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Given a point $P = (-1,6)$. $2P$ is found to be $(3,-2)$.

But how? I have used the standard formula and can never get this

With equation $y^2 = x^3 - 15x + 22$

$$m' = (3(x_o)^2+a)/2y_o,\\ x_1 = (m')^2 - 2x_o ,\\y_1 = y_o + m'(x_1-x_o).$$ Then $2P = (x_1 -y_1).$ No P was given it is a torsion subgroup question.

Reposted with more of the information I missed.

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  • $\begingroup$ It’s not $(3x_0)^2+a,$ it is $3(x_0^2)+a.$ $\endgroup$ Commented Jun 2, 2021 at 0:48
  • $\begingroup$ Thank you I did not realise. $\endgroup$
    – user799790
    Commented Jun 2, 2021 at 1:36
  • $\begingroup$ Your curve is LMFDB label 36.a2 with torsion generator $P =(-1,6)$. Given that $2P =(3,-2)$ the line determined by $P$ and $2P$ intersects the curve at $(2,0)$ which is $3P=-3P=(2,0)$. $\endgroup$
    – Somos
    Commented Jun 2, 2021 at 2:00

2 Answers 2

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If the elliptic curve is $y^2=x^3-15x+22,$ then, differentiating, $$\frac{dy}{dx}=\frac{3x^2-15}{2y}.$$ So the slope at $(-1,6)$ is $\frac{-12}{12}=-1.$

Then you want the other point on the curve on the line $y=-x+5$ that goes through $P$ and is tangent to the curve.

So you are trying to solve:

$$(-x+5)^2= x^3-15x+22$$

This will give a cubic equation with repeated roots $x=-1.$ since the constant term of the cubic is $-3,$ the third root is $x_1=3,$ and thus $y_1=-x_1+5=2.$

But then $(x_1,-y_1)=(3,-2)=2P.$

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  • $\begingroup$ Thank you but how would I proceed for 3P? $\endgroup$
    – user799790
    Commented Jun 2, 2021 at 1:36
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The multiples of a point on an elliptic curve $y^2=x^3+Ax+B$ can be computed using the division polynomials. See https://en.wikipedia.org/wiki/Division_polynomials.

For example, if $P=(x,y)$ is not of order $2$ (i.e. $y\neq 0$) we have

$$2P=\left(\frac{x^4-2Ax^2-8Bx+A^2}{4y^2}, \frac{x^6+5Ax^4+20Bx^3-5A^2x^2-4ABx-8B^2-A^3}{8y^3} \right).$$

Of course, the $4y^2$ can be replaced by $4(x^3+Ax+B).$ Plugging in $P=(-1,6),A=-15,B=22$ verifies that $2P=(3,-2)$. On the Wikipedia page, you can find a formula for $3P$ as well.

As a bonus, another useful formula is the following, valid for $P=(x_P,y_P), Q=(x_Q,y_Q)$ such that $P\neq \pm Q$. $$x(P\pm Q)=\frac{x_P^2x_Q+x_Px_Q^2+A(x_P+x_Q)\mp 2y_Py_Q+2B}{(x_P-x_Q)^2}.$$ Setting $Q=2P$ yields a formula for $x(3P)$.

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