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can someone please give me a hint on an exercise 8.4 (a) taken from Washingtons book on elliptic curves?

I have an elliptic curve in Weierstrass form $y^2 = x^3 + Ax + B$ with $A, B \in \mathbb{Z}$ and a point $P = (x,y)$ on the curve. $2P = (x_2, y_2)$.

I need to show that:
$y^2(4x_2(3x^2+4A)-3x^2+5Ax+27B)=4A^3+27B^2$

Unfortunately I don't see the solution here. The right hand side obviously is the discrimant and I'll tried to substitute $x_2$ by $m^2 -2x = \frac{3x^2+A}{2y}-2x$ and $y^2$ by $x^3+Ax+B$, however (maybe due to lots of errors in my calculation) I don't see a way to transform the left hand side in the right hand side. Any hints?

Thanks for help!

Edit: using division polynomials still I do not get further:
Multiplying the second half $-3x^2+5Ax+27B$ of the left hand side (substitute $y^2$ by $x^3+Ax+B$):
$-3x^5+5Ax-(3A-27B)x^3-(3B-5A^2)x^2+32ABx+27B^2$
Multiplying the first half $x^4-2Ax^2-Bx+A^2$ of the left hand side (substituting $x_2$, cancel the denominator $4y^2$):
$3x^6-2Ax^4-3Bx^3-5A^2x^2-4ABx+4A^3$

I obviously do get the right terms for the right hand side ($4A^3$ and $27B^2)$, however the degree 5 and 6 obviously do not match, so how to cancel out the rest?

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Just use the duplication formula for $P$ and the Weierstrass form you have, you should get that $x_2=\frac{x^4-2Ax^2-8Bx+A^2}{4x^3+4Ax^2+4Ax+4B}$ (modulo typos). Now substitute this into you expression, substitute $y^2$ with $x^3+Ax+B$ and there is no reason you shouldn't get an identity.

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