# How to find 2P and 3P of an elliptic curve?

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.

• It’s not $(3x_0)^2+a,$ it is $3(x_0^2)+a.$ Commented Jun 2, 2021 at 0:48
• Thank you I did not realise. Commented Jun 2, 2021 at 1:36
• 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 $3P=-3P=(2,0)$. Commented Jun 2, 2021 at 2:00

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

• Thank you but how would I proceed for 3P? Commented Jun 2, 2021 at 1:36

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 $$x(3P)$$.