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I have the following exercise.

If $ abc \equiv b \pmod{m} $ and $ \gcd{(a,m)} = 1 $, then $ ac \equiv 1 \pmod{m} $.

My attempt.

From $ abc \equiv b \pmod{m} $, by definition, $ abc - b = b ( ac - 1 ) = mh $.

Now: if $ b \mid h $, then $ ac - 1 = mk \implies ac \equiv 1 \pmod {m} $; if $ b \mid m $... I don't know what to do...

Ideas? Other solutions? Thanks.

user1988
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  • You need $,\gcd(\color{#c00}m,b) = 1,$ (not $,\gcd(\color{#c00}a,b)= 1$) to cancel $b$ modulo $,m.,$ Probably a typo. – Bill Dubuque Dec 24 '20 at 14:19
  • You are partially right. Is is $ \gcd{(a,m)} = 1 $. Edit done. – user1988 Dec 24 '20 at 14:22
  • That's still incorrect, e.g. let $, m = b(ac-1).\ \ $ – Bill Dubuque Dec 24 '20 at 14:23
  • I don't know... I'm newbie with number theory. I checked and the text says $\gcd{(a,m)}=1$ as hypotesis. I don't know... – user1988 Dec 24 '20 at 14:25
  • Then it's a typo, as I surmised in the first comment. In any case the correct version is a diupe of many other questiuons. – Bill Dubuque Dec 24 '20 at 14:26
  • See the linked questibns for the general theory (which applies when $,\gcd(m,b) = 1,$ the only hypothesis that makes the result generally true). – Bill Dubuque Dec 24 '20 at 14:31
  • I don't know what to say. As I said, I'm newbie and I don't know how ti undestand if there Is a typo in text or not. All I can say is that the text says "$\gcd{(a,m)}=1$". If you say that it is not correct, I believe you. – user1988 Dec 24 '20 at 14:32
  • I don't think "believing an expert" is the right attitude of learning math. – GNUSupporter 8964民主女神 地下教會 Dec 24 '20 at 14:34
  • @GNU There is no "proof by authority" since I gave a simple counterexample a few comments above. Further, the linked dupes give the general solvability / cancelability criterion - which also makes it clear that the typo version is not correct. – Bill Dubuque Dec 24 '20 at 14:36
  • I agree. You are right. I'm sorry. – user1988 Dec 24 '20 at 14:39
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    @user1988 No need to apologize - it's not your fault there is a typo in your text. – Bill Dubuque Dec 24 '20 at 14:40
  • @Bill Dubuque With your counterexample $m = b ( ac - 1 )$, I take numbers like $60 = 4 ( 8 \cdot 2 - 1 )$ and similar: I was not able to chose numbers such that $\gcd{(a,m)}=1$ starting from $m = b ( ac - 1 )$. I'm sorry. My fault. – user1988 Dec 24 '20 at 14:44
  • Note that if $,\color{#c00}{m\mid ac!-!1},$ and $,m = b(ac-1),$ then $,1 = b\color{#c00}{(ac-1)/m},,$ so $,b = \pm1,,$ so choosing $b$ otherwise yields infinitely many counterexamples. Conversely, it is clearly true when $,b = \pm1,$ (no gcd hypothesis is needed) – Bill Dubuque Dec 24 '20 at 14:53
  • @Bill Dubuque Starting from $ m \mid ac - 1 $, I have $ ac - 1 = mk $, so from $ 1 = b ( ac - 1 ) / m $ I have $ 1 = bk $. At this point it should be $ b = \pm 1 $ and $ k = \mp 1 $. Right? – user1988 Dec 24 '20 at 15:12
  • Note that $,\gcd(a,m) = \gcd(a,abc!-!b) = \gcd(a,b),$ so you need to choose $,a,b,$ coprime to satisfy your constraint $,\gcd(a,m)=1;\ $ e.g. for $,a=1,,b=2,$ your claim says that $\bmod 2c!-!2!:\ 2c=2 \Rightarrow c=1,,$ true iff $,c=1,,$ e.g. for $,c=2,$ it says $\bmod 2!:\ 4\equiv2\Rightarrow, 2\equiv 1,,$ so $,2\mid 1,,$ contradiction. (this is a reply to your 2nd last comment) – Bill Dubuque Dec 24 '20 at 15:16
  • I can't understand why $ \gcd(a,m) = \gcd(a,abc-b) = \gcd(a,b) $. Is that true in general or I need the hypothesis $ \gcd(a,m) = 1 $? – user1988 Dec 24 '20 at 16:03
  • By Euclid $,n\equiv \color{#c00}{\bar n} \pmod{!a}\Rightarrow \gcd(a,n) = \gcd(a,\color{#c00}{\bar n}).,$ Above $,n = abc!-!b\equiv \color{#c00}{-b}\pmod{!a},$ so Euclid $\Rightarrow,\gcd(a,abc!-!b) = \gcd(a,\color{#c00}{-b})=\gcd(a,b)\ \ $ – Bill Dubuque Dec 24 '20 at 16:21
  • OK. I understand. Thank you. – user1988 Dec 24 '20 at 20:41

1 Answers1

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This is incorrect. Take $(a,b,c,m) = (1,2,3,4)$ as a counterexample: We have $\gcd(a,m) = \gcd(1,4) = 1$ and $$abc \equiv 3! \equiv 6 \equiv 2 \equiv b \pmod4,$$ but $$ac \equiv 3 \cdot 1 \equiv 3 \not\equiv 1 \pmod 4.$$

GNUSupporter 8964民主女神 地下教會
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