New Largest Prime Found: Over 7 Million Digits

Jeff Gilchrist writes "On May 15, 2004, Josh Findley discovered the 41st known Mersenne Prime, 2 to the 24,036,583th power minus 1. The number is nearly a million digits larger than our last find and is now the largest known prime number! Josh's calculation took just over two weeks on his 2.4 GHz Pentium 4 computer. The new prime was verified by Tony Reix in just 5 days using only half the power of a Bull NovaScale 5000 HPC running Linux on 16 Itanium II 1.3 GHz CPUs. A second verification was completed by Jeff Gilchrist of Elytra Enterprises Inc. in Ottawa, Canada using eleven days of time on a HP rx5670 quad Itanium II 1.5 GHz CPU server at SHARCNET. Both verifications used Guillermo Ballester Valor's Glucas program." Read on for more on the discovery, including how you can help find more primes.

Gilchrist continues "If you want to see the number in written in decimal, Perfectly Scientific, Dr. Crandall's company which developed the FFT algorithm used by GIMPS, makes a poster you can order containing the entire number. It is kind of pricey because accurately printing an over-sized poster in 1-point font is not easy! Makes a cool present for the serious math nut in your family.

For more information, the press release is available.

Congratulations to Josh and every GIMPS contributor for their part in this remarkable find. You can download the client for your chance at finding the next world record prime! A forum for newcomers is available to answer any questions you may have.

GIMPS is closing in on the $100,000 Electronic Frontier Foundation award for the first 10-million-digit prime. The new prime is 72% of the size needed, however an award-winning prime could be mere weeks or as much as few years away - that's the fun of math discoveries, said GIMPS founder George Woltman. The GIMPS participant who discovers the prime will receive $50,000. Charity will get $25,000. The rest will be used primarily to fund more prime discoveries. In May 2000, a previous participant won the foundation's $50,000 award for discovering the first million-digit prime."

Comments

Not in this case...

[kevinatilusa]

since 2^(odd number)+1 is always a multiple of 3

Re:Not in this case...

[Anonymous Coward]

since 2^(odd number)+1 is always a multiple of 3

Theorem For any positive odd integer n, 3 divides 2^n+1

Proof We will use the Principal of Mathematical Induction.

Basis When n=1, we have 2^n+1=2^1+1=3. Furthermore, when n=3, we have 2^n+1=2^3+1=9.

Induction Now suppose n is a positive odd integer, and that 3 divides 2^n+1. We will now show that 3 divides 2^(n+2)+1.

Since 3 divides 2^n+1, there exists an integer q such that 2^n+1=3*q

2^(n+2)+1=2^(n+2)+4-3
=2^2*2^n+4-3
=4*(2^n+1)-3
=4*3*q-3
=3*(4*q-1)
=3*r, r=4*q-1

Where r is an integer by the closure properties of multiplication and subtraction.

QED

Re:Not in this case...

[chgros]

More directly (without induction):
if T = 2^(2p+1) + 1:
T = 2^(2p+1) - 2 [mod 3]
T = 2(2^2p - 1) [3]
T = 2(4^p - 1) [3]
T = 2(1^p - 1) [3]
T = 0 [3]
qed

hmmm

[challahc]

I guess these people trust the accuracy of these programs.

Personally I think someone should work this out on paper. Any volunteers/nominations?

Re:What kind of data structures are used?

[Anonymous Coward]

A Fortran vector could handle it, I guess.

Re:What kind of data structures are used?

[Anonymous Coward]

You could use the GMP library, but it's GNU code so you might want to be careful about licensing.

For a real free bignum library, try LibTomMath. It was written by a guy named Tom.

Yeah, but

[Anonymous Coward]

EVERYONE will think of that.. "oh I'll just make my encryption key the largest known prime"... it's like setting 12345 on your luggage

Re:I hate to be a pushover...

[Anonymous Coward]

Nothing. And we all die, too. Might as well kill yourself right now, avoid all the unnecessary hassle of life. Ya know.

Verification

[timeOday]

Both verifications used Guillermo Ballester Valor's Glucas program

Wouldn't it be more prudent to do verification with various algorithms?

Want a simple proof?

[product byproduct]

2^(odd number)+1
= (-1)^(odd number)+1 [mod 3]
= -1 + 1 [mod 3]
= 0 [mod 3]

Slashdot Earns It's Name

[POLAX]

This story is perhaps the most pure example of "News for Nerds. Stuff that matters."

I love it! :- )

Re:I hate to be a pushover...

[jayhawk88]

They're not useful now, but in another 200 years when we're all carrying around pocket quantum machines it may be useful.

Re:Persuit of uselessness != profit

[eidechse]

So, just how long have you been waiting for an excuse (no matter how flimsy) to post that rant?

I ask because your tirade, although vigorous and interesting, is entirely unrelated to my post.

Since you seem to be articulate and well read I'm giving you the benefit of the doubt and assuming that you have some sort of agenda...

What is it exactly?

Re:I hate to be a pushover...

[ComaVN]

Don't say that out loud on a star trek convention.