Find all solutions of the equation $x^2+x=0$ in $\mathbb Z_6$

Question

As in the title, I'm trying to find all the possible solutions to the equation: $x^2+x=0$ in $\mathbb Z_6$.

I know that the solution for the congruence equation is valid only if $m |(b-a)$. I assume that the equation is already in that form, so the solutions will be $\{0,2,3,5\}$. Is that correct? If not, I would appreciate any form of help.

Thank you.

Answer 1

Preliminary: $\mathbb Z_6$ = {0,1,2,3,4,5} where each $i$ is understood to represent an entire set of integers such that $i = \{n \in \mathbb Z| n = i + k*6; \text{ for some } k \in \mathbb Z\}$. That may be a little abstract and obtuse. In practical terms we can think of $\mathbb Z_6$ as the 6 digits with "circular" addition. But that's not really correct.

So $x^2 + x \equiv x(x+1) \equiv 0 \mod 6$. Thus $x$ and $x+1$ are the "zero divisors" of $\mathbb Z_6$ . $jk = 0 \mod 6$ means $6|jk$. Either $j$ or $k$ equal 0 (or 6) is an obvious solution but also as $6 = 2*3$, either $2|j$ and $3|k$ or $2|k $ and $3|j$ are ohter possibilities.

So we have either:

$x = 0$ or $x+1 = 0$. That is $x = 0$ or $x = 5$

or $2|x$ and $3|x+1$. That is $x =0,2,4$ and $x+1 = 0,3$ or in other words $x = 2$.

or $3|x$ and $2|x+1$. That is $x=0,3$ and $x+1=0,2,4$ or in other words $x =3$.

So the solutions are x = {0,2,3,5}. The only non-solutions are x $\ne$ {1,4}.

Answer 2

Maybe when you mention $b$ and $a$ you are thinking of something you learned about linear congruences? There are no $a$, $b$ mentioned in your post.

One salient observation is that you need only consider congruence classes; that is, $x=1$ will be a solution if and only if $7,\,13,$ etc. are. Thus, you only need to check finitely many integers (in fact, just six of them).