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I was reading the general form of Euclid's lemma which states:

If $a \mid bc$ and $a$ is relatively prime to $b$ then $a \mid c $

I don't really understand why the "relative prime" condition is there.
If for instance we take as $a = 49$ and $b = 91$ then $\gcd(a,b) = 7$
So then we would have:
$ax + by = 7 \Leftrightarrow cax + cby = 7c$
We know that $a \mid cax$
We also know that $a \mid bc \Rightarrow a \mid cby$
So $a \mid 7c$
But we already know that $a \nmid 7$ because $\gcd(a, b) = 7$ so $a\mid c$

So it seems to me that we can prove the lemma without specifying prime relativity.

What am I doing wrong here?

Michael Hardy
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Jim
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  • Consider $b=a$ and $c$ coprime to $a$. Or let $a = pq$ and then let $b=p$ and $c=q$. These should suffice to show why we want them to be coprime. – Cameron L. Williams Jun 21 '21 at 21:46
  • If you drop "coprime", then the statement says "if $a|bc$ then (it will always be the case that) $a|c$." That's patently false: take $a=2$, $b=2$, $c=1$. (The parenthetical addition is elided in the statement; you can always find specific examples where the conclusion holds, but the point of the theorem is that the conclusion must always hold). – Arturo Magidin Jun 21 '21 at 21:47
  • Your question is a bit confusing. First, what did you define $c$ as? Next, your conclusion is that $a|7c$, not $a|c$, so you haven't actually proven the lemma. – Alex R. Jun 21 '21 at 21:48
  • Also $y$ disappeared in favor of $x$ for some reason. – Cameron L. Williams Jun 21 '21 at 21:49
  • @AlexR.: $c$ is any number, I just used a specific example for $a$ and $b$ to show case the approach. And if $a|7c$ and we know that $a$ does not divide $7$ doesnt it mean it has to divide $c$? – Jim Jun 21 '21 at 21:50
  • @CameronWilliams: I corrected the typo, thank you! – Jim Jun 21 '21 at 21:51
  • @ArturoMagidin: I thought it would imply that only if $gcd(a,b) \ne a$ – Jim Jun 21 '21 at 21:52
  • @Jim: Still wrong; $a=4$, $b=c=6$. – Arturo Magidin Jun 21 '21 at 21:52
  • @ArturoMagidin: So which step in my thought process is broken? I am confused where the logic sidetracks – Jim Jun 21 '21 at 21:53
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    @Jim: If you want to substitute "$\gcd(a,b)=1$" with something weaker, then that weaker statement would need to be explicitly stated, and not just say "why do we need to assume coprime?" – Arturo Magidin Jun 21 '21 at 21:54
  • You can't go from $a|7c$ to $a|c$. I mean, your $a$ is $49$. What if $c=7$? Also, you assertion that "we know that $a\nmid 7$ because $\gcd(a,b)=7$" makes absolutely no sense to me. I don't see how the "because" supposedly justifies the previous assertion. – Arturo Magidin Jun 21 '21 at 21:54
  • @ArturoMagidin: that is why I left out replacing $c$ to focus on the logic than on a specific example (may be that was not a good idea) – Jim Jun 21 '21 at 21:55
  • @ArturoMagidin: I am not sure what that weaker statement would be. I was just not sure why does it have to be so strong based on what I wrote in the post – Jim Jun 21 '21 at 21:56
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    It doesn't matter. You can't go from $a|7c$ to $a|c$ when $7|a$. It just isn't true. You have no justification for that step at all. And nothing short of coprime will work. That's just a fact. If $\gcd(a,b)\gt 1$, then there always exist $c$ such that $a|bc$ but $a\nmid c$. Even if $a\nmid b$. – Arturo Magidin Jun 21 '21 at 21:56
  • @ArturoMagidin: that is because if a number does not divide two others it might divide their product which is the converse of addition/subtraction of numbers it divides? – Jim Jun 21 '21 at 21:57
  • I have no idea what "which is the converse of addition/subtraction of numbers it divides" even means. To me, it is nonsense. – Arturo Magidin Jun 21 '21 at 21:58
  • @ArturoMagidin: If $a|b \And b|c \Rightarrow a|(b+c) \And a|(b-c)$ is what I had in mind as analogy – Jim Jun 21 '21 at 22:58
  • The error is: $,a\mid 7c\Rightarrow a\mid c,$ holds iff $,7\nmid a ,$ (not $,a\nmid 7$ as you wrote). Divisibility dyslexia is a common oversight. – Bill Dubuque Jun 21 '21 at 22:58
  • Still doesn't make any sense to me in the context you invoked it; but, please, don't bother to try to explicate it: I don't think there is any point in the effort. – Arturo Magidin Jun 21 '21 at 23:02

4 Answers4

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Yes, you need coprime.

Theorem. Let $a$ and $b$ be integers such that $a\nmid b$. If $\gcd(a,b)\gt 1$, then there exists $c$ such that $a\mid bc$ but $a\nmid c$.

Proof. Let $d=\gcd(a,b)$; then we can write $a=da'$ and $b=db'$, with $\gcd(a',b')=1$. Let $c=a'$. Then $ba'= db'a' = da'b' = ab'$, so $a|bc$. However, $a$ does not divide $a'$, since $a'\lt a$ (because $d\gt 1$). $\Box$

So absolutely nothing short of "$a$ and $b$ coprime" will allow you to go from $a|bc$ to $a|c$. If $a$ and $b$ are not coprime, then there always exist values of $c$ that make the antecedent true (that is, $a|bc$), but the consequence false (that is, $a\nmid c$).

As to what you present: First, that $7$ is a linear combination of $a$ and $b$ does not, by itself, mean $\gcd(a,b)=7$; it means $\gcd(a,b)$ divides $7$. And you cannot go from $a|7c$ and $a\nmid 7$ to $a|c$; that is, after all, what you claim to be trying to prove, and you cannot when $\gcd(a,7)\neq 1$ (see above).

What you have correctly established the following:

If $a|bc$, then $a|\gcd(a,b)c$.

You may think of that as a generalization of Euclid's lemma (since it yields Euclid's Lemma when $\gcd(a,b)=1$); but you cannot go from that to deduce $a|c$.

As to proving that statement, what you have works: let $d=\gcd(a,b)$. There exist $x,y\in\mathbb{Z}$ such that $ax+by=d$, hence $axc+bcy=dc$. Since $a|axc$ and $a|bcy$, then $a|dc=\gcd(a,b)c$, as desired.

But that's where you get stopped. You cannot go further if $\gcd(a,b)\neq 1$.

Arturo Magidin
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The mistake you're making is that you chose an $a$ to be the greatest common denominator of the numbers. The requirement that they be relatively prime is easier to see if you look at the prime decomposition of the numbers involved. Say $bc= p_1p_2p_3$ for $p_k$ distinct primes and take $a=p_1$. Now if $b=p_1p_2$ it is not relatively prime to $a$ anymore but we're forced to have $c=p_3$ and since the primes are distinct $a_1$ cannot divide $p_3$.

If instead we take $b=p_2p_3$ then $a$ is relatively prime to $b$ and in this case $c=p_1$ and indeed $a$ divides $c$ in this case, since we're assured they have a prime in common. The relatively prime requirement ensures that the primes are distributed correctly between $b$ and $c$ in the product $bc$ to ensure $a$ divides $c$.

CyclotomicField
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$$a\nmid 7\implies a\neq 7\implies \gcd(a,b)\neq \gcd(7,b)=7$$ So you can't implies it doesn't divide into the gcd, ( which can actually be 1 or 7 in the $\gcd(7,b)$ ). The main point is, if it's not coprime it can take the factor it has in common from $b$ , and the ones it doesn't have in common from $c$ . If the gcd is 1 however, then it takes all factors from $c$ necessarily.

Roddy MacPhee
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$$ 6\mid 4\times3 \text{ but } 6\nmid4 \text{ and }6\nmid 3. $$ It one drops the requirement of coprimality, then $6\mid4\times3$ would imply that $6\mid4$ or $6\mid3.$

Michael Hardy
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  • I rolled back your edits on Arturo's answer. Choosing $,a|b,$ vs. $,a\mid b,$ is a matter of (personal) style and edits should not alter the writer's style. Please respect that. I seem to recall we had this discussion in the past over the same matter, so I am surprised that it is occurring again. – Bill Dubuque Jun 21 '21 at 22:37
  • @BillDubuque : In your case it is obvious that you have a deliberately chosen personal style, but with most posters here it is just as clear that they don't understand MathJax very well and don't know either what it does or what it can do. – Michael Hardy Jun 22 '21 at 18:32