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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.

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  • $\begingroup$ $$p^aq^b$$ 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 $\endgroup$ – lab bhattacharjee Jul 30 '16 at 7:32
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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.

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  • $\begingroup$ $p=ab$? – – – – $\endgroup$ – Kenny Lau Jul 30 '16 at 7:41
  • $\begingroup$ @KennyLau reformulated, should be understandable now. $\endgroup$ – Paul K Jul 30 '16 at 7:46
  • $\begingroup$ 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! $\endgroup$ – CoffeeCrow Jul 30 '16 at 8:17
  • $\begingroup$ @CoffeeCrow added a longer version. $\endgroup$ – Paul K Jul 30 '16 at 9:24
  • $\begingroup$ 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 $ ) $\endgroup$ – john Jan 12 at 17:28
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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?

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  • $\begingroup$ I don't see how this proves the converse. $\endgroup$ – john Feb 7 at 10:34
  • $\begingroup$ @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 $p\mid b$. $\endgroup$ – egreg Feb 7 at 10:42
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By assumption, $p|ab$ and either $p|a$ or $p|b$. Let $$ ab = p_{1}p_{2}\cdots p_{n} $$ be the prime decomposition of $ab$. We know that $p|ab$ and so, $p|p_{1}p_{2}\cdots p_{n}$. Then $p$ must divide one of $p_{i}$ for $1 \leq i \leq n$. Since $p_{i}$ are primes, it follows that $p$ must be prime.

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