Prove prime $p\mid ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\,$ without using Fundamental Theorem of Arithmetic

Question

Let: $p$ $\in \mathbb{P}$ $\wedge$ $n_{1},n_{2}\in \mathbb{Z}$. Then: $p|(n_{1}n_{2})\implies p|n_{1} \vee \space p|n_{2} $

This little hypothesis is straightforward while using fundamental theorem of arithmetic. I also know that this can be proved directly by the use of the contraposition for the above implication. However, I wonder how to do this without referring to the fundamental theorem of arithmetic or to contraposition. I think that this must be very easy, but I can't see it right now. Thanks for help in advance.

See here for a few proofs, including a direct proof by descent using the Division algorithm. This also includes further elaboration on the proofs using the GCD Distributive Law in Xam's answer, its Bezout form in Leox's answer. — Bill Dubuque, Mar 01 '17 at 18:43

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

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Let suppose that $p\not\mid n_1$, so $\gcd(p,n_1)=1$. Now, since $p\mid n_1n_2$ and $p\mid pn_2$, then by the definition of $\gcd$ we have $$p\mid \gcd(pn_2, n_1n_2)=n_2\gcd(p,n_1)^{*}=n_2.$$

In (*) we've used the property $\gcd(ac,bc)=c\gcd(a,b)$.

edited Apr 13 '17 at 12:21

Community

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answered Mar 01 '17 at 17:29

Xam

6,119

So easy????????????????????????????????? – Mar 01 '17 at 17:54
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Btw, thanks for help. – Mar 01 '17 at 17:54
@MIT it's easy if you have practice. I suggest you that in order to get familiarity with these simple proofs try to prove that property (*) that I've used. Good luck. – Xam Mar 01 '17 at 18:05
I would start the same way, then use Gauss theorem : since $p\mid n_1n_2$ and $\gcd(p,n_1)=1$, we see that $p\mid n_2$. – Adren Mar 01 '17 at 18:13

score 0 · Answer 2 · answered Mar 01 '17 at 18:36

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Suppose that $p\not\mid n_1$. Then there exist integers $x,y$ such that $px+n_1 y=1.$ Multiple both sides by $n_2$ $$ p (x n_2)+ n_1 n_2 y =n_2. $$ $p$ divides LHS thus $p \mid n_2.$

answered Mar 01 '17 at 18:36

Leox

8,120

This is the Bezout form of the proof in Xam's answer. See here where I prrecisely highlight the analogy. See also the answer linked in my comment on the question. – Bill Dubuque Mar 01 '17 at 18:38
Are you suggesting to me that I should delete my answer? – Leox Mar 01 '17 at 18:43
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No, why would you think that? Rather, I gave a link so that readers can learn how these common proofs are related. – Bill Dubuque Mar 01 '17 at 18:45

Prove prime $p\mid ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\,$ without using Fundamental Theorem of Arithmetic

2 Answers2