I have a 4-torsion point on elliptic curve

$$E : y^2=x^3-3267x+45630$$ given by $(x,y)=(15-36\sqrt{-2},27\sqrt{256\sqrt{-2} - 160})$ which I have checked that it satisfies my elliptic curve. Let $\beta=\sqrt{-2}$. So we have $$(x,y)=(15-36\beta,27\sqrt{256\beta - 160})$$

But than, I was told that the coordinate can be written as $$(x,y)=(15-36\beta,27\alpha(\alpha^2-4\beta-5))$$ where $\alpha^4 - 5\alpha^2 - 32=0$ and $\beta=\sqrt{-2}$.

My question is how is the y-coordinate $27\sqrt{256\beta - 160}$ equals to $27\alpha(\alpha^2-4\beta-5)$ ? How can I obtain that?

  • $\begingroup$ you have 4 possible values for $y$ : $\pm\sqrt{\pm256\beta - 160}$, and those should correspond to $\alpha_k(\alpha_k^2-4\beta-5)$ where $\alpha_1,\ldots,\alpha_4$ are the $4$ roots of $X^4-5X^2-32$ $\endgroup$ – reuns Nov 6 '16 at 23:03
  • $\begingroup$ @user1952009 But how did you know that? I mean it is not obvious that $y=\pm\sqrt{\pm256\beta-160}$ = $\alpha_k(\alpha_k^2-4\beta-5)$. Is there like a specific calculation I should do to obtain that? Because now I have another coordinate $$(x,y)=\bigg[-\dfrac{27}{2}\beta-\dfrac{1}{2}\sqrt{2430\beta+24768}-\dfrac{15}{2}, \sqrt{-264627\beta-\dfrac{1}{2}\sqrt{1180628067960\beta+5672869080264}-557685}\bigg]$$ where $\beta =\sqrt{17}$. How am I to write this in the form of $\alpha, \beta$ like above? $\endgroup$ – shahrina ismail Nov 7 '16 at 4:58

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