Solve $x^2 - 5x + 6 \equiv 0 \pmod{18}$

Question

Solve $x^2-5x+6\equiv 0 \pmod{18}$.

Here is what I did: I rewrote the equation into $(x-2)(x-3) \equiv 0\pmod{18}$ And I know if (mod p is a prime number) $x \equiv 2\pmod{18}$, $x \equiv 3\pmod{18}$ However, it is not in this case, how do I find other solutions within the range of $1$ to $17$. ($11$ and $12$)?

score 3 · Accepted Answer · answered Mar 13 '21 at 22:12

3

An integer is divisible by $18$ iff divisible by each of the coprime prime powers $2,\,9$. Your function of $x$ is always divisible by $2$, but is divisible by $9$ iff at least one of its factors is, as these are also coprime. So we have two cases $9|x-2,\,9|x-3$.

answered Mar 13 '21 at 22:12

J.G.

115,835

Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Mar 13 '21 at 22:37

score 1 · Answer 2 · answered Mar 13 '21 at 22:18

1

So you have $(x - 2) (x - 3) \equiv 0 \pmod{18}$. Thus the two factors (modulo $18$) multiply to $18 \equiv 0$, i.e., one is $1, 2, 3, 6, 9, 18$ and the other is the other factor. This gives the set of values.

answered Mar 13 '21 at 22:18

vonbrand

27,812

Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Mar 13 '21 at 22:37

Solve $x^2 - 5x + 6 \equiv 0 \pmod{18}$

2 Answers2