marcookker

3 ms·

1.
by marcookker 5y ago
"6469693230": 6469693230 % 163123711 = 107868501 "%" is modulo in software it means 6469693230 = 107868501 MOD 163123711. So basicly you find out that 6469693230 - 163123711 * k = 107868501. so you get: (a+c) * a - (b+c)
2.
by marcookker 5y ago
The normal approach would be the following: a = ceil(sqrt(N)) = 12772 b2 = (12772+1)*(12772) - N = 13045 <- factorize = 5*2609 <- does not fit in (b+1)*(b) b2 = (12773+1)*(12773) - N = 38591 <- factorize = 149*259 <- does not fi
3.
by marcookker 5y ago
Say we want to factor 163123711, N=163123711. Before we start the algorithm, we have to choose a number to start our algorithm with. A good pick for the first number that we want to start our algorithm with, is a number that is higher than
4.
by marcookker 5y ago
Imagine you start a number that has many possible c's, for example 2*3*5*7*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71 = 557940830126698960967415390 after that you do 557940830126698960967415390 MOD 1649. The number you are left w
5.
by marcookker 5y ago
If you want to factorize 1649 for example, and we want to use c=1. What you do is you squareroot it first. The squareroot of 1649 is around 41. than you do (41+1)*41 = 73 MOD 1649. After that you find the factors of 73, 1 * 73= 73. Than you
6.
by marcookker 5y ago
Okay, I am not sure if this answers your question, but I discover the methods the following way: imagine 5*5 - 2*2: 00000 00000 00000 000 000 Now imagine (5+1)*5 - (2+1)*2: 000000 000000 000000 000 000 Now imagine (5+2)*5 - (2+2)*2: 000000
7.
by marcookker 5y ago
When analysing fermat's integer factorization method, I found 3 new factorization methods, with rectangles, cubes and cuboids.
8.
Integer factorization with rectangles, cubes and cuboids(en.wikipedia.org)
2 points by marcookker 5y ago|15 comments