How can these two equations be solved by elimination?

Question

In this question, the following two equations were solved using elimination. With a google crash course I sort of get how elimination works, but it seems like these are much too complex to add the left sides together in a way that cancels out x or y.

$$ \frac{3x(3x^2+9)}{2y} - \frac{(3x^2+9)}{8y^3}^3 - y = 6 \pmod {23} $$

$$ \frac{(3x^2+9)^2}{(2y)^2} - 2x = 12 \pmod {23} $$

Ultimately I'm trying to figure out any way (the easiest preferably) to reduce these such that x=18, y=10 or x=19, y = 3 in order to convert it to a function in a scripting language.

If elimination is the way to go, what were the steps involved in arriving at the following equation? $$x^4-48 x^3-18 x^2+13968 x=-86481 \pmod {23}$$

Answer 1

If you would like to see how to get the equation by hand:

Introduce a third variable $$z = \frac{3x^2+9}{2y}$$ We have three equations

$$2yz - 3x^2 = 9\\3xz - z^3 - y = 6\\z^2 - 2x = 12$$

Multiplying the second equation by $2z$ and adding the first eliminates $y$: $$6xz^2 - 2z^4 - 3x^2 = 12z + 9$$

Substituting for $z^2$ from the final equation gives $$6x(2x + 12) - 2(2x +12)^2- 3x^2 = 12z + 9\\x^2 - 24x - 297 = 12z$$ Squaring both sides, then substituting for $z^2$ again $$(x^2 - 24x - 297)^2 = 144(2x + 12)\\x^4 - 48x^3 - 18x^2 +13968x + 86481 = 0$$ Modulo $23$, this reduces to $$x^4 - 2x^3 + 5x^2 + 7x + 1 \equiv 0 \mod 23$$

Answer 2

If you want to implement the resultant yourself, you can use the Sylvester matrix (see wikipedia).

In M2:

R=ZZ/23[x,y]
f=3*x*(3*x^2+9)*(4*y^2)-(3*x^2+9)^3-(y+6)*8*y^3;g=(3*x^2+9)^2-(2*x+12)*(2*y)^2;
sylvesterMatrix(f,g,y)

yields

$${\tiny \begin{pmatrix}{-8}& {-2}& -10 x^{3}-7 x& 0& -4 x^{6}+10 x^{4}+7 x^{2}+7& 0\\ 0& {-8}& {-2}& -10 x^{3}-7 x& 0& -4 x^{6}+10 x^{4}+7 x^{2}+7\\ -8 x-2& 0& 9 x^{4}+8 x^{2}-11& 0& 0& 0\\ 0& -8 x-2& 0& 9 x^{4}+8 x^{2}-11& 0& 0\\ 0& 0& -8 x-2& 0& 9 x^{4}+8 x^{2}-11& 0\\ 0& 0& 0& -8 x-2& 0& 9 x^{4}+8 x^{2}-11\\ \end{pmatrix} }$$

determinant sylvesterMatrix(f,g,y) gives the same output as resultant(f,g,y).

As for factoring the output

$$-11 x^{16}-x^{15}-3 x^{13}-2 x^{12}-3 x^{11}+7 x^{10}-10 x^{9}+x^{8}+8 x^{7}-6 x^{4}+4 x^{3}+10 x^{2}+10 x+8$$

to $$(x+5) (x+4) ({x^{2}+3})^{6} (x^{2}-11 x-8) ({-11})$$

you probably don't need the whole functionality, just check on $x=-11,\ldots,11$ whether the value is divisible by $23$.