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Find all of the solutions to $[x]^{2}+[9][x]-[10]=[0]$ in the ring $\Bbb Z _{12}$

So i know there are 4 of them.

So i started out with 1 is a solution: $1^{2}+9(1)-10=0$ $\Rightarrow$ 10-10=0 $\Rightarrow$ 0=0

but then i realized you have to use mod's..

and now i am confused.

i am not sure how to use a mod with the equation.

MRI
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    The ring has $12$ elements! – egreg Dec 10 '17 at 21:49
  • Those elements are {0,1,2,3,4,5,6,7,8,9,10,11} then? – MRI Dec 10 '17 at 21:55
  • Yes, that's it. – egreg Dec 10 '17 at 21:56
  • Rather: you can name the elements 0,...,11, but they behave differently as in $\mathbb{Z}$, as for instance $2 \cdot 6 = 0$. – 57Jimmy Dec 10 '17 at 22:01
  • @57Jimmy Can you explain to me how they behave? and how do i relate that to the equation? – MRI Dec 10 '17 at 22:03
  • this is mod12 then? so then 3*4=0? – MRI Dec 10 '17 at 22:10
  • Exactly, see the answer below. If you write them without brackets it might be confusing at the beginning – 57Jimmy Dec 10 '17 at 22:20
  • Another remark: you wrote $2-6=0$ and $2 \wedge 6 \neq 0$. The first is false, the second: I don't know what it means – 57Jimmy Dec 10 '17 at 22:23
  • Alternative hint: $;x^2+9x-10 \equiv x^2-3x+2 \pmod{12}$ and the latter factors as $(x-1)(x-2),$. In how many ways can a product of two consecutive numbers be $0$ in $\mathbb{Z}_{12},$? – dxiv Dec 10 '17 at 22:26
  • Ok, Thank you! I just deleted it to not confuse anyone else – MRI Dec 10 '17 at 22:26
  • If the equation you wrote is really what you mean, then you're overlooking part of the problem -- once you find the four possible values for $[x]$, you still need to find the possible values for $x$ (and you haven't told us what domain $x$ ranges over; the integers, I'd presume). This last step generally isn't a hard one, I just wanted to make sure you're aware that it's there. –  Dec 10 '17 at 23:31

2 Answers2

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You could put it like this (there are many different but "equivalent" ways to express it): the mod(12) of an integer $k$ is the unique number $m$ between $0$ and $11$ that can be the rest of the division of $k$ by $12$. For instance, the mod(12) of any number between $0$ and $11$ is just the number itself, and then it starts again: the mod(12) of $12$ is $0$, the mod(12) of $13$ is $1$,..., the mod(12) of $136$ is $4$ since $136 = 12 \cdot 11 + 4$. For negative numbers, you still take a positive mod. For instance, the mod(12) of $-1$ is $11$, since $-1 = -1 \cdot 12 + 11$.

For your equation, you can compute every step in mod(12), for example $7^2 = 1$ since in $\mathbb{Z}$ the number $7^2$ is $49$ which has $1$ as mod(12).

Another way would be to do the whole computation and then do mod(12) only at the end (but you should explain why this works).

In the case of $1$, your computation is actually almost correct: you never reach the number $12$, so everything is fine and you can compute as usual. You should just write $[1]^2 + [9]\cdot[1]-[10] = [1]+[9]-[10]=[10]-[10]=[0]$, as desired.

Now you just need to check all the other 11 possibilities. For instance, 7 is not a solution, since $[7]^2 + [9]\cdot[7]-[10] = [1]+[3]-[10] = [6] \neq [0]$.

57Jimmy
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Well, if you already found $x=1$ as solution for $x^2+9x-10=0$, it means we can pull out $(x-1)$ from $x^2+9x-10$, yielding $(x-1)(x+10)$.
So another solution of the equation (without using mod12 anywhere) is $x=-10$.

The hint rather says $2\cdot 6=0$, which is true modulo $12$, since $2\cdot 6=12\equiv 0$. And this (and perhaps also $3\cdot 4=0$) leads to the other solutions.

Berci
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