Oh, I see.  What about the idea to use boinc to study the relationships between semiprimes and their factors?

Is it possible? (I mean I know that it is, but could WE do it?)

Yes, Matthias, what method did you use?

Oh, and ninja'd in my last post.

Ah that would be the reason.  Mathias what language did you use to do this? 

If it was Basic i would have definetaly wanted to see *that* code

Nah ignore my post, I took a entirely wrong aproach. It was asm btw. I am now studying Euclides algorithm. Shouldn't be too hard to write a asm/vb/c program for. I only have trouble with the last step of the RSA encryption/decryption. I dont understand mod(). Well hopefully someone can explain in layman's language. What I did in my program was just brute forcing every possibilty until a match was found. That match however wasn't encrypted at all. Sorry for the false excitiment. If you don't get my rough explanation of my program just leave it be, it didn't do the trick anyway. I really want to crack this code!

eternal fame awaits

edit:

If it was Basic i would have definetaly wanted to see *that* code

Input "Secret Code:", A
Disp "Code cracked!", "isn't it:",A

somtimes things are really that easy

@graphmastur: I will try to explain the algorithm:

Shanks-Tonelli modular square root algorithm:
The problem: given n and p, with p prime, find x such that x^2 = n (mod p)
First, a digression. The Legendre symbol (n|p) (actually written in a different manner, but the forum does not support typesetting) is defined as follows: (n|p) = 1 if such an x exists, 0 if p divides n, and -1 otherwise. It can be computed as n^((p-1)/2) mod p.
Now we return to the algorithm:
Step 1: Let p - 1 = 2^s * q , with q odd. This can be done by dividing powers of 2 out of p - 1 until we cannot do so any further.
Step 2: Select z such that (z|p) = -1. We do this simply by randomly selecting z in the set {1, 2, ... p - 1} and computing (z|p) as described above.
Step 3: Let:
                    c = z^q mod p
                    r = n^((q + 1) / 2) mod p
                    t = n^q mod p
                    m = s
Note that all of these values can be computed rapidly using the repeated squaring algorithm.
Step 4: We execute the following loop:
                    While t != 1
                       i = 0
                       tmp = t mod p
                       While tmp != 1
                         tmp = tmp * tmp mod p
                         i = i + 1
                       EndWhile
                       b = c
                       For k = 1 to m - i - 1
                         b  = b * b mod p
                       EndFor
                       r = r * b mod p
                       t = t * b^2 mod p
                       c = b^2 mod p
                       m = i
                    EndWhile
Step 5: Return r

1024 bits, on the other hand, goes up to 5e30, or 5 nonillion.

Actually, ~1.8e308, with factors up to ~1.3e154.

e is fixed, 17 for the 512-bit RSA keys of TI-Z80 and TI-68k calcs and 65537 for the 1024-bit RSA keys of Nspire calcs.