Florida man stumbles on biggest prime number after working plucky i5 CPU for 12 days straight

The largest known prime number, made up over 24 million digits, has been discovered by a lone IT professional quietly crunching numbers with an Intel-powered computer in December. Patrick Laroche, a 35-year-old from Ocala, Florida, chanced upon the goody by running the free Great Internet Mersenne Prime Search (GIMPS) software …

[John 110]

waiting patiently...

...for the first person to find Optimus Prime...

[Pat Att]

Mersenne

Let's make the most of it. With a name like that they'll be harder to find after Brexit.

[Buzzword]

Re: Mersenne

Godwin's law has been updated. Every online discussion now ends up with a tenuous link to Brexit.

[MiguelC]

Re: Mersenne

Or Trump....

[Swarthy]

Re: Mersenne

Nah, that's still covered by the original Godwin.

[Andytug]

Re: Mersenne

No, that should be a separate law...how about "Farage's Law"?

[vir]

Hold Please

I am verifying with pen and paper.

[JDX]

If they're ann 2^n - 1, surely we have a list of which ones are found?

[Anonymous Coward]

We do...

https://en.wikipedia.org/wiki/Mersenne_prime#List_of_known_Mersenne_primes (though you may want to verify them).

[Chris Miller]

I can already recite this number from memory (but only in hexadecimal).

[F Seiler]

Well, AFAICT it begins with a 1, but the tricky part ist probably getting the length of the remainder FFFFFF.... right

[Anonymous Coward]

> If they're ann 2^n - 1, surely we have a list of which ones are found?

I think the author meant to say that there are unknown primes that are smaller than the larger, known Mersenne primes (because some will have been skipped).

[Chris Miller]

I think the point is that GIMPS doesn't search strictly sequentially (it can't, with millions of instances being run independently). So it's conceivable (but highly unlikely) that some as yet unchecked number could turn out to be a smaller Mersenne prime.

[JDX]

Yes but we can have a list of which have been checked and of those which are primes.

IIRC this one is 8 digits so the list is well under a billion items, quite manageable (if you just want to store flags not the actual number written out longhand)

[PyLETS]

Unknown primes

You're generate these every time (with a probability very close to 1) every time you create an RSA keypair. The primes you're likely to generate are probably unknown in the sense there are very many more of these, which are very easily discoverable, than the number of atoms in the observable universe - let's say that's 10**82. Start with 2048 bits of random noise output from /dev/random seeded with a good noise generator , make the last bit 1 so it's odd, then test it a few times with Fermat's Little Theorem and Rabin Miller tests, and if not prime add 2, and retest it until it is prime. There are approximately 2**2037 primes lower than 2**2048, which is a very large number compared to the number of atoms in the universe. So unknown primes are very numerous and easy to find and if useful for crypto are very much smaller than the largest known primes. The latter have millions of bits, while those useful for cryptography are likely to have thousands of bits.

If the number of atoms in the universe is only a few hundred bits long, it follows that the primes you're likely to generate for cryptography couldn't be stored if every atom in the universe were to be turned into a single memory cell which could be used to store one of these.

[Ken Shabby]

Obligatory xkcd https://xkcd.com/10/

[Inventor of the Marmite Laser]

Is that like Amazon Prime but less unpleasant?

[Anonymous Coward]

"{...] and it's unknown if there are a finite or infinite number of Mersenne primes."

If there are infinite possible numbers - then it seems intuitive that there will be infinite numbers of these primes?

[DavCrav]

"If there are infinite possible numbers - then it seems intuitive that there will be infinite numbers of these primes?"

What about primes of the form 2^n+1? These are called Fermat primes, and it is believe that there are only five of these.

So your intuition is way off.

[Dr Paul Taylor]

Fermat primes

correction: 2^(2^n) + 1 3, 5, 17, 257, 65537

Gauss showed that regular polygons with this many sides can be constructed with ruler and compasses.

[Anonymous Coward]

Re: Fermat primes

"Gauss showed that regular polygons with this many sides can be constructed with ruler and compasses."

For some value of ruler and compasses.

[PhilipN]

Re: Ruler and compass

Someone is going to have to tell me how long this may take.

I am genuinely intrigued because I learnt just enough maths at school to understand there is a point to such exercises without knowing WTF it is.

[DavCrav]

Re: Fermat primes

"correction: 2^(2^n) + 1 3, 5, 17, 257, 65537"

That's not a correction, that's an addendum. You are not correcting what I said, merely adding to it. I didn't feel the need to state that the exponent needed to be a power of 2, because doing the pre-factorization into cyclotomic polynomials isn't necessary to demonstrate the point.

[Richard Pennington 1]

Re: Fermat primes

The proof that the exponent of a Fermat prime is a power of 2 is (almost) a one-liner.

If k has an odd factor m (say k = m*n, where m≥3 is odd) then (by an algebraic factorisation) it is trivial to show that 2^n+1 divides 2^k+1 = 2^(m*n)+1 [e.g. 2^2+1 divides 2^10+1]. So if 2^j+1 is prime, then j has no odd factors ≥3, hence j is a power of 2.

[Spikey-Mike]

Intuitive possibly; the trick is to prove it;-)

[o p]

There is also an infinite number of even numbers.

Prime is in P. I do not find it intuitive. Maybe you do?

[YetAnotherLocksmith]

Nonsense. The definition of a prime is that it can't be divided by any number other than itself and 1. The definition of an even number is that it ends in a number divisible by 2.

There is only one even prime. By definition.

[jonathan keith]

$3k prize?

That might just about cover his electricity bill then.

[Timbo]

Re: $3k prize?

"That might just about cover his electricity bill then."

Given it's only a fairly common place CPU, 12 days of crunching on an i5 wouldn't have cost that much.

Might be a lot different if it had been crunched on a 1st or 2nd generation Butterfly Labs Bitcoin miner, which were extremely inefficient...they were pretty good as central heating fans in fact !!

https://www.amazon.co.uk/Butterfly-Labs-ASIC-Bitcoiner-Miner/dp/B00I3U99RY

[Simon Brady]

Re: $3k prize?

This raises an interesting question: what does BTC need to trade at before it's more profitable to spend your compute power hunting for primes? (Choosing an appropriate unit of compute to base the comparison on, e.g. general-purpose CPU/GPU throughput or FPGA gate count.)

[89724102172714182892114I7551670349743096734346773478647892349863592355648544996312855148587659264921]

Don't gamble what you can't afford to lose - $3,000 is about two years of electricity

[JeffyPoooh]

There is a much more efficient algorithm to find primes (and factor very large composite numbers)...

I have a truly marvelous demonstration of this proposition which this comment box is too narrow to contain.

[Ken Shabby]

NURSE...

Nurse, Mr Fermat is scribbling in his book again! Can we make sure it is the last time he does that?

[GBE]

When written out in full?

Mersenne primes, when written out in full, equal 2^n - 1 where n is the exponent needed to generate the prime and is used to form the codename.

Can anybody explain what the qualifier phrase "when written out in full" means? Aren't Mersenne primes always equal to 2^n - 1 (for some integer n) regardless of how they're written?

[diodesign]

"when written out in full"

As in, the full prime number, equals (2^n)-1 where n is the exponent that generates the prime. There are two numbers at play here: the prime number, calculated by (2^n)-1, and n, the exponent. 82,589,933 is not the prime number, it is the exponent that produces the final prime, which is what we were trying to drive at.

C.

[GrumpenKraut]

Re: "when written out in full"

> 82,589,933 is not the prime number, ...

Actually, it is a prime number. Exponents of Mersenne primes need to be prime.

Apologies for being a bit of a Klugscheisser. ----->

[emmanuel goldstein]

Re: "when written out in full"

If you're going to be a pedant, at least grasp the point being made. He didn't say 82,589,933 wasn't a prime number, he said it wasn't the prime number (i.e. it is not the prime number we are actually interested in here).

[GrumpenKraut]

Re: "when written out in full"

OK, pendant fail on my side.

[YetAnotherLocksmith]

Re: When written out in full?

If you wrote out the full number, all 23 million digits of it.

A dozen = 12

A googol = 10^100. Written in full: 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.

2^23 million - 1 = 1...............................................1

[Gene Cash]

Ocala

Interesting. That's the podunk hellhole I was born & raised in. I'm surprised they can do math.

[John F***ing Stepp]

Downvote time

I just checked, it does not calculate correctly.

OK, devide the exponate by 4 this gives you the number in hex with the fractional part (.25 or .75) determening the leading diget (a 1 or a 7) the rest being the hex diget F.

This one has a fractional part of .3 so it is not a Mersenne.

(I could show the math on that but it is kind of long)

[John F***ing Stepp]

Re: Downvote time

Oh scratch that. Damn calculater was wrong.

One followed by 2239733 Fs in hex.

(Goddamned smart phone

[skalamanga]

Sounds like Florida man finally got his shit together...

10chars

[Francis Boyle]

Don't worry

He'll be back to doing drugs and trying to eat people's heads in a jiffy.

[John Brown (no body)]

Oh crap!

Now I have come up with ANOTHER new password.

I'd have been ok if not for that meddling kid and his computer!

[harmjschoonhoven]

Math is hard

If the number of Mersenne primes is finite, there is a largest Mersenne prime, call this Mlargest. This is a prime number from which the number (2^Mlargest)-1 can be derived. Because this can not be a prime, it should be the product of two or more primes. At least one of its factors is expected to be larger than Mlargest.

Happy computing, or better: Do some hard thinking.

[DavCrav]

Re: Math is hard

"Mlargest. This is a prime number from which the number (2^Mlargest)-1 can be derived. Because this can not be a prime, it should be the product of two or more primes. At least one of its factors is expected to be larger than Mlargest."

Congratulations, you have proved that large numbers exist.

Problems with your proof:

1) All factors could be smaller than Mlargest.

2) The factors need not be Mersenne primes.

There's a reason that some problems are hard. It's not that mathematicians are bumbling idiots.

[harmjschoonhoven]

Re: Math is hard

Computing the n-th Mersenne prime does not add a iota to the the proof that the set of Mersenne primes is finite, infinite or that this is an undecidable conjecture.

[MonkeyCee]

Re: Math is hard

"Computing the n-th Mersenne prime "

I'm not certain, but my recollection is that they are not calculating the sequence of Mersenne primes, and that if we had a method for calculating the n-th one then there is a possible inductive proof of the finite/infinite nature of them.

What we have is a method of creating potential Mersenne primes (as described in the article) using existing primes and methods to test if these are in fact prime.

"does not add a iota to the the proof that the set of Mersenne primes is finite, infinite or that this is an undecidable conjecture."

Wasn't your "proof" attempting to show that they are in fact finite? Perhaps I misunderstood.

Being able to compute continually larger Mersenne primes may not prove that they are infinite, but may be close enough for practical purposes. In the same way you can't prove linear optimizations are efficient, but in application they are, the set of Mersenne primes may be large enough to be close enough to infinity for the purpose at hand.

[harmjschoonhoven]

@DavCrav Re: Math is hard

Congratulations, you have proved that large numbers exist.

No, I did not (although they exist), because the proposed large number critically depends on the proposition that the set of Mersenne primes is finite.

BTW, The association of mathematicians with idiots, bumbling or otherwise, is entirely yours.

[MonkeyCee]

Re: Math is hard

"This is a prime number from which the number (2^Mlargest)-1 can be derived. Because this can not be a prime"

Please submit some proof of this. If you start with the assumption that Mlargest is the largest, then simply asserting that there are no larger Mersenne primes is a fallacy.

Essentially the argument presented is: Assume p is true: Therefore p is true.

Ignoring this, lets move on to the next part:

"Because this can not be a prime, it should be the product of two or more primes. At least one of its factors is expected to be larger than Mlargest."

I'll just let the bad math slide, and stick with conclusions. Even if one of the factors is prime and larger than Mlargest (note, you've not actually proved that such a factor exists or is in fact prime), then you still need to show the factor is not only prime, but a Mersenne prime.

I'm looking forward to your proof of P = nP

Apparently it just involves some hard thinking :D

[Cynic_999]

Pretty sure the answer's 42

Isn't it?

[wayne 8]

"$3,000 GIMPS research discovery award cash prize"

Should not that prize have increasing value as the difficulty increases?

That is what makes Bitcoins so valuable, I hear.

[Miekey]

Written in English. From Russian, About the theory that'is prime numbers.

I talk, Hello everyone and good afternoon. The question, do you think, is, it easy to calculate a simple number without a computer, if the total sum of numbers is 10 ^ 32, and the number of prime numbers is only one and a half to two percent. Do you think this can be done on a calculator with a probability of 10%. So one of the numbers = 2 ^ 104 + 25. Please check and write the answer on the site ..