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Forum: Factoring 2019-10-09, 19:46
Replies: 1
Views: 289
Posted By baih
Condition on composite numbers easily factored (Challenger)

q prime numbre

p prime numbre
and q>p
lets c = qp


e= 2^p mod q

if we know e we can factore (c)
Forum: Factoring 2019-09-29, 15:05
Replies: 16
Views: 676
Posted By baih
my solution are :qp =c lets n is integer ...

my solution are :qp =c


lets n is integer

e= (c+1)/4

s= n^e Mod c

q= gcd(s-n,c)
Forum: Factoring 2019-09-28, 22:46
Replies: 16
Views: 676
Posted By baih
Smile yes i know but step by step:wink:

yes i know

but step by step:wink:
Forum: Factoring 2019-09-28, 20:02
Replies: 16
Views: 676
Posted By baih
purpose a large number c more than 1024BIT ...

purpose a large number c more than 1024BIT
with p and q also very large p and q (private key)


c is public
i can find pq from c
Forum: Factoring 2019-09-28, 16:44
Replies: 16
Views: 676
Posted By baih
Condition on composite numbers easily factored

Choose two large distinct prime numbers p and q

p = prime
q = prime

Compute c=pq


such that:
c=3 Mod 4
Forum: Soap Box 2019-08-14, 18:31
Replies: 71
Views: 4,945
Posted By baih
I discovered this (n+1)/2 years ago Maybe...

I discovered this (n+1)/2 years ago
Maybe others have also discovered it before (its very esy to notice it )
But No one will care about you because mathematics has become saturated and they do not...
Forum: Miscellaneous Math 2019-08-14, 13:54
Replies: 2
Views: 246
Posted By baih
lol

lol
Forum: Miscellaneous Math 2019-08-14, 13:24
Replies: 2
Views: 246
Posted By baih
Note that Every Even perfect numbers (except 6 ) are

Note that Every Even perfect numbers (except 6 ) are :


2pāˆ’1(2p āˆ’ 1) = 1 Mod (9*p) but not necessarily alternately.

p ======== perfect NUMBRE

3 ======== 28 ...
Forum: Homework Help 2019-08-07, 21:33
Replies: 1
Views: 1,103
Posted By baih
Question about Exponentiation modulaire

how to calculate pow(2,23571176971393,1523213.17)

I mean
pow(a,e,b)
with b non integer = 123213.17

Have a nice day. :)
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