Comments on: The Longest-Standing Math Problem http://www.damninteresting.com/the-longest-standing-math-problem/ A collection of Damn Interesting things Sat, 11 Feb 2012 01:49:52 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: Benschop http://www.damninteresting.com/the-longest-standing-math-problem/#comment-26583 Benschop Tue, 26 Jul 2011 12:48:02 +0000 http://www.damninteresting.com/?p=246#comment-26583 Correction: ‘Goldbach-for-Residues’ (mod m_k) : each even residue is the sum of two units. and : Taking principle values of residues, restrict summands . . . . Correction: ‘Goldbach-for-Residues’ (mod m_k) : each even residue is the sum of two units.
and : Taking principle values of residues, restrict summands . . . .

]]> By: Benschop http://www.damninteresting.com/the-longest-standing-math-problem/#comment-26582 Benschop Tue, 26 Jul 2011 12:37:20 +0000 http://www.damninteresting.com/?p=246#comment-26582 In http://de.arxiv.org/abs/math/0103091 you will find an 11-page elementary proof of GC. Prove first for residues mod m_k (where m_k = \prod first k primes) that in the group G1(k) of units mod m_k 'Goldbach-for-Residues' holds: each even unit is the sum of two prime units, taking principle values of residues. Restrict summands to unit principle values u < q_k= [p_{k+1}]^2 (which all are necessarily successive primes because q_k is the smallest composite in G1(k) ), and using Bertrand's Postulate ( p_{k+1} < 2p_k ) yield a proof of GC by induction over k, with a proof by complete inspection for k=3 ( 4 < 2n < 30 ) as induction base. ---- Best regards, Nico Benschop. PS: this proof is too simple to be published via the known peer-review process, as I learned...;-( In http://de.arxiv.org/abs/math/0103091 you will find an 11-page elementary proof of GC.
Prove first for residues mod m_k (where m_k = \prod first k primes) that in the group G1(k)
of units mod m_k ‘Goldbach-for-Residues’ holds: each even unit is the sum of two prime units,
taking principle values of residues. Restrict summands to unit principle values u < q_k= [p_{k+1}]^2
(which all are necessarily successive primes because q_k is the smallest composite in G1(k) ),
and using Bertrand's Postulate ( p_{k+1} < 2p_k ) yield a proof of GC by induction over k,
with a proof by complete inspection for k=3 ( 4 < 2n < 30 ) as induction base.
—- Best regards, Nico Benschop.
PS: this proof is too simple to be published via the known peer-review process, as I learned…;-(

]]> By: mathematics101 http://www.damninteresting.com/the-longest-standing-math-problem/#comment-26396 mathematics101 Thu, 17 Feb 2011 07:44:30 +0000 http://www.damninteresting.com/?p=246#comment-26396 [quote]Ravage said: "Arcangel said: “Well actually ………. yes. Thanks for asking! Back in March of 2000 a group of London publishers were offering the equivalent of 1.5 million Canadian dollars to anyone who could prove the theory that all even numbers greater than 4 were the sum of 2 prime numbers. Prime numbers for those not knowing is a number that is only divisible evenly by itself and one (examples are 1, 3, 5, 7, 11, 13, etc.). So for example, the even number 16 according to Goldbach would be the result of adding the two prime numbers 3 and 13. They gave participants until March 15, 2002 to come up with a solution. The publishers were not in the least bit worried about having to pay it out and they didn’t. This puzzle is known as Goldbach’s first conjecture and was first raised by the Prussian mathematician, Christian Goldbach way back in the year 1742. It was discovered in correspondence between Christan Goldback and another mathematician by the name of Leonhard Euler from Switzerland. According to all accounts this theory has not been proven in the past 263 years but has been verified up to 4×10 to the 14 power by using an optimized segmented sieve and an efficient checking algorithm. Enough already! It is easy to think of any even number and come up with 2 prime numbers that when added together total that even number. The problem though is trying to prove that it works for every even number imaginable. Now you know my dilemma so don’t get me started on Goldbach’s other conjecture! What is Goldbach’s other conjecture you ask? Check it out for yourself at the link below: http://www.andrews.edu/~calkins/math/biograph/biogoldb.htm Now, sorry you asked?” I also work on Goldbach’s Conjecture. I would like to point out that the other conjecture is nothing more than a rewording of the 2 primes conjecture. If one has 3 primes, the sum of two of them are an even number by the primary conjecture. This means an odd (prime) and an even number equal an odd number. What this means is that we can take an arbitary prime (eg 3) and then using the primary conjecture create any odd number by selecting the correct two primes to create the requisite even number. There is only one conjecture. The real answer to the conjecture is to prove that the prime numbers are clustered close enough together such that they can cover the binary partition of all even numbers less than double the largest odd prime."[/quote] i solved the problems you were looking at it was relativley easy [quote]Ravage said: “Arcangel said: “Well actually ………. yes. Thanks for asking!

Back in March of 2000 a group of London publishers were offering the equivalent of 1.5 million Canadian dollars to anyone who could prove the theory that all even numbers greater than 4 were the sum of 2 prime numbers. Prime numbers for those not knowing is a number that is only divisible evenly by itself and one (examples are 1, 3, 5, 7, 11, 13, etc.). So for example, the even number 16 according to Goldbach would be the result of adding the two prime numbers 3 and 13. They gave participants until March 15, 2002 to come up with a solution. The publishers were not in the least bit worried about having to pay it out and they didn’t.
This puzzle is known as Goldbach’s first conjecture and was first raised by the Prussian mathematician, Christian Goldbach way back in the year 1742. It was discovered in correspondence between Christan Goldback and another mathematician by the name of Leonhard Euler from Switzerland. According to all accounts this theory has not been proven in the past 263 years but has been verified up to 4×10 to the 14 power by using an optimized segmented sieve and an efficient checking algorithm. Enough already!
It is easy to think of any even number and come up with 2 prime numbers that when added together total that even number. The problem though is trying to prove that it works for every even number imaginable. Now you know my dilemma so don’t get me started on Goldbach’s other conjecture! What is Goldbach’s other conjecture you ask? Check it out for yourself at the link below: http://www.andrews.edu/~calkins/math/biograph/biogoldb.htm Now, sorry you asked?”

I also work on Goldbach’s Conjecture. I would like to point out that the other conjecture is nothing more than a rewording of the 2 primes conjecture. If one has 3 primes, the sum of two of them are an even number by the primary conjecture. This means an odd (prime) and an even number equal an odd number. What this means is that we can take an arbitary prime (eg 3) and then using the primary conjecture create any odd number by selecting the correct two primes to create the requisite even number.
There is only one conjecture. The real answer to the conjecture is to prove that the prime numbers are clustered close enough together such that they can cover the binary partition of all even numbers less than double the largest odd prime.”[/quote]

i solved the problems you were looking at it was relativley easy

]]> By: Benschop http://www.damninteresting.com/the-longest-standing-math-problem/#comment-25437 Benschop Wed, 21 Oct 2009 07:53:18 +0000 http://www.damninteresting.com/?p=246#comment-25437 An elementary FLT proof (16 pgs, using semigroup Z(.) mod p^k, odd prime p) was published in the Acta Mathematica of Univ. Bratislava (Nov.2005) see : http://pc2.iam.fmph.uniba.sk/amuc/_vol74n2.html (pp 169 - 184). For some intro : http://home.claranet.nl/users/benschop/marg-abs.htm An elementary FLT proof (16 pgs, using semigroup Z(.) mod p^k, odd prime p)
was published in the Acta Mathematica of Univ. Bratislava (Nov.2005) see : http://pc2.iam.fmph.uniba.sk/amuc/_vol74n2.html (pp 169 – 184).
For some intro : http://home.claranet.nl/users/benschop/marg-abs.htm

]]> By: mabs239 http://www.damninteresting.com/the-longest-standing-math-problem/#comment-22730 mabs239 Mon, 15 Sep 2008 07:20:13 +0000 http://www.damninteresting.com/?p=246#comment-22730 Very interesting column. But I donot know how to prove/derive the Euclid's formula :( Very interesting column. But I donot know how to prove/derive the Euclid’s formula :(

]]> By: catra7000 http://www.damninteresting.com/the-longest-standing-math-problem/#comment-20739 catra7000 Thu, 27 Mar 2008 21:17:42 +0000 http://www.damninteresting.com/?p=246#comment-20739 whenever i had a really horrible math problem that neither my mom or dad could figure out, i just said EFF IT and i'll get it wrong. cuz really, who cares? why waste the time and energy getting mad and stressed out by not getting the answer? when i decided to just pass on it, i felt so much more relieved and free whenever i had a really horrible math problem that neither my mom or dad could figure out, i just said EFF IT and i’ll get it wrong. cuz really, who cares? why waste the time and energy getting mad and stressed out by not getting the answer? when i decided to just pass on it, i felt so much more relieved and free

]]> By: george9589 http://www.damninteresting.com/the-longest-standing-math-problem/#comment-20406 george9589 Tue, 11 Mar 2008 10:28:20 +0000 http://www.damninteresting.com/?p=246#comment-20406 [quote]Eric Leeson said: "I hate math."[/quote] I agree with you one hundred percent haha [quote]Eric Leeson said: “I hate math.”[/quote]

I agree with you one hundred percent haha

]]> By: Ravage http://www.damninteresting.com/the-longest-standing-math-problem/#comment-17430 Ravage Thu, 27 Sep 2007 15:49:15 +0000 http://www.damninteresting.com/?p=246#comment-17430 [quote]Robert-H-Goretsky said: "Robert H. Goretsky says I always found the mathematics behind RSA encryption to be quite interesting. (Wikipedia Link ) What's crazy is how easy it is to generate a new public / private key pair, but how impossible it is to do the math 'backwards' to solve for the private key, given the public key… No computer in the world can solve these problems in our lifetime. Comment by Robert H. Goretsky of Hoboken, NJ"[/quote] This depends on the size of the key. It used to be that 1024 or 2048 was considered strong. Now techniques are in use that allow cracking 2048 keys in reasonable amounts of time. At the current time a key length of less than 4096 is probably only moderately strong. And I expect that key length to crack in 2-3 years. There was an announcement today for example about two new quantum computing chips that were released. Once this technology comes to market about 2015 RSA, knapsack, and similar algorithms based on modulo and prime factoring will no longer be usable. [quote]Robert-H-Goretsky said: “Robert H. Goretsky says I always found the mathematics behind RSA encryption to be quite interesting. (Wikipedia Link ) What’s crazy is how easy it is to generate a new public / private key pair, but how impossible it is to do the math ‘backwards’ to solve for the private key, given the public key… No computer in the world can solve these problems in our lifetime. Comment by Robert H. Goretsky of Hoboken, NJ”[/quote]

This depends on the size of the key. It used to be that 1024 or 2048 was considered strong. Now techniques are in use that allow cracking 2048 keys in reasonable amounts of time. At the current time a key length of less than 4096 is probably only moderately strong. And I expect that key length to crack in 2-3 years. There was an announcement today for example about two new quantum computing chips that were released. Once this technology comes to market about 2015 RSA, knapsack, and similar algorithms based on modulo and prime factoring will no longer be usable.

]]> By: Robert-H-Goretsky http://www.damninteresting.com/the-longest-standing-math-problem/#comment-17397 Robert-H-Goretsky Tue, 25 Sep 2007 02:56:09 +0000 http://www.damninteresting.com/?p=246#comment-17397 <i>Robert H. Goretsky says</i> I always found the mathematics behind RSA encryption to be quite interesting. (<a href="http://en.wikipedia.org/wiki/RSA" rel="nofollow">Wikipedia Link</a> ) What's crazy is how easy it is to generate a new public / private key pair, but how impossible it is to do the math 'backwards' to solve for the private key, given the public key... No computer in the world can solve these problems in our lifetime. <b><i> Comment by Robert H. Goretsky of Hoboken, NJ</b></i> Robert H. Goretsky says I always found the mathematics behind RSA encryption to be quite interesting. (Wikipedia Link ) What’s crazy is how easy it is to generate a new public / private key pair, but how impossible it is to do the math ‘backwards’ to solve for the private key, given the public key… No computer in the world can solve these problems in our lifetime. Comment by Robert H. Goretsky of Hoboken, NJ

]]> By: Ravage http://www.damninteresting.com/the-longest-standing-math-problem/#comment-17376 Ravage Sun, 23 Sep 2007 00:30:37 +0000 http://www.damninteresting.com/?p=246#comment-17376 [quote]Random5 said: "Maybe they will solve Goldbach's first when someone figures out how to predict prime numbers, but almost certainly not before."[/quote] Predicting prime numbers isn't the issue, it's determining if an arbitary odd number is prime. Given a prime i the next prime will be within 2i. The problem is testing that i-1 range, if you already know the list of primes below i even that's not real hard. [quote]Random5 said: “Maybe they will solve Goldbach’s first when someone figures out how to predict prime numbers, but almost certainly not before.”[/quote]

Predicting prime numbers isn’t the issue, it’s determining if an arbitary odd number is prime. Given a prime i the next prime will be within 2i. The problem is testing that i-1 range, if you already know the list of primes below i even that’s not real hard.

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