Point addition on an elliptic curve

Question

I have an elliptic curve $y^2 = x^3 + 2x + 2$ over $Z_{17}$. It has order $19$.

I've been given the equation $6\cdot(5, 1) + 6\cdot(0,6)$ and the answer as $(7, 11)$ and I'm unsure how to derive that answer.

I have $6\cdot(5, 1) = (16,13)$ and $6\cdot(0,6)=(0, 11)$ however when I use point addition to add them together I get $(16,13)+(0, 11)=(14,11)$ which isn't even a point on the curve...

Could someone help me identify where and why I've gone wrong?

For further information here's each of the points:

And here's the curve plotted out:

Answer 1

Hints:

• Your calculation of $6(5,1) = (16,13)$ is correct.
• Your calculation of $6(0,6) = (0,11)$ is incorrect, you should get $6(0,6) = (3,1)$.

Maybe if you show how you did that calculation, I can spot the issue.

Once you fix that, I verified the author's result is correct, that is:

$$(16,13)+(3,1) = (7,11)$$

Update

Here are some additional hints to help you with intermediate calculations:

• $P = (0,6)$
• $2P = (9,1)$
• $3P = (6,3)$
• $4P = (7,6)$

Answer 2

You can aid (read: double-check) your calculations with the computer, for instance using Sage or Magma. For example, you can use for free the Magma online calculator. The following code defines your curve $E$ over $\mathbb{Z}/17\mathbb{Z}$, and calculates $6\cdot (0,6)$.

g:=GF(17).1;

E:=EllipticCurve([0,0,0,2*g,2*g]);

P:=E![0,6];

6*P;

The output is $[3:1:1]$ which are projective coordinates for the point $(3,1)$, as it should be.