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November 18, 2009

Unsolved math problem turns 150

Posted: 12:14 PM ET

Happy 150th anniversary to the Riemann Hypothesis, one of the most important math problems ever!

Proposed by Bernhard Riemann in 1859, the Riemann Hypothesis deals with prime numbers. You may recall that a prime number is a positive whole number that has only two positive whole number divisors: one and itself. The first of them are 2, 3, 5, 7, 11, 13, in order.

This hypothesis would be able to provide a better estimate than ever before of a special function denoted as Pi(x). Pi(x) represents the number of prime numbers that are no bigger than x, where x is a positive number. For example, Pi(14) would be 6, because there are six prime numbers (2, 3, 5, 7, 11, 13) no bigger than 14. That's probably the most understandable explanation you're going to get that doesn't involve "zeta functions" and other technical terms.

Given that many of the best mathematicians have tried and failed to provide a solution, the proof is probably not easy or obvious, says Peter Sarnak, professor of mathematics at Princeton University and an authority on the subject. “Most experts expect that a proof will require a major new insight into the structure of whole numbers and the prime numbers,” he said.

But if you can solve it, the Clay Mathematics Institute will give you $1 million.

A proof would have implications not only for mathematics, but also for cryptography and computer science, says Ramin Takloo-Bighash, associate professor at the University of Illinois at Chicago. Internet security protocols, after all, are largely based on prime numbers. Experimental and theoretical evidence has supported the truth of the Riemann Hypothesis, although there are a small number of naysayers who say it can’t be proven, Takloo-Bighash said.

Still, there’s enough confidence in the truth of the Riemann Hypothesis that mathematicians have established “conditional” theorems, which can never be validated until someone proves the 150-year-old problem, says Kenneth Ribet, professor of mathematics at the University of California, Berkeley.

Riemann's paper on the subject was first published in November 1859, but no one knows the day. So, the American Institute of Mathematics picked a Wednesday in the middle of November to celebrate the 150th anniversary, said Brian Conrey, executive director.

Intrigued? Stop by one of these lectures today.

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Filed under: Mathematics

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Michael Harris   November 18th, 2009 1:05 pm ET

Very interesting proposition, however, you did not include how to follow up about winning the prize??? Where is this information? I checked at the Claymath website and there is nothing about this topic?

Please let me know.


Mom   November 18th, 2009 1:23 pm ET

It should be more specific, Can' t understand what the problem is ?

User   November 18th, 2009 5:17 pm ET

Prize info located here:

Raman   November 18th, 2009 5:19 pm ET

Here is the link to the seven millennium problems, each worth $1m prize:

Note that Poincare conjecture has already been proven by Perelman. You can have a stab at the remaining six. And here is the link to a description of Riemann hypothesis:

boo   November 18th, 2009 5:51 pm ET

even if you solve the problem if there is a solution that depicts a prime whole as the number you have two years in question to waite to see if your answer is even consisdered as being correct or close to solving the problem then its up the the scolars to pick it apart and then considerif it be even up to standards for the institute seems like you do all the work they get all the rewards ???

Fun with numbers   November 18th, 2009 6:44 pm ET

Dear Mom, if you don't understand what the problem is any further explination will fly completely over your head. As for Michael Harris, the reason there is no information on how to follow up with winning the prize is simple. No one has solved it in 150 years, therefore the chances that a blog on CNN will draw out the solution are effectivly zero.

Ubik   November 18th, 2009 7:06 pm ET

Actually, I have discovered a truly marvelous proof of this, which this text box is too narrow to contain. Sorry! 😉

Katerina   November 18th, 2009 7:22 pm ET

Amy   November 18th, 2009 7:59 pm ET

Read more about the Riemann Hypothesis here:

More information about the Millennium Prize Problems can be found here:

A Friend   November 18th, 2009 8:30 pm ET

I got dizzy just reading the first couple of lines...Not smart beware!

Elena P   November 18th, 2009 8:43 pm ET

Here is the link on the claymath website:

Dr. Mathq   November 18th, 2009 10:21 pm ET

If youre smart enough to solve the problem, you should have no problem figuring out how to contact them

Steve   November 18th, 2009 11:28 pm ET

0 is number 1, move everything 1 digit and theirs your answer. (0) is always number 1! Ask the banks!

Mike in NYC   November 19th, 2009 12:17 am ET

So what exactly is the Riemann hypothesis? A little oversight there, Liz.

That pic is pretty funny, BTW.   November 19th, 2009 3:32 am ET

Here are useful sites

3. (7 problems @ prizes of $1M / each!)
4. (email inquiries to )

Hopefully, your math skills surpass my Googling skills! 🙂

Good luck!

Greg   November 19th, 2009 3:57 am ET

There is more information about the Millenium Prizes (one of which a solution to the Riemann Hypothesis would earn) at the following Clay Math Institute page:

There are lots of descriptions of the Riemann Hypothesis, of various lengths and level of detail, all over the web. The Wikipedia page is as good a place to start as any:

Jason   November 19th, 2009 4:54 am ET

Mom: You didn't understand what the problem is because this article never explained what it is. And then they complement themselves for giving such an "understandable explanation"! I expect more from CNN; that this kind of sloppiness is offered and accepted feels like a symptom of meager math and science education in this country.

The Reimann hypothesis, unlike Fermat's last theorem, takes a little bit of time to explain carefully. (Although nowhere as much time as it takes to explain some of the other major mathematical propositions, like the recently solved Poincare conjecture.) Here's a link to an explanation that may be appropriate:

Stephen H   November 19th, 2009 8:21 am ET

I believe the answer is 42.

JT   November 19th, 2009 8:37 am ET

great, but what is the problem? you never say what it is, only an example of what Pi(x) is.

Charlie   November 19th, 2009 8:52 am ET

M's Landau obviously did not get her degree in Math. Her explanation of Riemann's Hypothesis indicates a complete lack of understanding of the problem. Those who are interested in a better explanation might try this link:

Bobby G.   November 19th, 2009 8:57 am ET

All you need to understand is that 1+1=2 and 1-1=0.

Also understand that just because you have checks doesn't mean you have money.

H A Helfgott   November 19th, 2009 9:18 am ET

The problem is as follows. We've got an approximate formula for
Pi(x) (it's roughly of size x/(log x), or more precisely the integral of
1/(log t) for t from 2 to x). We know that this is a pretty accurate formula.
Experiments show that it is actually extremely accurate – to a much greater extent than it has be proven to be.

The Riemann hypothesis states that this formula (without being exact) is essentially as accurate as it could possibly be (given that one can show that is isn't perfect).

All the talk about the Riemann zeta function is very important, but it's just a way to rewrap what I've said above.

David   November 19th, 2009 9:51 am ET

Ha! Yet another problem that Obama hasn't been able to solve... how predictable...

DMR   November 19th, 2009 9:54 am ET

Jason, it's very typical for CNN–I can usually find one grammatical error a week with minimal effort–they are only a touch better than fox, and far, far behind bbc...

Prof. Gerald Lambeau   November 19th, 2009 9:59 am ET

Quick, someone email Will Hunting and get the answer ASAP. He is probably roaming around the southside of Boston with his buddy Ben trying to stir up trouble with Math geeks at Cambridge.

For the love of God my head hurts from not knowing the answer to ridiculous theories in math. Get his number now!!

Will Hunting   November 19th, 2009 10:17 am ET

Hey guys I saw this problem up on the chaulk board last night when I was mopping the floor and I wanted to let you all in on the answer. No thanks necessary and feel free to keep the prize. Here goes:

enter the cubed root of 150,399,696,179,997,056,512 on your calculator and turn it upside down.

How ya like them apples?

Brian   November 19th, 2009 10:34 am ET

They randomly picked Wednesday? How unimaginative.
Should have gone with Nov. 13th at 17:19 and 23 seconds.

Brian   November 19th, 2009 11:28 am ET

I like the whole idea about this but I don't even get what the problem is either. Pi(14) would be 6 it definately makes sense. The higher number x is, the more prime numbers there are. So what is the problem they want to solve?

specter   November 19th, 2009 12:04 pm ET

Where do black holes go? What is dark matter? How did we get here? What is the meaning of it all? Ignorance is truly bliss. I should have payed more attention in school. To borrow from a fellow ignoramis,"there are known known's and known unknown's...." I know I don't know the answer to a lot of these questions. /shrug

S. M.   November 19th, 2009 1:43 pm ET

I know the answer to this but I don't need a million dollars right now.

shutthehellup   November 19th, 2009 2:37 pm ET

How bout you write it up first, then find out how to claim your prize.

ioan   November 19th, 2009 2:52 pm ET

Took me 5 minutes to solve it 🙂
There you go:

function Pi(x: integer): integer;
i: integer;
J: Integer;
isPrime: boolean;
if x >= 2 then
result := 1
result := 0;
for i := 3 to x do
isPrime := true;
for J := 2 to I – 1 do
if (i mod j) = 0 then
isPrime := false;
if isPrime then
result := result + 1;

sam   November 19th, 2009 3:00 pm ET

Simple (English) Wikipedia might be easier to understand for some folks

MathPhd   November 19th, 2009 3:21 pm ET

Yeah, I completely agree with FunwithnNumbers. As a PhD student at Penn, I've worked on solving the Riemann Hypothesis with other ivy-league students and Professors on the East Coast for a year. Something tells me a middle aged Mom who posts on CNN that she doesn't "understand the problem" wouldn't provide too much help in the solution to the one of the most cryptic mathematical equations of all time.

Jorgen   November 19th, 2009 4:18 pm ET

Anything past basic math be moot. Its the same as a language. People make up languages all the time such as elvish and Klingon, and they make sense. The same thing applies to math, it was invented, people began to follow it and as you get into things like imaginary numbers it just becomes stupid. Stop the madness, admit that this problems have no real application on anything useful and use math just enough to balance your check book, because lets face it, it has to be done.

onlycontent   November 19th, 2009 7:41 pm ET


Todd Timberlake   November 19th, 2009 8:09 pm ET

Thanks Justin. Definitely an error in my earlier post – not sure how that happened, I must have been half asleep when I wrote it. The REAL part of the nontrivial zeros is 1/2, the imaginary part can be lots of different things (Google Andrew Odlyzko to find datasets with hordes of nontrivial zeta zeros, all of which fit the hypothesis).

It's nice that CNN is even including a link to a story about the Riemann Hypothesis – but I agree with several other posters that the article could give a little more explanation. It's a difficult subject, but if we can't describe difficult science and math then we can't talk about almost anything that is a current topic of research. For anyone who wants to learn more about this fascinating topic I recommend Derbyshire's "Prime Obsession" (especially if you like some history with your math).

dtc   November 19th, 2009 10:03 pm ET

why is there a picture of a black guy trying to solve a math problem.

Hard Tymes   November 20th, 2009 10:51 pm ET

Right about now I wish we had Mr Data from Star Trek!
I'd make book that android could solve it!
One other item, folks.
Why don't we give the problem to that monster computer in Brussels, Belgium? Then we could go on to solve the reason why we can't balance a checkbook! 🙂

Hard Tymes ~~ HardTymes.Com

Mark C   November 29th, 2009 8:29 pm ET

**** The same thing applies to math, it was invented, people began to follow it and as you get into things like imaginary numbers it just becomes stupid. Stop the madness, admit that this problems have no real application on anything useful and use math just enough to balance your check book, because lets face it, it has to be done. ****

Wow, could you possibly be a bigger moron?

Kindly explain how you get a spaceship to Saturn using "basic math" that you use to balance your checkbook. Or how you do the fluid dynamics necessary to design an airplane. Or understand semiconductors (which involve quantum mechanics) well enough to build the computer you posted this drivel from.

Louise   November 30th, 2009 1:08 pm ET

If you think this is hard; just try explaining the federal tax code.

Jessica*   December 2nd, 2009 1:05 am ET

It cracks me up to see how many people post on the CNN blog that they have the answer – and even funnier that people are quoting Wikipedia – since when is Wikipedia a reliable source of information?

RingerMe   December 5th, 2009 9:00 am ET

FTAA proof would have implications not only for mathematics, but also for cryptography and computer science,

Okie in Exile   December 6th, 2009 8:44 pm ET

Wikipedia is usually pretty good on mathematical topics. Geeks take stuff pretty seriously.

Double Action   December 8th, 2009 1:55 am ET

Wolfram Alpha is my internet math professor.

ebrown2112   December 11th, 2009 1:53 am ET

The answer is 42, duh.

anirban saha   April 6th, 2011 10:41 am ET

this hypothisis is greatest beauty in mathematics. i want to solve this. i want the e-mail id of the institute to contact.

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Unsolved problems in science/math? - Page 2 - Forums   September 22nd, 2011 1:58 am ET

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thriller   June 3rd, 2013 4:47 pm ET

Obama on Internet Security – Riemann Hypothesis

Peter L. Griffiths   August 30th, 2013 2:43 pm ET

In discussing the Riemann hypothesis, why does nobody mention that the non-trivial zeta evaluations are all terms in the cotangent series whose first term is 1/2 which seems to have been interpreted as being a critical line.

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Peter L. Griffiths   November 2nd, 2013 1:48 pm ET

Euler's product formula for primes quoted in Riemann's 1859 paper is not quite correct, the Harmonic series needs to be divided by the composite fractions including 1 followed by 1/4 contained in the Harmonic series.

Peter L. Griffiths   November 3rd, 2013 12:36 pm ET

Further to my comment of 2 November 2013, this division of the Harmonic series by the composite series has up to now been overlooked because of the initial transfinite location of the finite composite fractions. These composite fractions need to be brought together in a finite location to divide the Harmonic series.

Peter L. Griffiths   November 8th, 2013 12:24 pm ET

Further to my comment of 3 November 2013, I have come to the conclusion that it is not only the primes but also the composites which are deducted from the original Harmonic series (H), leaving 1 equal to H. product (p-1)/p which is Euler's formula. The great attraction of this formula is that the fractions in both
H and product (p-1)/p can be raised to the same power or root and still equal 1. I have however come to the conclusion that Riemann was on the wrong track for identifying primes. To identify primes, the number of factors in the composites needs to be studied. Nobody seems to have done this.

Peter L. Griffiths   November 25th, 2013 11:56 am ET

This comment I hope delivers the fatal thrust. Near the beginning of his 1859 paper Riemann incorrectly assumes that the complex variable s =1/2 +ti is a Zeta power. Riemann fails to recognise that an expression containing an imaginary number such as 1/2 +ti cannot be a power unless the base is a log base such as e. The best known example of this is Cotes's formula
cosu + isinu equals e^(iu) where it is not possible for e to be meanfully replaced by other values. This means that Riemann is badly wrong in applying as a power s =1/2 +ti. It also means that practically all the arguments in his 1859 paper are fallacious.

Peter L. Griffiths   December 17th, 2013 1:18 pm ET

I wish to clarify the observation in my comment of 25 November 2013 that it is not possible for e to be replaced by other values. I now consider that it is possible for e to be replaced by other values but only because the general formula for any variable base n and power u should be the constant n^(u/logn). When n equals e, log n will be clearly 1. The fact that n^(t/logn) has to be a constant particularly when applied to i that is n^(ti/logn) undermines virtually all the arguments in Riemann's 1859 paper.

Peter L. Griffiths   December 20th, 2013 2:10 pm ET

The e number mentioned in my previous comments has an equivalent n^(1/logn) where n can be any number. This does not affect my conclusions.

Peter L. Griffiths   August 23rd, 2014 11:54 am ET

One possible approach to identifying primes is to apply Euler's discovery H(1-[1/2])(1-[2/3])(1-[4/5])(1-[6/7]) ... =1, where H is the harmonic series 1+(1/2) +(1/3) +(1/4)...., particularly if neither series is taken to infinity.

Peter L. Griffiths   September 7th, 2014 1:32 pm ET

One mystery which needs to be solved is why is it that one half the Harmonic series being (1/2) +(1/4) +(1/6) ..being deducted from the full Harmonic series being 1 +(1/2) +(1/3) +(1/4) ...leaves 1 +(1/3) +(1/5) + (1/7) ...and not the original half. It appears that the positioning of the terms is important as well as their values. The 1 needs to be retained to arrive at the primes product series.

shyam sundhar   August 25th, 2015 1:43 am ET

hii everyone,ihave one proof i.e any possitive integer is equal to another any possitive integer with explanation and i think its not applicable in present generation ,can anyone help me where we use this and what is the purpose or this,(i.e x value = y value ,,where x and y are integers),iam not well in communication sooo ,if you understand what it is then reply to my comment plz with name

thanking you everyone

Peter L. Griffiths   April 10th, 2017 10:04 am ET

I have proved Fermat's Last Theorem in less than 300 words in the February 2017 issue of M500

Galen Krumbholz   June 7th, 2018 3:28 pm ET

Me too.

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