# Tag Info

4

There are various ways to do this, but I will use the method you show. We are given the elliptic curve $$x^3+17x+5 \pmod{59}$$ We are asked to find $8P$ for the point $P = (4,14)$. I will do one and you can continue. We have: \lambda = \dfrac{3 x_1^2 + A}{2 y_1} = \dfrac{3 \times 4^2 + 17}{2 \times 14} = \dfrac{65}{28} = 65 \times 28^{-1} ... 3 You are right. The computation is just too inefficient. The best known attack on ECDLP, the pollard rho attack, would be useless against elliptic curves over the rationals. Consider this, if you were to do the computations over a finite field of say 512-bits, you will only have to deal with 512-bit intermediate values along the way. Considering the same ... 2 If f_P(S) is a function with divisor m[P]-m[\mathcal{O}], then f_P has a zero at P and a pole at \mathcal{O}. Then, for a fixed constant T, the function h_{P,T}(S)=f_P(T+S) is simply a translation of f_P by T. Thus, h_{P,T} has a zero when T+S=P, i.e., at S=P-T and a pole at T+S=\mathcal{O}, i.e., at S=-T. Moreover, the ... 2 Just compute the powers: \begin{align} 2^2&\equiv_{13}4 &2^3&\equiv_{13}8 & 2^4=4^2&\equiv_{13}16\equiv_{13}3& 2^5&\equiv_{13}6&2^6&\equiv_{13}12\equiv_{13}-1\\ 3^2&\equiv_{13}9\equiv_{13}-4 &3^3&\equiv_{13}-12\equiv_{13}1\\ 5^2&\equiv_{13}-1\\ ... 2 Based on Moore's Law, RSA predicts that 1024 bit keys will be broken in the near future and 2048 bit keys are sufficient until 2030. A 3072 bit key is about equivalent to a 128 bit symmetric key, which should not be broken for a very long time (current technology would take 10^{18} years to crack). 2 Ifn=\prod_{p\,{\rm prime}}p^{\alpha(p)}\ ,$$then you need$$\prod_{p\,{\rm prime}}p^{\alpha(p)-1}(p-1)=110\ . Since $11\mid110$ you must have $p=11$ as one of the factors on the LHS. (It can't be $p-1=11$ as $12$ is not prime.). Then the exponent must be $\alpha(11)-1=1$ since $11^2\not\mid110$, and the $p-1$ factor is $10$. This accounts for all ...

2

http://en.wikipedia.org/wiki/Cantor%E2%80%93Zassenhaus_algorithm with $x^3-a.........................$ Notice that there is one cube root when $p \equiv 2 \pmod 3,$ and either three cube roots or none when $p \equiv 1 \pmod 3.$ For example, $2$ has three cube roots $\pmod p$ when $p = u^2 + 27 v^2$ in integers, otherwise none (for $p \equiv 1 \pmod 3,$ in ...

1

Theorem: Let #$E(Fq) = 1 - a + q$ Write $X_2 − aX + q = (X − α)(X − β)$. Then # $E(F_{q^n}) = 1 − (α^n + β^n) + q^n$ for all n ≥ 1. Now for the problem: It is easy. Write $x^2 + 2$ = $(x+i \sqrt2)(x-i \sqrt2)$ and so #$E(F_{2^n})=$ $2^n+1 -(x+i \sqrt2)(x-i \sqrt2)$ and from there it is then easy to use a phase argument to duduce the answer. Note: ...

1

RSA-768 has been factored and currently most systems are using at least 1024 bits. For example, Microsoft has an update for minimum certificate key length in 2012, requiring keys be at least 1024 bits. As of now, 1024 bits is still considered secure. How secure? An old estimate for the cost of factoring 1024 bits is about US50 million for just one of the ...

1

I am assuming you mean strictly doing everything in $\Bbb Q$, including eventual transmission. It is not sensible to do computations in $\Bbb Q$ and reducing to $\mathbb F_p$ before sending out since that is harder to do and you do not gain anything. The summary of the following discussion is that you cannot use any generated points to keep secrets. ...

Only top voted, non community-wiki answers of a minimum length are eligible