0

I am reading the following exercise:

Find all solutions to the congruence $3x \equiv 5 \mod m $ when $m$ equals $2$

The solution states:

Since $3 \equiv 5 \equiv 1 \mod 2$ the congruence is equivalent to $x \equiv 1 \mod 2$. There is one solution which is $x \equiv 1 \mod 2$

The problem I have is I don't really understand the specific steps taken to find the solution.
I understand that:
$3x \equiv (3 \mod m)\cdot (x \mod m)$, I also know that $3x - 5$ is a multiple of $m$ but I am not sure what (other) rules exactly were used to form the solution.

I mean what rule was used to directly do $3 \equiv 5$ and ignore $x$?

Could someone please break down the exact steps the solution is using to reach the conclusion of one solution?

Update:
I read the links and comments and I came up with the following steps.
$3x \equiv 5 \mod2 \Leftrightarrow 3x - 5 = 2k\space$ for some $k$
$3x - 5 = 2k \Leftrightarrow 2x + x - 5 = 2k \Leftrightarrow x - 5 = 2k - 2x \Leftrightarrow x - 5 = 2(k - x) \Leftrightarrow x - 5 = 2j\space $ for some $j$

Hence:
$x - 5 = 2j \implies x \equiv 5 (\mod 2) \implies x \equiv 1(\mod 2)$

Jim
  • 1,729
  • @BarryCipra: Typo, corrected it – Jim Jul 06 '21 at 21:32
  • Ah, I wondered. In answer to your question, they're not really ignoring the $x$. The coefficients $3$ and $5$ are both odd, so if $m=2$, then $3\equiv1$ and $5\equiv1$ mod $m$. I.e., $x\equiv3x\equiv5\equiv1$ mod $2$. Does that make sense? – Barry Cipra Jul 06 '21 at 21:37
  • @BarryCipra: Why do you name $5$ a coefficient? It is the congruence class right? Also the confusing part is that it is $3x \equiv 5 \mod 2$ so how do we go directly to $3 \mod 2$? There must be some rule I am not aware – Jim Jul 06 '21 at 21:40
  • See also here on mod as a congruence relation vs. operation. – Bill Dubuque Jul 06 '21 at 21:41
  • Sorry, a better word choice might have been simply numbers. – Barry Cipra Jul 06 '21 at 21:46
  • @BarryCipra: I understand that $3 \equiv 1 \mod 2$ and that $5 \equiv 1 \mod 2$. What I am not clear is what rule do we use to split these $2$ if we start with $3 \equiv 5 \mod 2$ – Jim Jul 06 '21 at 21:50
  • Where the solution says $3\equiv5\equiv1\mod2$, it's just combining the two separate facts that $3\equiv1\mod2$ and $5\equiv1\mod2$. – Barry Cipra Jul 06 '21 at 21:57
  • @BarryCipra: So $3$ and $5$ are in the same congruence class. And $3 \equiv 3x$ ? That is why we don't take $x$ into account? – Jim Jul 06 '21 at 22:02
  • I'm sorry if my attempts to explain things haven't helped. Hopefully something at the links Bill Dubuque has provided will do a better job. – Barry Cipra Jul 06 '21 at 22:11
  • @BarryCipra: thank you for your comments and trying to help me! I need to re-read the links provided and your comments more carefully to clarify this in my mind. I am sure your explanations do not have a problem at all – Jim Jul 06 '21 at 22:22
  • @BarryCipra: I updated my post, is my logic correct? – Jim Jul 07 '21 at 10:27
  • @BillDubuque: I updated my post, is my logic correct? – Jim Jul 07 '21 at 10:27
  • 1
    @Jim, yes, your update looks OK. One possible improvement would be to write the $x-5=2(k-x)$ step as $x-1=2(k-x+2)$ instead. Then you could go straight from $x-1=2j$ for some $j$ to $x\equiv1$ mod $2$. – Barry Cipra Jul 07 '21 at 10:34
  • @BarryCipra: Ah thank you for that! – Jim Jul 07 '21 at 10:41
  • @Jim, if you can get your hands on a copy of Gauss's Disquisitiones Arithmeticae, you might take a look at its Section I. In four pages, it lays out the basic theory of modular arithmetic. – Barry Cipra Jul 07 '21 at 11:21
  • @BarryCipra: Do you recommend some specific translation? Does it require a specific/advanced background? – Jim Jul 07 '21 at 11:29
  • 1
    The logic is correct (though to be completely rigorous you should write in each equivalent statement the implicit quantifiers "there exists an integer $k$ such that ...". But you should instead use the linked Congruence Rules, which, make this much intuitive. They are the congruence analog of the well known fact that equalities are preserved if we substitute equal values in arithmetic expressions (congruences are generalized equality relations and you should learn to view them just like you do equalities). – Bill Dubuque Jul 07 '21 at 15:56
  • @BillDubuque: Your last comment, the way you have phrased it, it is very helpful to me. Thank you very much for that! – Jim Jul 07 '21 at 20:16

1 Answers1

0

With a polynomial modulo $m$ you can reduce the coefficients of the polynomial modulo $m$ directly because $(ax^k)\mod m = (a)(x^k) \mod m$. This means that for $\varphi: \mathbb{Z} \rightarrow \mathbb{Z}_n$ with $\mathbb{Z}_n$ the integers modulo $n$ that $\varphi (ab) = \varphi (a) \varphi (b)$. It's instructive and accessible to try prove this for yourself given the computational definition of the integers modulo $n$.

Now since you're working modulo $2$ every number will be congruent to $0$ or $1$. This means we can reduce $3x \equiv 5$ to $1x \equiv 1$ which is $x \equiv 1$. Since we only have two choices for $x$ it's easy to see that $1$ is the only solution since $0$ is not.

CyclotomicField
  • 11,925
  • 1
  • 14
  • 32
  • I didn't understand how did you use $(ax^k)\mod m = (a)(x^k) \mod m$ and how do we go from $3x$ to $1x$ – Jim Jul 06 '21 at 21:24
  • @Jim I have expanded my answer to give a you more clarity without spoiling the fun of letting you prove this yourself. It's a common and useful exercise – CyclotomicField Jul 06 '21 at 23:53
  • I still don't understand how $(ax^k)\mod m = (a)(x^k) \mod m$ is helping/is relevant with the particular question I posted. I think I am misunderstanding something in the logic of your answer – Jim Jul 07 '21 at 08:04
  • @Jim everything else is just reduction modulo $2$. – CyclotomicField Jul 07 '21 at 08:52
  • What does $\varphi$ represent? The $\mod $ operation? – Jim Jul 07 '21 at 09:39
  • I updated my post. Is my logic correct? – Jim Jul 07 '21 at 10:26
  • @jim yes $\varphi$ is the mod operation written as a function. Your work is correct but unnecessarily complicated. – CyclotomicField Jul 07 '21 at 15:11
  • It's instructive and accessible to try prove this for yourself given the computational definition of the integers modulo I don't doubt that it is, but I havent really understood what it is that should dig into further to try to prove and understand it. The first paragraph is highly more advanced and rigorous and I am kind of stuck there to be honest – Jim Jul 07 '21 at 15:28