Finding the roots of $x^2+x-6=0$ in $\mathbb{Z}_{14}$

Question

I am currently studying ring theory and encountered the following problem $x^2+x-6=0$ In $\mathbb{Z}_{14}$. My instructor informed us to look out for zero divisors, but I cant see why

$x-2=7$

Or

$x+3=7$

Is two of the four options for the roots (the other being when those two equations $=0$)? I understand that if $xy=0$ in $\mathbb{Z}_{14}$ then x or y must be a multiple of 14, but wouldn’t that mean

$x-2=14$

Or

$x+3=14$

Instead of the former?

If $x-2 \equiv 7 \bmod{14}$ then $x+3$ will be even so $(x-2)(x+3) \equiv 0 \bmod{14}$. — xxxxxxxxx, Dec 07 '19 at 22:32
suggest you simply take each, say, $-6 \leq x \leq 7$ and calculate $x(x+1) - 6$ for each and see what you have $\pmod {14}$ for each — Will Jagy, Dec 07 '19 at 22:38
Lets see, $x=-6,$ giving $-6(-5) - 6 = 30-6=24,$ subtract $ 14 $ and get $10.$ Good start. — Will Jagy, Dec 07 '19 at 22:40
Next $x=-5,$ giving $-5(-4) - 6 = 20 - 6 = 14 \equiv 0 \pmod {14}.$ Who Knew? — Will Jagy, Dec 07 '19 at 22:42
@WillJagy Suggesting brute-force search is not very helpful from a pedagogical viewpoint. — Bill Dubuque, Dec 07 '19 at 23:07
@BillDubuque it is if I suspect a student has never done such a calculation. — Will Jagy, Dec 07 '19 at 23:36

score 3 · Answer 1 · answered Dec 07 '19 at 23:18

Chinese remainder theorem says that the equation decomposes into two equations: mod $2$ and mod $7$.

The first one, mod $2$, is trivial since $x^2 + x - 6$ is always $0$ in $\mathbb Z / 2 \mathbb Z$.

The second one, mod $7$, gives you $(x - 2)(x + 3) = 0$ in $\mathbb Z / 7 \mathbb Z$. That ring being a field, the solutions are exactly $x \equiv 2, -3\pmod 7$.

Thus the roots of that polynomial in $\mathbb Z/14\mathbb Z$ are all the elements which are congruent to $2$ or $-3$ mod $7$. These are then $2, 4, 9, 11$ mod $14$.

score 1 · Accepted Answer · answered Dec 07 '19 at 23:31

You wrote

I understand that if xy=0 in Z14 then x or y must be a multiple of 14

That's not true. The product $2\times7\equiv0\bmod14,$ but neither $2$ nor $7$ is a multiple of $14$.

Thus, for example, $4$ is a zero of $x^2+x-6=(x-2)(x+3)\bmod14,$

even though $4$ is not a zero of $x-2$ nor of $x+3$.

Euclid's lemma $p|xy\implies p|x$ or $p|y$ works for prime $p$.

Finding the roots of $x^2+x-6=0$ in $\mathbb{Z}_{14}$

2 Answers2