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 Showing results 1 to 25 of 84 Search took 0.02 seconds. Search: Posts Made By: mgb
 Forum: Lounge 2022-10-13, 11:25 Replies: 4 Views: 608 Posted By mgb Thanks for your answers. It seems there cannot... Thanks for your answers. It seems there cannot be any pattern, for the following reasons- Let p be a prime < 2^n Let z = 2^n - p so z = -p (mod 2^n) pz^{-1} \equiv -1 (mod 2^n), so pz^{-1}...
 Forum: Lounge 2022-10-12, 10:28 Replies: 4 Views: 608 Posted By mgb k*2^n - 1 Anyone know if there is anything I can read up on concerning the divisors of numbers of the form k2^n - 1 or general information on numbers of this form?
 Forum: Factoring 2021-03-28, 13:58 Replies: 9 Views: 4,355 Posted By mgb It seems you have to calculate these x values one... It seems you have to calculate these x values one by one as there is no known formula for 'looking ahead' with such a random algorithm. This might interest you...
 2021-03-28, 12:18 Replies: 3 Views: 5,559 Posted By mgb Yes. It follows then that the roots are u =... Yes. It follows then that the roots are u = 2-1(a + a-1r)(mod p) v = 2-1(a - a-1r)(mod p) for u^2 - v^2 = q1 - q2 = r (mod p)
 Forum: Miscellaneous Math 2020-06-18, 11:03 Replies: 4 Views: 2,622 Posted By mgb Sorry about that. Here you go- The function... Sorry about that. Here you go- The function ai+1 = (ai^2 + 1) (mod n) is typical of the kind used in Pollard's Rho Method. The sequence is divided into two parts. The 'tail' and the 'cycle'. But...
 Forum: Miscellaneous Math 2020-06-17, 21:11 Replies: 4 Views: 2,622 Posted By mgb Some Notes on Pollard's Rho I have been examining the output from Pollard's method for different seeds and an interesting picture has emerged which you might have use for. Comments welcome as always. ...
 Forum: Lounge 2020-06-12, 21:54 Replies: 10 Views: 2,442 Posted By mgb Our science teacher once showed us how phosphorus... Our science teacher once showed us how phosphorus fizzes in water. Then we got a new science teacher and we asked him to do the phosphorus trick for us. He put in WAY too much and it exploded with...
 Forum: Lounge 2020-06-12, 21:28 Replies: 10 Views: 2,442 Posted By mgb LOL Great analogy. LOL Great analogy.
 Forum: Lounge 2020-06-12, 21:23 Replies: 10 Views: 2,442 Posted By mgb You caught all the essential points. When I read... You caught all the essential points. When I read it first I thought this guy was really on to something but then I thought (p + q)/N? Where did he get p and q? He's pulling rabbits out of hats here....
 Forum: Lounge 2020-06-12, 19:04 Replies: 10 Views: 2,442 Posted By mgb Interesting idea on factoring This is interesting but how does he find Nf(N)? https://mae.ufl.edu/~uhk/QUICK-SEMI-PRIME-FACTORING.pdf
 Forum: Miscellaneous Math 2020-05-20, 20:33 Replies: 31 Views: 19,147 Posted By mgb See 'Safe Primes' and Sophie Germain primes:... See 'Safe Primes' and Sophie Germain primes: https://en.wikipedia.org/wiki/Safe_prime
 2020-02-02, 20:24 Replies: 3 Views: 5,559 Posted By mgb Yes, I noticed that it works for odd sums as... Yes, I noticed that it works for odd sums as well. Yes, this would be equivalent to writing v = (a - a-1c)/2 as v = (a - a-1R)/2 giving a distance of R between q1 and q2 Many thanks for...
 2020-02-01, 18:44 Replies: 3 Views: 5,559 Posted By mgb Creating Consecutive Quadratic Residues While working on another problem I stumbled across this if anyone has a use for it. To create all consecutive quadratic residues modulo p, p an odd prime. For all a < p find a-1(mod p) If (a...
 2019-12-23, 22:21 Replies: 28 Views: 14,780 Posted By mgb gcd([sqrt(n)]^2 - t^2, n) for t = 1, 2, 3,... is... gcd([sqrt(n)]^2 - t^2, n) for t = 1, 2, 3,... is faster than Fermat's. [sqrt(n)] = t (mod p)
 2019-10-26, 19:21 Replies: 2 Views: 4,261 Posted By mgb Yes but at step 2 the task is reduced to a... Yes but at step 2 the task is reduced to a smaller pair of numbers, to which the extended Euclidean algorithm can be applied. This is an immediate saving in processor time. Step 2 = Find r -1 mod a...
 2019-10-25, 19:48 Replies: 2 Views: 4,261 Posted By mgb Calculating inverses quickly. Do you think this would be faster than the extended Euclidean Algorithm for finding inverses? For a given a find a-1 (mod m) 1. Let m = r mod a 2. Find r -1 mod a 3. Let x = a - r -1 (ie, x =...
 2019-08-05, 20:02 Replies: 45 Views: 19,833 Posted By mgb Yes, it is hard to see how Brent's ideas can... Yes, it is hard to see how Brent's ideas can apply. At present I can't see how I can find a way to make it leap forward, ie shorten the cycle. I'll work some more on it as I understand it better now...
 2019-08-04, 10:26 Replies: 45 Views: 19,833 Posted By mgb Many thanks for this. I had been looking for... Many thanks for this. I had been looking for something similar. I have found, with this version, that there is no lead in (no tail). The cycle begins at the start. I can see this because if a factor...
 2019-08-03, 16:18 Replies: 45 Views: 19,833 Posted By mgb Seemingly all this is quite general and... Seemingly all this is quite general and independent of z2 That is, if we choose any R in the period [1, n - 1] and let Q = n - R then Q = -R (mod n) and therefore mod p and mod q That is, if R...
 2019-07-21, 21:10 Replies: 45 Views: 19,833 Posted By mgb Yes I do but I'm not too worried until I can get... Yes I do but I'm not too worried until I can get it up to speed. Since m = p + z, m2 = z2 mod p and I'm trying to find a way to split z2 off m2. I thought using rho might do this but it just keeps...
 Forum: PARI/GP 2019-07-21, 20:47 Replies: 0 Views: 3,853 Posted By mgb Pari link Don't know if this Pari documentation has been posted but it is very extensive- https://buildmedia.readthedocs.org/media/pdf/cypari2/latest/cypari2.pdf
 2019-07-21, 20:28 Replies: 45 Views: 19,833 Posted By mgb Thanks, but how do I 'backtrack'? Nor sure what... Thanks, but how do I 'backtrack'? Nor sure what you mean.
 2019-07-21, 19:27 Replies: 45 Views: 19,833 Posted By mgb Well well...I finally got it to go as fast as the... Well well...I finally got it to go as fast as the traditional rho. Here is the code. r = n%m; R = r; Q = r; for(i = 1, Lim, R = (R^2 + r)%n; R = (R^2 + r)%n; Q = (Q^2 + r)%n; ...
 2019-07-21, 13:26 Replies: 45 Views: 19,833 Posted By mgb You are quite right. There doesn't seem to be a... You are quite right. There doesn't seem to be a difference. I thought the different ways of writing z2 would make a difference...
 2019-07-21, 10:25 Replies: 45 Views: 19,833 Posted By mgb That is, for some c and d I want (cp - zs) - (dp... That is, for some c and d I want (cp - zs) - (dp - zs) = kp for some k. The idea is to get a match at zs Alternatively (cp - hz2) - (dp - hz2)
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