CurveEnthusiast
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A very concrete example of why we should not stop studying something once we have isomorphisms, although going away from vector spaces. A lot of modern (asymmetric) cryptography is based on discrete ...

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I think Jyrki's answer is great, and I completely agree with it. It focuses on public key cryptography, which is probably most interesting from a mathematical point of view. Let me try to give what I ...

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Short answer: the point at infinity is an easy case. For any affine point, notice that $\det H=8\psi_3$, where $\psi_3$ is a division polynomial. From standard properties of division polynomials, the ...

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Let $E/\mathbb{F}_p$ be the elliptic curve which you describe. I'm going to assume that $E(\mathbb{F}_p)=\langle P\rangle$, because you mention that $P$ is a generator. Therefore $\#E(\mathbb{F}_p)=n$,...

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I'm not sure what you mean by a standard way, here's an answer from someone working with elliptic curves related to cryptography. Steven Galbraith has a nice chapter on Elliptic Curves in his book ...

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To add to my comment: the problem is that in $\mathbb{F}_p$, $p$ prime, we'll have $x^p=x$ for all $x\in\mathbb{F}_p$. Therefore your group law is almost trivially commutative. Instead you can ...

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Let $E/\mathbb{F}_2:y^2+y=x^3+x$ be an elliptic curve. It is easily checked that it is supersingular, and that $\#E(\mathbb{F}_2)=5.$ That means that $a=p+1-\#E(\mathbb{F}_2)=-2$. Suppose that $\... View answer 1 answers 3 votes 234 views Accepted answer 2 votes Disclaimer: I don't consider myself a mathematician, but I have some background in it. I am currently working on topics related to elliptic-curve cryptography, so I hope my perspective might be useful.... View answer 1 answers -2 votes 135 views Accepted answer 2 votes Let$n=14$. Then$(n-2)n=12\cdot 14=168$, and$n^2-1=14^2-1=195$. Take$X=2$. Then $$2^{168}\equiv1\bmod{195}.$$ But$n-1=13$, while$n+1=15$, and 15 is clearly not prime. Your statement seems to be ... View answer 1 answers 1 votes 139 views Accepted answer 2 votes By Lagrange$\#E(\Bbb F_p)=kp$for some integer$k. By Hasse's theorem $$kp<p+1+2\sqrt{p}<2p\quad (\text{since }1+2\sqrt{p}<p\text{ for }p>7)$$ and kp>p+1-2\sqrt{p}>0\quad (\text{... View answer 1 answers 1 votes 154 views Accepted answer 2 votes The hint seems to just point you to using Pohlig-Hellman. We can use this approach whenever the order of our base element is smooth, in this case g which has smooth order 2\cdot11\cdot19\cdot281. ... View answer 1 answers 1 votes 67 views Accepted answer 2 votes It seems that we are browsing the same SE pages :-). I think the idea of the result has not sunk in yet, so let me try to explain what we're actually trying to do. We are assuming a k-bit RSA-... View answer 2 answers 0 votes 225 views Accepted answer 2 votes Since we can pay A cents with B coins, there are non-negative integers b_1,b_2,b_5,b_{10},b_{20},b_{50} and b_{100} such that \begin{align*} A &= b_1+2b_2+\ldots+100b_{100}, \\ B &= ... View answer 1 answers 3 votes 246 views 1 votes I've managed to solve it, posting it here for whoever is interested. Let E/\mathbb{F}_q be an elliptic curve, and assume that \operatorname{tr}(\phi)=0\pmod{p}. Then by (a), we know that E is ... View answer 1 answers 1 votes 1k views 1 votes Hints: a) A point P has order 2 if P+P=\mathcal{O}, and therefore P=-P. What is -P? Then equate the y-coordinates. b) A point Q has order 3 if Q+Q+Q=\mathcal{O}, and therefore Q+Q=-Q.... View answer 1 answers 1 votes 94 views 1 votes Any \omega\in\Omega_C is holomorphic if \operatorname{ord}_P(g)\geq0 for all P\in C, where \omega=g\,dt and t is a uniformizer at P. Note that g\in\bar{K}(C). Explanation: On page 31 ... View answer 1 answers 0 votes 246 views Accepted answer 1 votes Shank's algorithm can be used for any group, it does not use any specific properties. The same is true for the Pohlig-Hellman algorithm. Suppose we have a group of orderr=\prod_ip_i^{e_i},then ... View answer 1 answers 0 votes 68 views Accepted answer 1 votes Multiplying the second equation by 2 we have the system \begin{align*} 6P+2Q+R=\infty, \\ 4P+2Q+4R=\infty. \end{align*} Subtracting the second from the first gives2P-3R=\infty$, i.e.$2P=3R$. Since$...

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The points $P_1,P_2$ together form a basis of $E[m]$. Hence for any $\alpha,\beta\in\mathbb{Z}/m\mathbb{Z}$ such that $\alpha P_1+\beta P_2=\mathcal{O}$ we can conclude that $\alpha=\beta=0$. As ...
Not sure if I understand correctly, but here's my viewpoint on where you go wrong. In the first step to obtain $z_0$, you check whether $g^z$ is a quadratic residue modulo $p$. You then reduce to $... View answer 1 answers 1 votes 344 views 1 votes As far as I know, answering my own question is considered okay. Here is my very computational approach: there is likely still a much more elegant solution. Suppose$(x,y)\in E_{\text{ns}}(L)$such ... View answer 1 answers 3 votes 921 views 0 votes Let$n=\#E(\mathbb{F}_q)$, which is finite because$\mathbb{F}_q$is. Then$E(\mathbb{F}_q)\subset E[n]\cong \mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$. View answer 1 answers 2 votes 583 views 0 votes Let's write$y\equiv g^x\bmod{p}$, so that the goal is to find$x\$. By exponentiating both sides we have the equalities \begin{align*} y^{33}\equiv (g^{33})^{x}\bmod{p}, \\ y^{22}\equiv (g^{22})^{x}\...