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How do I solve 2x=4 in $Z_{12}$

I know the $gcd(2,12) = 2$ and $2|4$ therefore there are 2 solutions, but I'm not sure how to solve this. I tried using the euclidean algorithm but it doesn't seem to work with numbers this small.

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$\begin{align}{\bf Hint}\qquad\ \ \ 2x&\equiv 2a\!\pmod{\!2n}\\[.2em] \iff \color{#c00}2x &= \color{#c00}2a + \color{#c00}2nk \ \ \ {\rm for\ some}\ \ k\in\Bbb Z\\[.2em] \iff\ \ x &= \ \ a +\ \ nk \ \ \ {\rm for\ some}\ \ k\in\Bbb Z\\[.2em] \iff\ \ x&\equiv\ \ a\!\pmod{\!n} \end{align}$

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That's very simple: since $d$ divides all coefficients and the modulus, you can simplify, since $\mathbf Z$ is an integral domain: $$2x\equiv 4\mod 12\iff x\equiv 2\mod 6$$

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  • $\begingroup$ At this level you should explain why that is true (if the OP knew they probably wouldn't ask the question) $\endgroup$ – Bill Dubuque Feb 12 at 20:48
  • $\begingroup$ I've added a hint (integral domain). I think the O.P. think from this hint, and ask for more if he/she does see why. $\endgroup$ – Bernard Feb 12 at 20:50
  • $\begingroup$ Ok thank you, so just to be clear, you can only simplify like this when $Z_m$ is an integral domain? also why is Z an integral domain in this case as 12 is not prime?? $\endgroup$ – user520403 Feb 12 at 20:53
  • $\begingroup$ @user520403 I elaborated in my answer. As you can see, it amounts to cancelling $2$ in $\,\Bbb Z,\,$ which is possible since $2$ is a nonzero element of a domain, so not a zero-divisor. $\endgroup$ – Bill Dubuque Feb 12 at 20:58
  • $\begingroup$ No. What I meant is that, as $\mathbf Z$ is an integral domain, when you interpret the congruence in terms of integers, you can simplify and interpret the relation you obtain as another congruence. Is tat clear? $\endgroup$ – Bernard Feb 12 at 20:58

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