3

I'm currently reading the 'A Graduate Course in Applied Cryptography' paper written by Boneh and Shoup. More precisely, I'm reading the chapter about 'Elliptic Curve' and I'm stuck at the exercise about scalar multiplication.

enter image description here

I managed to answer equation 15.37 but I'm stuck for equation 15.38.

What I did so far.

First of all $(2 \alpha + 1)P$ can be expressed as $\alpha P + (\alpha + 1)P$. So as to compute $x_{2\alpha +1}$, I can use the first rule of the law of addition that says that in order to computer x coordinate I have to compute $s_c^2 - x_a - x_{a+1}$ where $s_c^2 = (\frac{y_\alpha - y_{\alpha+1}}{x_\alpha - x_{\alpha+1}})^2$. From this step, we can observe that we managed to obtain a part of the desired denominator. Concerning the above part $y_\alpha^2$ and $y_{\alpha+1}^2$ can be replaced using the Weierstrass equation. My only problem concerns $-2 \times y_\alpha \times y_{\alpha+1}$. My first idea was to replace $y_{\alpha + 1}$ by $\frac{y_\alpha - y_1}{x_\alpha - x_1} \times (x_\alpha - x_{alpha +1}) - y_\alpha$ but I will be stuck with $y_\alpha \times y_1$.

I'd be really grateful if someone could give me a tip on how to answer this question.

kelalaka
  • 48,443
  • 11
  • 116
  • 196
Hugo Peyron
  • 31
  • 1
  • You have given $x_1$ so you know $y_1$ and similarly for $y_\alpha$, too. Once you solve all, could you post them as an answer? – kelalaka Oct 26 '23 at 16:03
  • And, +1 for the effort. – kelalaka Oct 26 '23 at 16:12
  • Note that we don't use the term point multiplication, we use the term scalar multiplication. EC points addition doesn't form a ring, but rather a Z-Module with scalar multiplication. – kelalaka Oct 26 '23 at 19:28
  • 1
    Thank you @kelalaka for the change, you are totally right, the terminology employed was not correct. – Hugo Peyron Oct 27 '23 at 07:40
  • 1
    Btw, I finally managed to find the trick. Indeed we need three steps to solve the question. First of all, we need to compute $x_{2 \times \alpha + 1 }$ using the appropriate rule of addition. Reminder $(2 \times \alpha + 1)P = \alpha P + (\alpha + 1)P$. For simplicity, we will consider that the result obtained is of the form $(x_\alpha - x_{\alpha + 1})^2 \times x_{2 \times \alpha + 1} = A$. Second step, compute $x_1 = (\alpha + 1)P+(-\alpha)P$. Finally multiply $x_1$ with $x_{2 \times \alpha + 1}$, arrange the terms and divide by $x_1$. This explains why we find $x_1$ in the denominator part. – Hugo Peyron Oct 27 '23 at 07:46
  • Nice. Now can you post a full answer for both parts? – kelalaka Oct 27 '23 at 07:49
  • Can I post pictures on the website as I am limited with 600 characters for comments ? – Hugo Peyron Oct 27 '23 at 07:50
  • You can answer your questions, this is valid and we want it, especially like in your case. No need to photos, we have just $\LaTeX$/MathJax as you noted. Sometimes people do answer their question for the improvement of our common knowledge – kelalaka Oct 27 '23 at 08:13

0 Answers0