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Let $p$ be a prime number.

Let $r_1$, $r_2$ be integers such that $1\leq r_1 < p$ and $1 \leq r_2 < p$.

How to prove that $p \nmid r_1r_2$?

I know one way to prove this. It can be proved by the fact that $p \nmid r_1$ and $p \nmid r_2$ (because $r_1 < p $ and $r_2 < p$) and by Euclid's lemma that says that $p \mid ab \Rightarrow p\mid a\vee p \mid b$.

I'm asking if there is a simpler proof for this that proved this without using this lemma.

Another (related) question:

Let $p$ be a prime number and let $a$ be some integer. I want to prove that $p \mid a^2 \Rightarrow p \mid a$.

I can prove this by proving $p \nmid a \Rightarrow p \nmid a^2$ by: $p \nmid a \Rightarrow a = pq+r $for some $q,r$ integers. Than, $a^2=(pq+r)(pq+r)=p(pq^2+2qr)+r^2$. Now, I can use the proof in my first question to prove that $p \nmid r^2$ and thus $p\nmid a^2$. I want to know if there is a simpler proof to what I prove in question 2.

egreg
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George
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  • I don't think you'll find a proof that doesn't depend on some form of that lemma. – Mathmo123 Jun 25 '14 at 20:55
  • You could incorporate the proof of that lemma into your proof, and thus avoid its use. – Charles Jun 25 '14 at 20:58
  • $p$ prime. $ p \mid a^2 \Rightarrow p \mid a$. I want to know if there is a trivial or intuitive explanation for this rather than using the proof I wrote in my question. – George Jun 25 '14 at 21:24
  • I actually like your proof - it's a good application of what you've proven already.

    Intuitively, think of the primes as the smallest building blocks of an integer. Every integer can be factorised into primes... so if $p \nmid a^2$ that means that $p$ does not feature as a "building block" of $a^2$

     – Mathmo123 Jun 25 '14 at 21:49
  • How do you define "prime"? – vonbrand Jun 25 '14 at 22:53

3 Answers3

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Below are a few proofs of Euclid's Lemma: $ $ prime $\,p\mid ab\,\Rightarrow\,p\mid a\,$ or $\,p\mid b,\, $ which, for variety, are all a bit different from the standard Bezout-based proof.

$(1)\ $ By little Fermat, $\,{\rm mod}\ p\!:\ a\not\equiv 0\,\Rightarrow\, 1\equiv a^{p-1}\!\equiv a\,a^{p-2}\Rightarrow\,a^{p-2}\!\equiv a^{-1},$ therefore if $\,ab\equiv 0\,$ then scaling by $\,a^{-1}$ yields $\,b\equiv 0,\,$ hence $\,p\mid ab,\ p\nmid a\,\Rightarrow\,p\mid b.\,$

$(2)\ $ Use Gauss's algorithm, to deduce, $\,{\rm mod}\ p\!:\ a\not\equiv 0\,$ is invertible. Finish as in $\,(1).$

More generally it is the special case $\,c = p\,$ prime below, since $\,(p,a)=1\iff p\nmid a$

$(3)\ $ Theorem $\,\ (c,a) = 1,\ c\mid ab\ \Rightarrow\ c\mid b$

Proof $\ $ The set $\,S$ of naturals $\,n\,$ such that $\,\color{#c00}{c\mid nb}\,$ is closed under subtraction and contains $\, a,c\,$ therefore its least positive element $\,\color{}{d\mid a,c}\,$ so $\,(a,c)=1\,\Rightarrow\,\color{}{d=1}\in S\ $ so $\ \color{#c00}{c\mid d b} = b$

Community
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Bill Dubuque
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$r_1 r_2$ has a prime factorization. By uniqueness of this factorization, it is equal of the factorization of $r_1$ times the factorization of $r_2$. So, $p|r_1 r_2$ implies $p$ appears as one of the prime factors of $r_1r_2$, which implies $p$ is a prime factor of $r_1$ or $r_2$. Thus $p$ divides $r_1$ or $r_2$. This is impossible: if $p$ divides a positive integer, then that positive integer must be $\geq p$.

Forever Mozart
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    The Fundamental Theorem of Algebra barely seems like a simpler assumption than Euclid's Lemma (we need Euclid's Lemma to prove it). – whosleon Jun 25 '14 at 23:16
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Suppose you know that $1\le r_1<p$ and $1\le r_2<p$ implies $p\nmid r_1r_2$. Now, suppose $p\mid ab$. Write $a=q_1p+r_1$ and $b=q_2p+r_2$, with $0\le r_1<p$ and $0\le r_2<p$. Then $$ ab=q_1q_2p^2+(q_1r_2+q_2r_1)p+r_1r_2 $$ and so $$ p\mid r_1r_2 $$ By hypothesis, we must either have $r_1=0$ or $r_2=0$.

This shows that what you want to prove is actually equivalent to Euclid's lemma.

egreg
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  • But that's not saying much since all ways of proving this are essentially equivalent in $,\Bbb Z.,$ See this answer for a long list of statements which are all equivalent to Euclid's Lemma in $,\Bbb Z.\ \ $ – Bill Dubuque Jun 26 '14 at 00:20
  • @BillDubuque The question was whether the given statement can be proved without Euclid's lemma. The answer is no. – egreg Jun 26 '14 at 06:41
  • @vonbrand It depends on how one interprets "prove without". One often sees such requests for logically equivalent statements. Often it means without using the standard proof of the result that one wishes to avoid. Thus it amounts to asking for a proof "different" from the standard proof. – Bill Dubuque Jun 26 '14 at 13:11