# Adding two points on an elliptic curve. Where did I go wrong?

I'm trying to add the points $(3,2)$ and $(5,5)$ on the elliptic curve $E: y^2 \equiv x^3 - 2 \mod 7$, of course working in the field $\mathbb{F}_7$.

Knowing that the secant line connecting the two points is given by $$y \equiv mx + a$$ we can solve and find $$m \equiv 5$$ $$a \equiv 1$$

Now here's where I think something goes wrong. Using the above expression for $y$, I should have values satisfying $$(5x+1)^2 \equiv x^3 - 2 \mod 7$$ Unfortunately, no values of $x$ satisfy this equality.

Can someone point out where I've gone wrong?

$(x^3-2) - (5x+1)^2 = x^3-25x^2-10x-3 \equiv (x-3)(x-3)(x-5)$
so the third point corresponds to $x=3$.

It's correct there is no third point. This just means the line $y = 5x+1$ is tangent to $(3,2)$ or $(5,5)$, and then the sum is the inverse of the tangent point, in this case $(3-2)$. (if the tangent line at $P$ intersects the curve at $Q$ then $P + P = -Q$ which implies that $P + Q = -P$).

Alternatively, use the addition formulae: suppose $P = (x_P, y_P)$, $Q = (x_Q, y_Q)$ with different $x$-coordinates then compute $R = P + Q = (x_R, y_R)$ as follows (all in the finite field):

• $m = \frac{y_P - y_Q}{x_P - x_Q}$

• $x_R = m^2 - x_P - x_Q$

• $y_R = y_P + m(x_R - x_P)$

see this intro. In this case, if $x_P= y_P$, use $m = \frac{3x^2_P}{2y_P}$ to find the slope of the tangent. The latter also confirms the slope of the tangent at $(3,2)$ is indeed $5 = \frac{3 \cdot 9}{4}$ mod $7$.

• yes you are right. how do you know its not the point at infinity ? you check the tangent at $P$ ? – reuns Mar 28 '17 at 9:23
• @user1952009 the sum is $O$ iff the original points are inverses, of course. – Henno Brandsma Mar 28 '17 at 9:25