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Example: $7x\equiv 3 \mod 12$

solution: $x=9+12t$

What if I said $$2*[7x\equiv 3 \mod 12]$$ $$(14x\equiv 6 \mod 12)\div 2$$ $$7x\equiv 3 \mod \frac{12}{gcd(12,2)}$$ $$3*7x\equiv 3*3 \pmod 6$$ $$-3x\equiv 3 \pmod 6$$ $$-x\equiv 1 \mod {\frac{6}{\gcd(3,6)}}$$ $$x\equiv -1 \pmod 2$$ $$x\equiv 1 \pmod 2$$ $$x=1+2t$$

What's my misunderstanding/mistake about the rules that adds the solutions 1,2,3$ etc...

user5389726598465
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    You forgot to also multiply the $\rm\color{#c00}{modulus}$ by $2$, i.e. $,3/7\pmod{!12}, =, 6/14\pmod{\color{#c00}{!24}}.\ $ Eliminate the congruences and write everything as integer equations to understand it. – Bill Dubuque Jan 19 '17 at 17:04
  • See this answer for much further discussion. – Bill Dubuque Jan 19 '17 at 18:18
  • So is it not useful or allowed to use division in later steps as opposed to congruence equivalence reduction when the a and p are coprime in $a \equiv b \pmod p$? $$[4x\equiv13 \pmod {47}]12->48x\equiv1213\pmod {47}->x\equiv 15$$ Here $15*48\pmod{47}=15$. However $$18x\equiv42 \pmod {50}$$ does have gcd(18,50)>1 and division as the first step does give multiple values $$x\equiv19 ,or, 44 \pmod {50}$$. – user5389726598465 Jan 19 '17 at 18:36
  • Yes, and I made a mistake. It's almost already answered on your link. – user5389726598465 Jan 19 '17 at 19:29
  • @Joffan It's not a matter of acceptance but, rather, time optimization. Ok, I'll add an answer in a bit. Today is very busy so I keep getting pulled away by real-world matters. – Bill Dubuque Jan 19 '17 at 19:31
  • @user135711 Your edit is a bit puzzling. It would help if you could please explain precisely what properties you are using to derive each congruence from the prior one. This will help to pinpoint the conceptual errors. – Bill Dubuque Jan 19 '17 at 19:45
  • That's a good Idea. It's very likely that there are lots of mistakes since I'm just learning this. – user5389726598465 Jan 19 '17 at 19:54
  • What rules or logic are you using to multiply and divide congruences by $2$ as in the first $3$ displayed equations? – Bill Dubuque Jan 19 '17 at 20:08
  • I'm using a step that is avoided in the text book when solving formulas and is probably what is makes all this wrong. $ca\equiv cb \pmod m \iff a\equiv b \pmod {\frac{m}{gcd(m,c)}}$ Looked everywhere, can't find the proof in the book where I read it. Maybe that's wrong when $cax\equiv cb \pmod m$ – user5389726598465 Jan 19 '17 at 20:10
  • Oh, the multiplying by 2 is the origin of my question. I'm purposefully creating an example that doesn't follow the algorithm 1) reduce reduce by gcd(a,m), 2)reduce by gcd(a,b) 3)reduce by congruences. – user5389726598465 Jan 19 '17 at 20:16
  • Anyway, I think I can atleast sleep well following the formula now, knowing that more solutions isn't more information but less. – user5389726598465 Jan 19 '17 at 20:18

1 Answers1

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What is basically happening here is that you are discarding information about $x$, making the solution set appear wider than it really is.

Given that $7x\equiv 3 \bmod 12 $, it is certainly true that $7x\equiv 1 \bmod 2 $ or other statements that use a base that divides $12$. But if you only use these reduced equivalances, you will allow solutions excluded by the full condition (as well as including correct solutions).

Putting your statements side by side with the implied divisibility statements:

$$\begin{array}{c|cl} 7x\equiv 3 \bmod 12 & 7x-3 = 12k \\ 2\times[7x\equiv 3 \bmod 12] & 2(7x-3) = 2\cdot 12k \\ [ 14x\equiv 6 \bmod 12 ] / 2 & (14x-6)/2 = 12k /2 & \quad \small\color{brown}{\text{not equivalent to line above}} \\ 7x\equiv 3 \bmod \frac{12}{gcd(12,2)} & 7x-3 = (12/2)k\\ x\equiv 1\pmod 2 & x-1 = 2k'& \quad \small\color{brown}{12/2 \ne 2}\\ \end{array}$$

Joffan
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  • I made a mistake in my theoretical example. But that doesn't make your explanation about discarding information any less enlightening. A lot less information is now given if you use multiplication and then division reducing to mod 2. – user5389726598465 Jan 19 '17 at 19:29