On May 12, 2017, at 20:52 , Venkataraman S wrote:
> I vaguely remember that if one can quickly find a quadratic non-residue, one
> can find a primitive root fast. I don't remember the exact connection now.
> Does anybody in the group have any reference?
Internet search can be helpful in times
I vaguely remember that if one can quickly find a quadratic non-residue, one
can find a primitive root fast. I don't remember the exact connection now. Does
anybody in the group have any reference?
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One way or the other, the bottleneck is in the primitivity test.
On Friday, May 12, 2017 at 4:36:20 AM UTC+1, Venkataraman S wrote:
>
> The German school thinks differently. There is a different (well known)
> algorithm due to Gauss. Take an arbitrary number a coprime to p. Find its
> order. If
The German school thinks differently. There is a different (well known)
algorithm due to Gauss. Take an arbitrary number a coprime to p. Find its
order. If it is the primitive root, we are done. If not, choose another b and
check whether order of ab is greater than the order of a. If it is, repl
> Not really: generators of the additive group are coprime to p, not to p-1.
>
> Perhaps Johan was thinking of the fact that if g is one multiplicative
> generator (aka primitive root) then g^k is another if and only if
> gcd(k,p-1)=1.
I think I should just not answer sage-support questions before
> "primitive element" is meant as "generator for the multiplicative group
> GF(p)^*" and not the additive group GF(p). The OP question is about the
> former and Johan answer is about the latter.
Yes, I just realised that too - sorry about the noise. I'll think more
about the multiplicative group
On 11 May 2017 at 08:16, Vincent Delecroix <20100.delecr...@gmail.com> wrote:
> Hi,
>
> "primitive element" is meant as "generator for the multiplicative group
> GF(p)^*" and not the additive group GF(p). The OP question is about the
> former and Johan answer is about the latter.
Not really: gener
Hi,
"primitive element" is meant as "generator for the multiplicative group
GF(p)^*" and not the additive group GF(p). The OP question is about the
former and Johan answer is about the latter.
For very large p such as what you asked for is likely to be delicate
(but I am not a specialist).
Hi Panos
In GF(p) then an element g is primitive if its embedding into ZZ is
coprime with p-1. Since Euclidean algorithm is so fast, you can test
this:
sage: p = Primes().next(2^2048) #long
sage: g1 = 3
sage: gcd(g1, p-1)
3
sage: g2 = 5
sage: gcd(g2, p-1)
1
So 3 is not a primitive element in GF(
Hello everyone,
I am trying to calculate a primitive element (g) of a big Finite Field:
GF(p) where p is prime number > 2^2048
So then, i could share a secret integer (r) as: m=g^r, but it seems
impossible to calculate it with function primitive_element()
Is there another way i can use to calcu
On 4 May 2014 13:20, Jan Medina wrote:
> But the log() function works on a finite field too. So its not correct if a
> i use the log on finite fields?
OK -- you did not explain at all what is was that you were doing, na d
I could not guess!
>
> I wan to calculate log(\theta+i,\theta) for i in a
But the log() function works on a finite field too. So its not correct if a
i use the log on finite fields?
I wan to calculate log(\theta+i,\theta) for i in a finte field and theta a
primtive element
2014-05-04 6:07 GMT-05:00 John Cremona :
> log() is a function you would apply to numbers, say
log() is a function you would apply to numbers, say real or complex:
sage: a = RealField(100)(2)
sage: a.log()
0.69314718055994530941723212146
while discrete_log() is something to do in a finite cyclic group. For example:
sage: F = FiniteField(101)
sage: a = F(2)
sage: b = a^67
sage: discrete_l
Hi everybody.
I want to know what algorithm are implemented for calculate log() and
discrete log(). and what are the differences?
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sage: var('x,y')
(x, y)
sage: E = EllipticCurve(y^2 == x^3 - 36*x)
sage: P=E(-3,9)
sage: Q=E(12,36)
sage: discrete_log(Q, P, operation='+', bounds=(0,100))
---
ValueErrorTraceback (most recent c
Hi Dears,
I have the elliptic curve Y^2=X^3-36X and P=(-3,9) as the its
generator. Q=(12,36) is the other point on this curve. I would like to
solve Discrete Logarithm but I do not know.
Please tell me how I can find the "n" which nQ=p.
Best,
Raman
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