Question

Find all solutions of the equation \(x^2 +x-6=0\) in the ring \(Z_{14}\) by factoring the quadratic polynomial.
Please, help

Answer 1

\(x^2+x-6=x^2+3x-2x-6=x(x+3)-2(x+3)=(x-2)(x+3)\)

Answer 2

note \((x-2)(x+3)\equiv0\pmod{14}\) but \(14=1\cdot14=2\cdot7\) ergo we have multiple cases:

1. we have \(x+3\equiv0\pmod{14}\)
2. we have \(x-2\equiv 0\pmod{14}\)
3. we have \(x-2\equiv0\pmod2\) and \(x+3\equiv0\pmod7\)
4. we have \(x-2\equiv0\pmod7\) and \(x+3\equiv0\pmod2\)

Answer 3

notice that since \(14=2\times7\) is not prime we have two zero divisors \(2,7\) and therefore \(\mathbb{Z}/14\mathbb{Z}\) is not an integral domain -- this explains why the bottom two possibilities must now be considered

Answer 4

in familiar rings like \(\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}\) we have no zero divisors i.e. \(ab=0\implies (a=0)\lor (b=0)\) -- these are known as integral domains. our case here does indeed have zero divisors so \(ab=0\) actually gives MORE such possibilities

Answer 5

we know that in \(\mathbb{Z}/14\mathbb{Z}\) that \(a=0\implies ab=0\) but we also know that \(a=2,b=7\implies ab=0\); the existence of zero divisors means we have more possibilities

Answer 6

so, we just solve x =-3 (mod 14)
x = 2 (mod 14)
right?

Answer 7

that only covers the possibilities \((1),(2)\) -- there are more possibilities than these... P.S. \(x\equiv-3\equiv11\pmod{14}\) which is the element in \(\mathbb{Z}/14\mathbb{Z}\)

Answer 8

you must also solve \((3),(4)\)

Answer 9

Oh !! I understand the whole thing now. Since this problem relate to the previous one, which stated that although (x-a) (x+a) =0 , we cannot conclude that (x-a) =0 or (x+a) =0
i understood that concept.
But to this problem, I don't know how to solve with "in Z-14"

Answer 10

so, after solving 1,2,3,4
combine the solutions are the final answer, right?

Answer 11

dunno what you mean by combine... each of 1,2,3,4 gives distinct solutions

Answer 12

take the union of their solution sets sure

Answer 13

got you, union, that my "combine", Thank you very much.

Answer 14

1. we have \(x+3\equiv0\pmod{14}\)
as we noted, this gives \(x\equiv-3\equiv 11\pmod{14}\) so \(x=11\) is our solution in \(\mathbb{Z}/14\mathbb{Z}\)

2. we have \(x−2\equiv0\pmod{14}\)
as we noted, this gives \(x\equiv2\pmod{14}\) so \(x=2\) is our solution in \(\mathbb{Z}/14\mathbb{Z}\)

3. we have \(x−2\equiv0\pmod2\) and \(x+3\equiv0\pmod7\)
notice \(x-2\equiv x\equiv 0\pmod2\) ergo \(x\in\{0,2,4,6,8,10,12\}\subset\mathbb{Z}/14\mathbb{Z}\)
also we have \(x\equiv-3\equiv4\pmod7\) ergo \(x\in\{4,11\}\subset\mathbb{Z}/\mathbb{14Z}\)
the solutions that satisfy both are then \(\{0,2,4,6,8,10,12\}\cap\{4,11\}=\{4\}\)

4. we have \(x−2\equiv0\pmod7\) and \(x+3\equiv0\pmod2\)
notice \(x\equiv2\pmod7\) ergo \(x\in\{0,2,9\}\subset\mathbb{Z}/14\mathbb{Z}\)
also we have \(x\equiv-3\equiv1\pmod2\) ergo \(x\in\{1,3,5,7,9,11,13\}\subset\mathbb{Z}/\mathbb{14Z}\)
the solutions that satisfy both are then \(\{0,2,9\}\cap\{1,3,5,7,9,11,13\}=\{9\}\)

Answer 15

therefore the total set of solutions is the union of all of the above solutions \(\{11\}\cup\{2\}\cup\{4\}\cup\{9\}=\{2,4,9,11\}\)

Answer 16

Woaahahha . thanks a ton.

Answer 17

this idea is the heart of the Chinese remainder theorem:

http://en.wikipedia.org/wiki/Chinese_remainder_theorem#Statement_for_general_rings