Calculating point 2P on an elliptic curve

Question

The equation for the curve is $$y^2=x^3+ax+b$$ and the point in question is $P(x,y)$. We have to verify that the $x$ coordinate of $2P$ is $(x^4-2ax^2-8bx+a^2)/4y^2$. However, the value I get is $(9x^4+6ax^2-8xy^2+a^2)/4y^2$. I derived the algebraic formula for $2P$ ($x(2P)=m^2-2x)$ and used it to calculate the answer.How do I arrive at the actual answer?

Answer 1

We have:

$$\tag 1 y^2 = x^3 + ax + b$$

To add a point, we have:

• $\lambda = \dfrac{3x_1^2 + a}{2 y_1} = \dfrac{3x^2 + a}{2y}$
• $v = y_1 - \lambda x_1 = y-\dfrac{3x^2 + a}{2y}x$
• $x_3 = \lambda^2-x-x = \left(\dfrac{3x^2 + a}{2y}\right)^2-2x = \dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2}$
• $y_3 = -(\lambda x + v)$ (Calculate this out)

Next, we can substitute $(1)$ into the numerator of $x_3$, yielding:

$$\dfrac{9x^4+6ax^2-8xy^2+a^2}{4y^2} = \dfrac{9x^4+6ax^2-8x(x^3+ax+b)+a^2}{4y^2} = \dfrac{x^4-2ax^2-8bx+a^2}{4y^2}$$

Of course, you could have also done the same substitution in the denominator, but they chose not to.