Htykuut

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

I try from to find x using other equation since $2x equiv 4 (mod 6)$ so $x=6k+2$ or $x=6k+5$, where $kin mathbb N$ 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 $knot =7$ then $k=7m+l, lin {1,2,3,4,5,6,}$, $min mathbb N$ 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 $xin {74,116,...}$.

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 $k=3r+2$,$rin mathbb N$ it show that $knot =3r+1$and $knot =3r+2$.

So solution $xin {11,23,42m+32}$, $min mathbb N$.

Is this ok? Or you know some better way?

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:
$${21m+2,21n-10:m,nin Bbb Z}$$

As $(7,4)=1$

$$x^2+x+1equiv0pmod7iff(2x+1)^2equiv-3equiv4$$

$implies$

either $2x+1equiv2iff2xequiv1equiv8iff xequiv4pmod7 (1)$

or $2x+1equiv-2iff2xequiv-3equiv4iff xequiv2pmod7 (2)$

and we have $2xequiv4pmod6iff xequiv2pmod3 (3)$

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

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