Solve equations $x^2+x+1\equiv 0(\mod 7)$ and $2x-4\equiv 0(\mod 6)$

Question

Solve equations $x^2+x+1\equiv 0(\mod 7)$ and $2x-4\equiv 0(\mod 6)$

I try from to find x using other equation since $2x \equiv 4 (mod 6)$ so $x=6k+2$ or $x=6k+5$, where $k\in \mathbb N$ if $x=6k+2$ then if we put this in first equation i get $36k^2+30k+7$ so if we want nature number then $7|36k^2+30k$ if I put $k=7$ then I get one solution $x=44$. If I think that $k\not =7$ then $k=7m+l, l\in \{1,2,3,4,5,6,\}$, $m\in \mathbb N$ then it show that $7|36k^2+30k$ if $k=7m+5$ so then $x=6(7m+5)+2=42m+32$ then if I put in other equation and first it is true that $x=42m+32$, so $x\in \{74,116,...\}$.

In second option that $x=6k+5$. And if I put in first equation I get $36k^2+60k+25+6k+5+1=36k^2+66k+31$ so this we can write as $35k^2+k^2+63k+3k+28+3$ so I need to show that $7|k^2+3k+3$ from here we can see that one solution $k=1$, then $x=11$ for $k=2$ this is not solution, for $k=3$ this is solution then $x=23$ if we want to find another number then we can write $k=3r+1$ or $k=3r+2$,$r\in \mathbb N$ it show that $k\not =3r+1$and $k\not =3r+2$.

So solution $x\in \{11,23,42m+32\}$, $m\in \mathbb N$.

Is this ok? Or you know some better way?

Answer 1

We can take $x=3k+2$. So $$(3k+2)^2+3k+2+1 \equiv 0 \mod 7$$ \begin{aligned} \begin{align} 9k^2+12k \equiv 0 \mod 7&\iff k(k+4)\equiv 0 \mod 7 \\ &\iff k=7m \ \ \text{or} \ \ k=7n-4 \end{align} \end{aligned} Hence, all solution set: $$\{21m+2,21n-10:m,n\in \Bbb Z\}$$

Answer 2

As $(7,4)=1$

$$x^2+x+1\equiv0\pmod7\iff(2x+1)^2\equiv-3\equiv4$$

$\implies$

either $2x+1\equiv2\iff2x\equiv1\equiv8\iff x\equiv4\pmod7\ \ \ \ (1)$

or $2x+1\equiv-2\iff2x\equiv-3\equiv4\iff x\equiv2\pmod7\ \ \ \ (2)$

and we have $2x\equiv4\pmod6\iff x\equiv2\pmod3\ \ \ \ (3)$

Apply CRT on $(1),(3)$

and $(2,3)\implies$lcm$(3,7)\mid(x-2)$