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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/<PID>/status and/or...

try reading /proc/<PID>/status and/or /proc/<PID>/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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