How to complete my proof on $a \mid c$, $b \mid c$ and $HCF(a,b)=1 \implies ab \mid c$

Asked Mar 13 '24 at 10:57

Active Mar 13 '24 at 12:11

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How to complete my proof on $a \mid c$, $b \mid c$ and $HCF(a,b)=1 \implies ab \mid c$

My effort:

Since $a,b$ both divides $c$, we have for some integers $m,n$ $$c=am,c=bn \implies c^2=abmn \implies ab \mid c^2----(*)$$

Now on the contrary let us assume that $ab \nmid c$, so we have by division lemma $$c=abq+R, 0 <R<ab $$

Then squaring we get $$c^2=abq_1+R^2---(**)$$

Now from $(*)$, we should have $$R^2=abp, \: p \in \mathbb{Z}$$

Now I am unsure how to arrive at a contradiction. Any help is greatly appreciated.

edited Mar 13 '24 at 12:11

Bill Dubuque

asked Mar 13 '24 at 10:57

LifeIsMath

What is HCF? Do you mean the greatest common divisor? – 5xum Mar 13 '24 at 11:00
@5xum I thought it was greatest common divisor – Aig Mar 13 '24 at 11:01
Its $GCD$ alternatively called as $HCF$ – LifeIsMath Mar 13 '24 at 11:01
This is a very standard fact about divisibility and GCD, there is a lot of proofs of this on the Internet. Do you need to prove it exactly this way? – Aig Mar 13 '24 at 11:03
@Aig Yeah, typo – 5xum Mar 13 '24 at 11:03
Can you use Bezout's Lemma? – D S Mar 13 '24 at 11:12
After $c=am$, and knowing that $HCF(a, b) = 1$ you can deduce from $b|c$ that $b|m$. – mcd Mar 13 '24 at 11:13
Yes I am curious to know, how to continue in that way without Bezout's? I have already seen MSE forums on this. – LifeIsMath Mar 13 '24 at 11:13
Does this answer your question? Prove that if $a\mid c$ and $b\mid c$ with $\gcd(a,b)=1$, then $ab\mid c$ Or this? https://math.stackexchange.com/questions/1017556 Or simply: if $b\mid am$ and $\gcd(a,b)=1$ then $b\mid m$ hence $ab\mid am=c$. – Anne Bauval Mar 13 '24 at 11:20
@JMP How do you derive this? It’s rather $R^2\equiv 0 \pmod {ab}$. – Aig Mar 13 '24 at 12:52
@Aig; true; yet another wild goose chase! – JMP Mar 13 '24 at 12:58
Try if $a|c,b|c$ then $LCM(a,b)|c$. As $GCD(a,b)=1$, we have $ab|c$. – JMP Mar 13 '24 at 13:13

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