Solve congruence $x^2+x+9\equiv 0 \pmod{63}$

Question

$x^2+x+9\equiv 0 \pmod{63}$

I know that $63=3^2 \cdot7$, so I try to solve like this $x^2+x+9\equiv 0 \pmod 9$ and $x^2+x+9\equiv 0 \pmod 7$, from first equation $x^2+x \equiv -9 \pmod 9 $,$x^2+x \equiv 0 \pmod 9 $, then I find that $x=9m+2$ and $x^2=9k+7$, or $x=9m+5$ and $x^2=9k+4$ or $x=9m+8$ and $x^2=9k+1$, the same for $7$ but it does not help me, then I multiply 63 until I do not get a square equation that I can solve, so I then I get $x^2+x+9=315$ from here $x_1=-18$ and $x_2=17$, but is there some other way to do this?

Answer 1

This is equivalent to $7|x^2+x-12=(x+4)(x-3)\land 9|x(x+1)$, i.e. $7|x-3\land (9|x\lor 9|x+1)$, i.e. $63|x-45\lor 63|x-17$.

Answer 2

The congruence $x^2+x+9\equiv 0\pmod 9$ simplifies to $$x^2+x=x(x+1)\equiv 0\pmod 9. $$ Calculating for all values of $x\bmod 9$, it is easy to check there are two solutions.

Modulo $7$, the congruence becomes $$x^2+x+2\equiv 0.$$ As we're in a field, you may calculate the discriminant, which happens to be $0\bmod 7$. So there is a double root, which has to be a root of the derivative, so it satisfies the equation $$2x+1\equiv 0\pmod 7.$$