) – john Jan 12 '20 at 17:28
  • Why assume that p = ab? – ngc1300 Apr 19 '24 at 02:54
  • They are not assuming $a,b$ are any known numbers. And for all natural $m$ there are natural $j,k$ so that $m = jk$. If worse comes to worse we can let $j=1$ and $k=m$. So there is nothing wrong with the assumption. – fleablood Apr 19 '24 at 04:17
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A few years late but I don't believe the other proofs are as direct as they could be.

Actually the statement is false because $p=1$ has the property but $p$ is not prime. So I'm assuming the converse is for $p> 1$.

Let $p$ be such an number with that property. Let $d$ be a divisor of $p$ that is less than $p$. (This is valid. If $p$ is prime then $1$ is such a divisor. If $p$ is not prime we can find another.) Let $m = \frac pd$.

Then we have $md = p$ and $p\mid p$. So by the property $p\mid m$ or $p\mid d$. But $p$ can't divide a natural number smaller than it. So $p\mid m$. But $m\mid p$. We can only have $j\mid k$ and $k\mid j$ if $j=k$. So $p=\mid m$. And so $d=1$.

So the only divisor of $p$ that is less than $p$ is $1$ and the only divisors of $p$ are $1$ and itself, so it is prime.

....

I suppose that that I am assuming that (among natural numbers) we can't have a number be a divisor less than itself. And that $j\mid k$ and $k\mid j \implies k=j$.

But those are easily shown by noting no natural numbers exist between $0$ and $1$.

fleablood
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  • Ok so now I am realizing that the way I was using the assumption was throwing me off. We are syntactically accustomed to the fact that p|ab means n*p = ab for some n, so realizing that realizing that we can apply the assumption to p = md is tricky because it is syntactically missing that n. – ngc1300 Apr 19 '24 at 04:44
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Apr 19 '24 at 06:58
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Suppose $p=ab$. By assumption, either $p\mid a$ or $p\mid b$.

If $p\mid a$, then $a=pc$, for some $c$ and so $$ a=pc=abc $$ Since $a\ne0$, we get $bc=1$. Can you go on?

egreg
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  • I don't see how this proves the converse. – john Feb 07 '20 at 10:34
  • @john We have to prove that either $a=p$ or $b=p$. In the case $p\mid a$, we get $bc=1$, so $b=c=1$ and so $p=a$. Similarly if $p\mid b$. – egreg Feb 07 '20 at 10:42
  • why are we supposing p = ab? – ngc1300 Apr 19 '24 at 02:29
  • @ngc1300 The task is to prove that if $p$ satisfies Euclid’s property, then $p$ is prime, that is, it has no other divisors than $1$ and $p$. How do you write that $a$ is a divisor of $p$? – egreg Apr 19 '24 at 06:18
  • I see now. Before I thought that ab were the same ab that satisfied p|ab. In other words I thought you were claiming p|ab therefore p = ab. – ngc1300 Apr 19 '24 at 13:58