Solving a quadratic congruence $x^2-2x+4 \equiv 0 \pmod 9$

Question

Solving a quadratic congruence $x^2-2x+4 \equiv 0 \pmod 9$ where $x\in \mathbb{Z}_9$.

My idea:

$x^2-2x+4\equiv (x-1)^2+3 \equiv 0 \pmod{9} \iff (x-1)^2=6\pmod{9}$. How can I deal with mod 9 since it's not prime...

Answer 1

I claim there is no such $x$. Since $\mathbb{Z}_9$ is quite small, you can do brute force: you have only to check that for every $k \in \{0,1,2,..., 8\}$ no one square $k^2$ is congruent to $6$ mod $9$. Thus you cannot find a $x$ that solves your problem

Answer 2

Considering $\mod 3$, it must be $$ \Big[(x-1)^2\equiv 0\mod 3\Big]{\iff \Big[x-1\equiv 0\mod 3\Big] \\\iff \Big[ (x-1)^2\equiv 0\mod 9\Big] \\\iff \Big[ 6\equiv 0\mod 9\Big] \\\iff \text{Contradiction} } $$ hence, this equation has no integer solution.

Answer 3

It has no solution since $6\equiv -3\bmod 9$ is not a square mod. $9\mkern1mu$:

\begin{array}{r|*{4}{r}} n&0&\pm 1&\pm 2&\pm 3&\pm4 \\ \hline n^2&0&1&4&0&-2 \end{array}

Answer 4

HINT.- $(x-1)^2+3=0\iff(x-1)^2=6$. There are not solutions because $(\mathbb Z/9\mathbb Z)^2=\{0,1,4,7\}$.