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No, it doesn't need a server, and runs on almost every platform imaginable. You'll be able to specify your range, and claim it here.

Expect 500+ guests if they link to omni.This should be fun.

We've already had lots of talk about factoring the key, in another topic (but it's understandable that you missed it: there's lots of activity on the forum, and that is a good thing): http://ourl.ca/6236 I have summarized multiple times how impractical the factorization is (unless some ground-breaking algorithm comes to the rescue, which we should not hold our breath on): basically, three orders of magnitude harder than the state of the art.A linear extrapolation of the figures given in the paper detailing the factorization of RSA-768, consistent with what we know of 512-bit RSA factoring, gives the need of sifting through 10000-100000 TB (yes, I really mean terabytes) of data, after those have been produced by several dozen thousands of computers running for years (or ten times as many computers, running for one tenth of the time).This does not, however, mean that we can't spend up to several CPU-years trying to find a factor by Trial Factoring (beyond that would not be reasonable): it's extremely unlikely that we'll succeed by sifting through a search space which represents an vanishingly small part of a particle, compared to the whole universe - but such is the beauty of random.

Best case: It'll be solved by tomorrow.Worst case: The sun will have died by the time the solution is found using this method.

:PQ→N:1→F:int(√(N→X:{0→L1:L1→L2:For(θ,1,2:While (fPart(N/X)>fPart(N/(X+1)))+(θ=2:If X=Q-1:Return:X-1→X:F+1→F:If θ=2:Then:If X>0:Then:If fPart(N/X)<fPart(N/(X+1:Then:Disp {X,F:If L1(1)=0:Then:{X→L1:{F→L2:Else:X→L1(dim(L1)+1:F→L2(dim(L2)+1:End:X-1→X:1→F:End:End:End:End:X-1→X:End

I've been trying to find some function that could be used to find the next number that *might* be a factor in a semiprime. I've been using this program to get my test data, I can't really explain what it's doing so if someone with more math skill than me and can understand this program, if you would I would appreciate it Code: [Select]:PQ→N:1→F:int(√(N→X:{0→L1:L1→L2:For(θ,1,2:While (fPart(N/X)>fPart(N/(X+1)))+(θ=2:If X=Q-1:Return:X-1→X:F+1→F:If θ=2:Then:If X>0:Then:If fPart(N/X)<fPart(N/(X+1:Then:Disp {X,F:If L1(1)=0:Then:{X→L1:{F→L2:Else:X→L1(dim(L1)+1:F→L2(dim(L2)+1:End:X-1→X:1→F:End:End:End:End:X-1→X:End(not very effective because of the limited memory on the calculator, have to find almost a perfect semiprime to work with. Might try whipping up a C program that can be used to get/graph the data)EDIT: Right now this program assumes Q is the smaller of the two primes. Just add '(P>Q)Q+(Q>P)P→L' at the beginning and change 'If X=Q-1:Return' to 'If X=L-1:Return'.

EDIT: Right now this program assumes Q is the smaller of the two primes. Just add '(P>Q)Q+(Q>P)P→L' at the beginning and change 'If X=Q-1:Return' to 'If X=L-1:Return'.

Quote from: Tribal on March 07, 2011, 05:34:00 pmEDIT: Right now this program assumes Q is the smaller of the two primes. Just add '(P>Q)Q+(Q>P)P→L' at the beginning and change 'If X=Q-1:Return' to 'If X=L-1:Return'.usually P is used as the smaller. And if you already know P or Q, why do we need to find it?