# Modular Algebra in Finding Roots

I'm playing around with this problem.

$$x^{2} - 2x + 10 \equiv 7 \ \text{mod} \ 6$$ Find the equivalence class(es) in $\mathbb{Z_{6}}$ solving this.

The following doesn't work out:

$$x^{2} - 2x + 3 \equiv 0 \ \text{mod} \ 6$$

$$(x-1)(x-2) \equiv 0 \ \text{mod} \ 6$$

suggesting that the solution to this problem should be the equivalence classes under $\mathbb{Z}_{6}$, $\bar{1}$ and $\bar{2}$.

Why does this not work out and are there more general guidelines as to when approaches and problems in, say, algebra I and algebra II, can be similarly posed and solved in modular arithmetic?

Edit

The correct way to solve this problem is to realize that:

$$x^{2} - 2x + 3 \equiv 0 \ \text{mod} \ 6$$

implies

$$x^{2} - 2x + 3 = 6k , k \in \mathbb{Z}$$

This can be solved for and certain solutions obtained.

The first problem is that your factorization is incorrect: $x^2-2x+3\neq (x-1)(x-2)$ (as polynomials over $\mathbb{Z}$ at least).

A bigger problem is that $\mathbb{Z}/6\mathbb{Z}$ has zero divisors. So for instance if $x\equiv 4$ (mod $6$) then $x-1\equiv 3$ and $x-2\equiv 2$ are non-zero in $\mathbb{Z}/6\mathbb{Z}$, but their product is zero. If you replaced $6$ with a prime, however, it would be correct to conclude that $(x-1)(x-2)\equiv 0$ (mod $p$) implies $x\equiv 1$ or $2$ (mod $p$).

The factorization you wrote is not correct - if you multiply it out, you should find $(x-1)(x-2) = x^2 - 3x + 2$, not $x^2 - 2x + 3$ as you wrote.

The polynomial $x^2 - 2x + 3$ does not factorize over the rational numbers, so that is not the place to start. Instead you could try brute force (try all elements of $\mathbb{Z}_6$ since there are only 6 of them). By this method, you quickly find that only $x = \overline{3}$ and $x = \overline{5}$ work.

$$x^2-2x+3\equiv x^2-2x-3\pmod 6$$ and $x^2-2x-3=(x-3)(x+1)$

As $6=2\cdot3,$ we need $$x^2-2x+3\equiv0\pmod3\iff x(x-2)\equiv0$$ $$\iff x\equiv0\ \ \ \ (1a)\text{ or }x\equiv2\ \ \ \ (1b)$$

and $$x^2-2x+3\equiv0\pmod2\iff x^2\equiv1\pmod2\iff x\equiv1\pmod2\ \ \ \ (2)$$

Now apply CRT on $(1a),(2)$ and $(1b),(2)$