### y-coordinate of a 4-torsion point

I want to find the `y`

-coordinate of a 4-torsion point. I have the following code where I found the x-coordinate

```
E3=EllipticCurve([0,0,0,-3267,45630])
E=E3.division_polynomial(4);E
E.factor()
```

which gives

```
8*(x^4 + 30*x^3 - 18252*x^2 + 280530*x + 6465339)*(x^2 + 15*x - 3042)*(x^2 - 30*x + 2817)*(x - 15)
```

Then I factor $x^2 - 30x + 2817$ in $\mathbb{Q}(\sqrt{-2})$ and I got $[(-36\sqrt{-2} + 15, 1), (36\sqrt{-2} + 15, 1)]$ which is my x-coordinate. How to get the y-coordinate? If I denote $\sqrt{-2}=B$, then I have $x=15\pm 36B$. And I substitute it into my elliptic curve ~~gives me ~~gives

```
B = var('B')
y2=x^3-3267*x+45630;y2
y=y2.subs({x:15+36*B});y
Y=y.simplify_full();Y
YY=Y.subs({B^2:-2});YY
YYY=YY.subs({B^3:-2*B});YYY
```

which equals to $y^2=186624B-116640$. How to get the y-coordinate. By trying an error multiple times (using PARI-gp) I manage to get y-coordinate as $y=27a^3-135a-108aB$ with $a^4-5a^2-32=0$ and $B^2=-2$. But i got this by trying an error comparing LHs and RHS equation. Not straight away from $y^2=186624B-116640$. I tried the following but it did not work :

```
K.<a> = NumberField(x^4 -5*x^2-32); K
f = K.factor(186624*B-116640); f
```