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New Largest Known Prime Number: 2^57,885,161-1

timothy posted about a year and a half ago | from the sorry-won't-fit-on-vanity-plate dept.

Math 254

An anonymous reader writes with news from Mersenne.org, home of the Great Internet Mersenne Prime Search: "On January 25th at 23:30:26 UTC, the largest known prime number, 257,885,161-1, was discovered on GIMPS volunteer Curtis Cooper's computer. The new prime number, 2 multiplied by itself 57,885,161 times, less one, has 17,425,170 digits. With 360,000 CPUs peaking at 150 trillion calculations per second, GIMPS — now in its 17th year — is the longest continuously-running global 'grassroots supercomputing' project in Internet history."

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Wrong (5, Informative)

Anonymous Coward | about a year and a half ago | (#42798277)

Actually it would be 2 multiplied by itself 57,885,160 times, minus 1.

Re:Wrong (3, Interesting)

Anonymous Coward | about a year and a half ago | (#42798879)

I'm not trying to be inflammatory here... this is a genuine question.

Why? I mean... why bother with calculating this.. What has it added to study if prime numbers? And if it's added to the study of primes... then what use is that?

Re:Wrong (5, Informative)

chalsall (185) | about a year and a half ago | (#42799015)

As is well known, there is no direct mathematical benefit from finding these primes.

It is, however, a very useful "driving problem" to developing new algorithms, software, and distributed computing infrastructure which have wide ranging real-world applications.

Check out the Mersenne Forum [mersenneforum.org] where all types of interesting mathematical, software and computer issues are discussed.

Re:Wrong (1)

kaizendojo (956951) | about a year and a half ago | (#42799033)

+1. I was wondering the exact same thing; is this a "because it's there" type of thing or is there an actual use to calculating this number?

Re:Wrong (5, Insightful)

Anonymous Coward | about a year and a half ago | (#42799093)

Your first question: "What has it added to the study of prime numbers?", I'm not sure but...

Your second question: "What use is that? (the study of prime numbers)"

Well... Nearly all modern public key cryptography (SSL / TLS / SSH etc.) is based on the asymmetry in time between the multiplication of two prime numbers (very fast operation) and the factorization of the result back into these two primes (very very slow: the goal being to make so slow that it become impractical to do).

Actually, it's "worse" than that: the "proof" that most modern PKCS crypto works is: "It's hard to find the (prime) factors of the product of two primes".

In other words: the study of prime numbers is one of the most important area of mathematics when it comes to computer security.

Re:Wrong (So many Fucking Experts) (-1)

Anonymous Coward | about a year and a half ago | (#42798993)

Math Nazi's

Re:Wrong (3, Funny)

houghi (78078) | about a year and a half ago | (#42799075)

That would be 3:
2 multiplied by itself is 4. No matter if you do it once or 57,885,160 times. Then minus one.
Or it could be 4 if you only do it 57,885,159 times. But I think 4 is not a prime number.

Uhhh... (-1, Troll)

The MAZZTer (911996) | about a year and a half ago | (#42798295)

Considering any number 2^n-1 is prime [wikipedia.org] , this isn't too impressive, except for maybe that they bothered to expand it into a full number.

Oh look! I've discovered a new, larger prime! 2^57,885,162-1 What do I win?

Re:Uhhh... (5, Informative)

SuricouRaven (1897204) | about a year and a half ago | (#42798353)

2^4-1 = 16-1 = 15.

5 * 3 = 15.

Go read it again.

Re:Uhhh... (2)

SuricouRaven (1897204) | about a year and a half ago | (#42798423)

I see a problem though: According to wikipedia, "Some definitions of Mersenne numbers require that the exponent n be prime."

Some? Huh? Mathematics does not deal in disputed definitions! I've never heard before that Mersenne numbers need a prime exponent.

Re:Uhhh... (3, Informative)

Anonymous Coward | about a year and a half ago | (#42798509)

It's really just a matter of semantics. If n is composite, then 2^n - 1 cannot be prime.

Re:Uhhh... (4, Informative)

fredprado (2569351) | about a year and a half ago | (#42798537)

Mathematics does deal with a lot of "disputed" definitions. Mathematics deal even with a lot of "disputed" logic and "disputed" interpretations. Read about the axiom of choice, set Theory in general, Constructivism (mathematics) and Finitism and you will understand that things get quite more complicated than you thought.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798549)

Notation is one of the most hotly debated things in mathematics because you cannot write proofs that one system of notation is better than another!

Re:Uhhh... (4, Funny)

NoNonAlphaCharsHere (2201864) | about a year and a half ago | (#42798749)

Mathematics may not deal in disputed definitions, but Wikipedia certainly does.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798789)

AFAIK humans like to disagree on the most basic things (lets call it bikesheding) and from my limited exposure to math papers i can only assume that every mathematician has his own definition of any mathematical concept, which they happily define the first 2/3 of said paper in the most mindnumbing wway possible.

Re:Uhhh... (3, Informative)

happylight (600739) | about a year and a half ago | (#42798355)

You might want to read that link again. Not any number 2^n-1 is prime. Just that a prime in the form of 2^n-1 is called a Mersenne prime.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798539)

Any binary number that consists of a prime number of 1's is going to be prime. There's no way of factoring them by subdividing them evenly. You would need two numbers, one of which is like a set of fenceposts and another which is the actual bit of fence: 11111111 = 01010101 x 11

Re:Uhhh... (5, Insightful)

Anonymous Coward | about a year and a half ago | (#42798773)

111 1111 1111 == 2047 == 23 * 89

Funny how many assertions here that number disproves

Re:Uhhh... (1)

ircmaxell (1117387) | about a year and a half ago | (#42798979)

Definitely not true:

110

That has 2 1s, which is prime (2 is prime), but 6 definitely is NOT prime...

Likewise,

1001

That has 2 1s, which is prime, but 9 definitely is NOT prime...

Re:Uhhh... (1)

Aardpig (622459) | about a year and a half ago | (#42798357)

n=4: 2^4-1 = 15, which isn't prime.

You get nothing! You lose, GOOD DAY SIR!

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798737)

http://bit.ly/VW0mFj -- GOOD DAY SIR

Re:Uhhh... (1)

superdave80 (1226592) | about a year and a half ago | (#42799021)

I said GOOD DAY!

Read the article (1)

schneidafunk (795759) | about a year and a half ago | (#42798379)

You didn't read your link did you? There is no equation for prime numbers right now. There are some 'hacks' that work up to high prime numbers but no existing formula [wikipedia.org] .

Re:Uhhh... (1)

Skapare (16644) | about a year and a half ago | (#42798381)

Not all Mersenne numbers are prime. Consider 2^4-1.

Re:Uhhh... (1)

mark-t (151149) | about a year and a half ago | (#42798935)

Doesn't a mersenne prime require that the exponent also be prime? As 2^n-1, where n is composite is *never* prime, as far as I am aware.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798383)

Counterexample: 15.

Re:Uhhh... (2)

Shimbo (100005) | about a year and a half ago | (#42798393)

Sigh, you need to go back and read the link you posted:

"Though it was believed by early mathematicians that Mp is prime for all primes p, Mp is very rarely prime."

Re:Uhhh... (1)

Mateo_LeFou (859634) | about a year and a half ago | (#42798395)

There must be a lot of prime numbers in your world
2^n-1 is always prime?

n=4
(2**4)-1 = 15

Re:Uhhh... (1)

Mente (219525) | about a year and a half ago | (#42798403)

No. Any number 2^n-1 is not prime. It is only prime when n is also prime.
  Mp=2^p - 1 is prime where p is prime

Re:Uhhh... (1)

Skapare (16644) | about a year and a half ago | (#42798501)

... except when it isn't ... like 2^11-1.

Re:Uhhh... (1)

canadiannomad (1745008) | about a year and a half ago | (#42798535)

Now if that were true wouldn't we be able to say the largest known prime is:
2^((2^57,885,161)-1) ? That can't be true or they would have found it already.

Re:Uhhh... (2)

canadiannomad (1745008) | about a year and a half ago | (#42798671)

Though it was believed by early mathematicians that Mp is prime for all primes p, Mp is very rarely prime. In fact, of the 1,622,441 prime numbers p up to 25,964,951,[5] Mp is prime for only 42 of them. The smallest counterexample is the Mersenne number

Mv11 = 2^11 1 = 2047 = 23 × 89.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42799049)

I just cannot believe that the early mathematicians only tested their statement for
Mv2, Mv5, Mv7 and then decided, ok, we checked enough, no need to check the next one Mv11

It's not that it's sooo hard to factor 2047...

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798677)

Not even then. For example, if p=11, 2^11-1 = 2048-1 = 2047. But 2047 = 23*89. As p gets larger, less and less 2^p-1 are in fact prime.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798869)

I don't think so, 2^11-1 = 2047 which is 23 * 89.

so 2^p-1 is not prime for prime p=11
QED

besides, it would be quite easy otherwise to find a new bigger prime, as if this were true, I could give you
a bigger prime in mere seconds:

2 ^ (2^57885161 - 1) - 1 ,oh wait

2 ^ (2 ^ (2^57885161 - 1) - 1 ) - 1, oh wait ...

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798411)

Wrong, 2^(4-1) = 8, which is not prime.

Re:Uhhh... (1)

ender8282 (1233032) | about a year and a half ago | (#42798415)

Not for n = 4 2^4 -1 = 16 - 1 = 15 = 3 * 5 ==> not prime!

Re:Uhhh... (1)

Kjella (173770) | about a year and a half ago | (#42798427)

Considering any number 2^n-1 is prime

Why don't you try that with n=4...

Re:Uhhh... (1)

canadiannomad (1745008) | about a year and a half ago | (#42798429)

Um, no. read that wiki article again.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798431)

Seriously? Not every 2^n - 1 number is a prime.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798435)

2^8-1 is 255, which is divisible by three.

Re:Uhhh... (1)

oobayly (1056050) | about a year and a half ago | (#42798439)

Considering any number 2^n-1 is prime [wikipedia.org], this isn't too impressive, except for maybe that they bothered to expand it into a full number.

Nope: 2^6 - 1 = 63. Correct me if I'm wrong, but 63 isn't prime.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798441)

any number 2^n-1 is prime

2^4-1 is prime?

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798455)

n = 4
2^n - 1 = 15
15 is not prime.

Might wanna read that article one more time.

Re:Uhhh... (1)

Anonymous Coward | about a year and a half ago | (#42798457)

No, 2^n-1 is not prime for all n (example n = 4). A Mersenne prime is a Merseene number that is also prime.

Re:Uhhh... (1)

History's Coming To (1059484) | about a year and a half ago | (#42798461)

Wrong - you mean all Mersenne Primes take the form 2^n-1, not that all numbers of the form 2^n-1 are prime. 11, for example, is prime but cannot be rationally expressed in the form 2^n-1.

If you could find new primes that easily then internet banking wouldn't be secure (well...as secure as it currently is, which is "enough for the insurance companies").

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798619)

This post is painfully incorrect. Please reread carefully, and if you still don't understand why it's wrong, please ask. I'd rather not even try to formulate a response because it is hurting me.

Re:Uhhh... (3, Informative)

Kjella (173770) | about a year and a half ago | (#42798863)

If you could find new primes that easily then internet banking wouldn't be secure (well...as secure as it currently is, which is "enough for the insurance companies").

No, it relies on factoring being much more difficult than multiplication. That is, if I have two large primes p and q I can trivially calculate p*q = n, but you can not easily find p and q from n. Being able to generate primes quickly doesn't give you anything.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798473)

... 2^4-1 = 15 = 3*5 ...

Re:Uhhh... (1)

gewalker (57809) | about a year and a half ago | (#42798479)

2^4-1 = 15
Maybe you meant 2^n-1 where n is prime. Still no good 2^1-1 = 2047 = 23 * 89

Unless you meant this as a bad joke, you are mistaken.

Re:Uhhh... (1)

gewalker (57809) | about a year and a half ago | (#42798491)

Sorry, 2^11-1 = 2047

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798481)

You misunderstand what a Mersenne prime is.

2^n - 1 isn't necessarily prime (e.g., if n = 5, then 2^5 - 1 = 15).

If n is prime AND 2^n - 1 is prime, then 2^n -1 is a Mersenne prime.

Re:Uhhh... (1)

gman003 (1693318) | about a year and a half ago | (#42798497)

Uh, did you actually read the article you linked to?

Not all 2^n-1 numbers (Mersenne numbers) are prime. For instance, if n is not prime, then 2^n-1 is not prime (so no, 2^57,885,162-1 is not prime). But even if n is prime, 2^n-1 is not prime - take n=23, where 2^n-1 (8388607) is divisible by 47 and 178481. In fact, only 48 Mersenne primes are currently known, as listed in the article you linked.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798543)

So by that logic 15 = 2^4 - 1 is prime?

If 2^n - 1 is prime then it is also Mersenne prime. It doesn't mean that anything of the form 2^n - 1 is prime.

Re:Uhhh... (1)

adonoman (624929) | about a year and a half ago | (#42798621)

Um.... let's try n=4. 2^4-1 = 15, so not a prime. You might want to read that wikipedia article a little closer.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798647)

Considering any number 2^n-1 is prime [wikipedia.org] , this isn't too impressive, except for maybe that they bothered to expand it into a full number.

Oh look! I've discovered a new, larger prime! 2^57,885,162-1 What do I win?

(2^8)-1 = 255 is a prime number?

Then again I am kind of nitpicking. According to the first paragraph of the Wikipedia article a Mersenne prime is Mp = 2^p-1 where p = prime. While a Mersenne number is mn = 2^n-1.

Re:Uhhh... (5, Funny)

amicusNYCL (1538833) | about a year and a half ago | (#42798713)

Hey, has anyone told you that your post is wrong yet?

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798715)

Sir, I applaud your dedication to the age-old skill of actual trolling. Though this is indeed a short-form troll, rather than a long-form one that spans multiple threads, you have quite impressively gone for the sounds-intelligent-but-is-fatally-flawed style of troll, complete with a misleading citation for your obviously wrong information. Well done! It is rare that we see that sort nowadays, what with the sad prevalence of trolls simply going for the low-hanging fruit of easily-ignored racial slurs, misogynistic comments, and assertions regarding the sexual proclivity of our mothers.

But, you're still a troll and knew you were wrong when you posted that, so kindly piss off.

Re:Uhhh... (1)

vlm (69642) | about a year and a half ago | (#42798787)

Epic fail 57885162 = 2×3×9647527 and a Mersenne prime is only a Mersenne prime if p is prime, which it clearly isn't.

2 ** 57885167 - 1 might or might not be a prime, but at least you can't rule it out automagically because 57885167 is in fact a prime.

Re:Uhhh... (0)

Anonymous Coward | about a year and a half ago | (#42798827)

No.

A Mersenne prime is a special type of prime that is both prime and and can be written as 2^n-1.

Not all primes are Mersenne primes, and not all values of 2^n-1 are prime.

And what shall we call this number? (1)

schneidafunk (795759) | about a year and a half ago | (#42798299)

Post responses of cool names.

Re:And what shall we call this number? (0)

Anonymous Coward | about a year and a half ago | (#42798469)

The "who cares?" number.

Re:And what shall we call this number? (1)

vlm (69642) | about a year and a half ago | (#42798851)

The goatse number = number of seared eyeballs requiring mind bleach. 2 ** 57885161 - 1 is probably about right. Either that or its the decimal representation of the future ipv8 standard which expands the puny ipv6 addressing space from 2 ** 128 to 2 ** 1e100 aka when we replace the internet with the google-net. All hail the almighty GOOG! huzzah!

They would have more primes to choose from ... (1)

Skapare (16644) | about a year and a half ago | (#42798331)

... if they didn't limit the search to just Mersenne primes [wikipedia.org] .

Re:They would have more primes to choose from ... (2, Informative)

Anonymous Coward | about a year and a half ago | (#42798513)

There isn't a 100% correct primality proving method for general numbers that's anywhere near as efficient as the Lucas-Lehmer test for Mersenne numbers of the same size.

Re:They would have more primes to choose from ... (1)

Skapare (16644) | about a year and a half ago | (#42798589)

So Mersenne primes are just low-hanging fruit?

Re:They would have more primes to choose from ... (4, Interesting)

dwillmore (673044) | about a year and a half ago | (#42798953)

Yes. The LL test only works on Mersenne numbers--numbers of the form 2^p-1 where p is prime. The LL test is not statistical. It can determine if a given mersenne number is prime or not without any doubt.

To protect against errors, GIMPS (Great Internet Mersenne Prime Search) uses a variety of double checks to ensure no number if mistested. Any number that passes the LL test is double (and sometimes triple checked) to verify that there wasn't a hardware or software error that caused a false positive. I had the honor of performing the double check of a record Mersenne prime some time ago.

Re:They would have more primes to choose from ... (0)

Anonymous Coward | about a year and a half ago | (#42799027)

The low-hanging fruit was picked by the Greeks. The not-quite-so-low-hanging fruit was picked by the Romans. Etc. etc.

What we're at now is the decision wether to pick the fruit at the end of long branches than bends down a bit (Mersenne), or pick the stuff near the base of lower branches at about the same height. Either way, we're a few dozen stacked forklifts off the ground.

Re:They would have more primes to choose from ... (0)

Anonymous Coward | about a year and a half ago | (#42798583)

Mersenne primes are easier to search for if you want to find a prime as large as possible (as opposed to finding every prime), since they are somewhat likely primes and you get to skip quite a few candidates.

What I'm curious of is to what certainty is this proven prime? Most likely it's statistical and not definitive (to my limited knowledge the latter would be computationally impossible). Don't get me wrong - statistical methods are good - that's what we normally use for things like generating RSA keys.

Re:They would have more primes to choose from ... (1)

Skapare (16644) | about a year and a half ago | (#42798683)

Maybe we could have more success with certainty by searching for non-primes. If you got the factors (even one factor is enough), such as 23*89 for 2^11-1, which some people think has to be prime because 11 is prime, then we could have more numbers. Why are primes so great? I can get tons of very large non-prime numbers by generating Pascal's triangle.

Re:They would have more primes to choose from ... (1)

vlm (69642) | about a year and a half ago | (#42798949)

I can get tons of very large non-prime numbers by generating Pascal's triangle.

Given that the rate of primes drop as the number length goes up, you can just pick any ole random really large number and the odds eventually become ridiculously high that its composite. You can have a lot of fun Fing with people (aka Socratic method), at least if they don't know much number theory, by trying to convince them there exists a certain largest prime number above which they're all composite, I mean, just look at the graph of rate of prime numbers vs number of digits or however you wanna phrase it, obviously it must intersect zero at some huge number (what would hitting negative even mean?). This would be the kind of "stupid interview trick question" I'd enjoy, certainly more creative than yet another stupid fizzbuzz implementation.

Meta discussing it even further from the original post, the Socratic method is pretty much good training to become a really effective internet troll.

Re:They would have more primes to choose from ... (3, Informative)

billstewart (78916) | about a year and a half ago | (#42798665)

Mersenne primes have a structure that makes it possible to test primality for very large numbers; there's no way to test whether unrestricted numbers of that size are prime (it's theoretically possible, but there aren't enough computing resources on the planet.)

I used to run the GIMPS search application back in the 90s; you really really don't want to run it on a laptop on batteries, especially with the battery technology of the time, and eventually I decided that my laptop didn't have enough horsepower to bother, compared to desktops that could run GPU-based calculations.

stats? (1)

rwa2 (4391) | about a year and a half ago | (#42798401)

So.... is there an equivalent to rc5stats for GIMPS, so we can compare, uh, our harnessed computing prowess?

No, I'm not going to google it. I want your opinion on whether it's any good or not.

Re:stats? (1)

dwillmore (673044) | about a year and a half ago | (#42799005)

Yes, if you go to www.mersenne.org and look on the left hand side, you will see a variety of lists and reports. They cover the system as a whole and individual contributers.

Trolling editors (0)

Anonymous Coward | about a year and a half ago | (#42798437)

Someone posted recently that slashdot editors intentionally add mistakes to summaries in order to get the replies going. Now I think they were right, but I still can't help myself.

257,885,160 is not a prime number, 2^57,885,161-1 is. Also, who says "less one" to mean "minus one"?

Re:Trolling editors (1)

Aardpig (622459) | about a year and a half ago | (#42798519)

I think, rather, that Hanlon's razor applies here: Never attribute to malice that which is adequately explained by stupidity.

Write the whole number (0)

Anonymous Coward | about a year and a half ago | (#42798471)

I dare you to write the whole number, including every digit. It shouldn't take up more than a few megabytes.

There you go (1)

InsectOverlord (1758006) | about a year and a half ago | (#42798615)

This [isthe.com]

Re:Write the whole number (0)

Anonymous Coward | about a year and a half ago | (#42798673)

Can you print it out and send me a hardcover copy? that'd be soooo cool ...well, at least until the next prime number is found.

Re:Write the whole number (1)

vlm (69642) | about a year and a half ago | (#42799019)

Can you print it out and send me a hardcover copy? that'd be soooo cool ...well, at least until the next prime number is found.

Better idea: Tattoo

Why 2^n-1 (1)

bigsexyjoe (581721) | about a year and a half ago | (#42798485)

I know that that is called a Mersenne prime. Why do they always look for primes of that form? Are these numbers more likely to be prime? Why don't they start looking for primes of the form 2^n+1?

Re:Why 2^n-1 (4, Informative)

godrik (1287354) | about a year and a half ago | (#42798613)

number of the form 2^n-1 are Mersenne numbers which are much more likely to be prime than a randomly chosen odd number. Also, we have "simple" test for these number to weed out many Mersenne numbers that are not prime. Once you have a Mersenne number that passed the "simple" primality test, there is a good chance that it will really be a prime number.

Re:Why 2^n-1 (1)

Anonymous Coward | about a year and a half ago | (#42798971)

I doubt that numbers of the form 2^n-1 are more likely to be prime than a randomly chosen odd number (of the same length). Do you have a reference to back up your claim?

Re:Why 2^n-1 (0)

Anonymous Coward | about a year and a half ago | (#42798627)

Isn't 2^n-1 all ones in binary. Maybe that's why.

Re:Why 2^n-1 (4, Informative)

Anonymous Coward | about a year and a half ago | (#42798833)

There is a very fast primality test for Mersenne numbers, the Lucas–Lehmer primality test. [wikipedia.org]
2^n+1 is prime only if it's a Fermat prime, n=2^k. None of these are known to be prime for k>4, and there probably aren't any more, whereas there are probably infinitely many Mersenne primes.

Re:Why 2^n-1 (0)

Anonymous Coward | about a year and a half ago | (#42798841)

I know that that is called a Mersenne prime. Why do they always look for primes of that form? Are these numbers more likely to be prime? Why don't they start looking for primes of the form 2^n+1?

Arithmetic and test procedure for Mersenne primes is highly efficient considering the size of the resulting prime. I think it takes something like n squarings and n small additions (something like +4 or -4) modulo 2^n-1 in order to tell for sure whether 2^n-1 is composite or not, once one knows that n is prime. Look it up, it was something like the Lucas-Lehmer test or similar.

Re:Why 2^n-1 (2)

einyen (2035998) | about a year and a half ago | (#42798875)

Actually they are a lot less likely to be prime than any random odd number, only 48 are now known and all have been tested up to above 13 million digits. But yes they are very easy to prove prime compared to "random" numbers of no special form, I *think* it was proven mathematically that there is no faster proof possible, but don't hold me to that. The highest "random" number proven to be prime is only 26,642 digits vs 17 million for this new mersenne prime. There are numbers of other special forms that are also "easy" but not as easy to prove, highest one proven is 3.9 million digits. They don't look for primes 2^n+1 because they are always divisible by 3 so they are never prime. They are looking for primes (2^n+1)/3 however, they are called Wagstaff primes, but again they are harder to prove than mersenne numbers.

Re:Why 2^n-1 (1)

necro81 (917438) | about a year and a half ago | (#42798987)

Why do they always look for primes of that form

For one thing, it is very easy to write them down. For another, they lend themselves very easily to computation using binary math. Any number of the form 2^n - 1 is just a string of n 1's. E.g., 2^5 - 1 = 31 = 11111b.

As to your other question, regarding 2^n + 1, those would be of the form of a '1', followed by (n-1) '0's, followed by another '1'. E.g., 2^5 + 1 = 33 = 1000001b. I don't know if these are any more or less likely to be prime. I suspect less likely, otherwise those would be what are studied so much.

Re:Why 2^n-1 (1)

loufoque (1400831) | about a year and a half ago | (#42799059)

Because it is believed that there is an infinite amount of Mersenne primes, but no one has been able to prove it yet.

Even the scientists' press release wrong (1)

domulys (1431537) | about a year and a half ago | (#42798487)

I thought the Slashdot summary goofed on this, but the scientists did as well.

It's the largest known Mersenne prime. For instance, 1000000! - 1 is also known to be prime, and it is much larger than this number.

Re:Even the scientists' press release wrong (2)

domulys (1431537) | about a year and a half ago | (#42798575)

Yup, was totally wrong about this. Euclid's Proof of the Infinitude of Primes suggests doing something like this (multiplying all primes and then adding 1), but it doesn't work in generating new primes.

Re:Even the scientists' press release wrong (0)

Anonymous Coward | about a year and a half ago | (#42798937)

It sort of does; If you multiply all known primes together and add 1, and then factor that number, you will have at least one new prime. The trouble is factoring that number is really hard.

Re:Even the scientists' press release wrong (1)

swimboy (30943) | about a year and a half ago | (#42799057)

I don't think that if you multiply all the known primes and add one you get a number that's guaranteed to be prime. All you get is a number that isn't divisible by any known prime. You could very well get a composite number whose factors are all bigger than the largest known prime.

A Little More Perfection (5, Informative)

14erCleaner (745600) | about a year and a half ago | (#42798611)

Since the only known perfect numbers [wikipedia.org] are derived from Mersenne Primes, this means there are also now 48 known perfect numbers. Interestingly, this property of Mersenne Primes was discovered by Euclid about 2000 years before Mersenne was born (time machine, anyone?). Finding a non-Mersenne perfect number would be a huge accomplishment.

Re:A Little More Perfection (0)

Anonymous Coward | about a year and a half ago | (#42798845)

It has been proved that such perfect number would be odd,
a rather odd property for being a perfect number...

So... (1)

Anonymous Coward | about a year and a half ago | (#42798697)

I wonder if 2^(2^57,885,161-1)-1 is also prime.

Re:So... (2)

cellocgw (617879) | about a year and a half ago | (#42799069)

I wonder if 2^(2^57,885,161-1)-1 is also prime.

Stop wondering and start dividing. Call us when done.

Number of Digits (5, Interesting)

necro81 (917438) | about a year and a half ago | (#42798897)

The new prime number, 2 multiplied by itself 57,885,161 times, less one, has 17,425,170 digits

"There are 10 kinds of people in the world. Those who understand binary, and those that don't."

This new number is 2^57,885,161 - 1, so naturally it has 57,885,161 digits, all of them 1. A simpler example: 2^5 - 1 is a Mersenne prime. Written in binary it's 100000 - 1 = 11111.

Oh! You meant that it has 17,425,170 decimal digits. Booooooooring!

CPUs? why not GPUs? (0)

Anonymous Coward | about a year and a half ago | (#42798965)

wouldn't this be somewhat ideal to compute using GPUs, and substantially faster than CPUs?

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