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 Showing results 1 to 15 of 15 Search took 0.01 seconds. Search: Posts Made By: mpenguin
 Forum: Factoring 2005-09-29, 07:46 Replies: 10 Views: 2,203 Posted By mpenguin Yes, the square root of 2 is trivial mod Fn. ... Yes, the square root of 2 is trivial mod Fn. (2+2^(1/2))^2 and (2+(2+2^(1/2))^(1/2))^(1/2) mod Fn are a little more subtle.
 Forum: Math 2005-09-22, 14:19 Replies: 18 Views: 712 Posted By mpenguin The algorithm was extended to either solving ... The algorithm was extended to either solving A) x^2 = +/- k mod N where N=a^((2^m)*k)+1 or N=a^((2^m)*k)-1 with k odd, + or - depending on k mod 4
 Forum: Math 2005-09-21, 15:36 Replies: 18 Views: 712 Posted By mpenguin We don't care about fame much. What... We don't care about fame much. What surprised me that +/- p is guaranteed to be square residue (+ or - depending on p mod 4) for all p up to about 4000. Is there an analytic explanation?
 Forum: Math 2005-09-21, 14:55 Replies: 18 Views: 712 Posted By mpenguin Note that the algorithm does NOT find ARBITRARY... Note that the algorithm does NOT find ARBITRARY square roots, just a square root of +/- p mod 2^p-1
 Forum: Math 2005-09-21, 14:52 Replies: 18 Views: 712 Posted By mpenguin By definition: N=a^k-1; For Mersenne numbers... By definition: N=a^k-1; For Mersenne numbers the value of "a" is 2. Can't edit the original post, but t0=tanh(x/2); So: a=sinh(x)+cosh(x)=e^x Let t0=tanh(x/2) then:
 Forum: Math 2005-09-21, 14:17 Replies: 18 Views: 712 Posted By mpenguin I believe my algorithm works for composite... I believe my algorithm works for composite numbers with unknown factorization. Will try to explain. Lets try to do some trigonometry mod N. All variables mean tan(a/2) and are in Zn. So some...
 Forum: Math 2005-09-21, 13:35 Replies: 18 Views: 712 Posted By mpenguin There is an efficient solution to: x^2 = k... There is an efficient solution to: x^2 = k mod (a^k-1) or x^2 = -k mod (a^k-1) with k odd. The estimated running time is loop from 1 to (k-1)/2.
 Forum: Math 2005-09-21, 13:26 Replies: 18 Views: 712 Posted By mpenguin Actually i was asking about solving: x^2 =... Actually i was asking about solving: x^2 = 2969 mod (2^2969-1) Code will follow.
 Forum: Programming 2005-09-21, 09:39 Replies: 12 Views: 4,670 Posted By mpenguin try reading /proc//status and/or... try reading /proc//status and/or /proc//statm or /proc/self/status /proc/self/statm
 Forum: Math 2005-09-21, 07:28 Replies: 18 Views: 712 Posted By mpenguin x^2 = n mod Mn If Mn=2^n-1, n prime is there known efficient algorithm to compute: x^2 = n mod Mn if n mod 4 =1 or x^2 = -n mod Mn if n mod 4 = 3 (Mn is with unknown factorization). ? Thanks.
 Forum: Factoring 2005-08-21, 16:11 Replies: 10 Views: 2,203 Posted By mpenguin sqrt(2+sqrt(2+sqrt(2+sqrt(2+....sqrt(2))) are... sqrt(2+sqrt(2+sqrt(2+sqrt(2+....sqrt(2))) are computable from Z and iterating x -> x^2 -2 may reach zero in n+1 steps. mupad code: fn:=proc(n) begin 2^(2^n)+1;end_proc; n0:=7; n:=fn(n0);...
 Forum: Factoring 2005-08-19, 16:05 Replies: 10 Views: 2,203 Posted By mpenguin sqrt(2) follows from: fn(n) =... sqrt(2) follows from: fn(n) = fn(n-1)^2-2*(fn(n-2)-1)^2 fn(n) is the nth Fermat number
 Forum: Factoring 2005-08-18, 16:14 Replies: 10 Views: 2,203 Posted By mpenguin When one takes the smaller absolute value of sq2... When one takes the smaller absolute value of sq2 and sq2a ("signed mod") they have strange factorizations F[n] | sq2[n+i] and F[n] | sq2a[n+j].
 Forum: Factoring 2005-08-18, 15:23 Replies: 10 Views: 2,203 Posted By mpenguin I am theory lamer, but according to wikipedia: ... I am theory lamer, but according to wikipedia: >Let n ≥ 3 be a positive odd integer. Then n is a Fermat prime if and only if for every a >coprime to n, a is a primitive root mod n if and only if a...
 Forum: Factoring 2005-08-18, 12:18 Replies: 10 Views: 2,203 Posted By mpenguin Closed form solution of x^2 = 2 mod Fermat number Sorry if this is known, but didn't find it on the net. The solution of x^2 = I in C leads to solution to x^2 = 2 mod (2^(2^n)+1). It is also possible to compute sqrt(2+sqrt(2)) mod (2^(2^n)+1) and...
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