Comments on: The Longest-Standing Math Problem https://www.damninteresting.com/the-longest-standing-math-problem/ Fascinating true stories from science, history, and psychology since 2005 Thu, 24 Oct 2019 14:09:02 +0000 hourly 1 https://wordpress.org/?v=6.7.1 By: Cecil https://www.damninteresting.com/the-longest-standing-math-problem/#comment-39529 Wed, 14 Jan 2015 14:32:47 +0000 https://www.damninteresting.com/?p=246#comment-39529 The algebraic solution ends in 5 places(_ _ _ _ _) answer, however the process is only 4 places( _ _ _ _ ) long. it takes a lifetime to solve.

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By: Vestal https://www.damninteresting.com/the-longest-standing-math-problem/#comment-38953 Thu, 29 May 2014 12:02:28 +0000 https://www.damninteresting.com/?p=246#comment-38953

mrBlah said: “What about 1 not actually being prime, but special. A prime is divisible by itself and 1 only; two numbers. The number 1 is only divisible by itself; only one number.”

It took eight and a half years but here it is….finally.

That’s why 1 is the loneliest number.

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By: Benschop https://www.damninteresting.com/the-longest-standing-math-problem/#comment-26583 Tue, 26 Jul 2011 12:48:02 +0000 https://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 . . . .

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By: Benschop https://www.damninteresting.com/the-longest-standing-math-problem/#comment-26582 Tue, 26 Jul 2011 12:37:20 +0000 https://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…;-(

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By: mathematics101 https://www.damninteresting.com/the-longest-standing-math-problem/#comment-26396 Thu, 17 Feb 2011 07:44:30 +0000 https://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

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By: Benschop https://www.damninteresting.com/the-longest-standing-math-problem/#comment-25437 Wed, 21 Oct 2009 07:53:18 +0000 https://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

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

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By: catra7000 https://www.damninteresting.com/the-longest-standing-math-problem/#comment-20739 Thu, 27 Mar 2008 21:17:42 +0000 https://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

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By: george9589 https://www.damninteresting.com/the-longest-standing-math-problem/#comment-20406 Tue, 11 Mar 2008 10:28:20 +0000 https://www.damninteresting.com/?p=246#comment-20406 [quote]Eric Leeson said: “I hate math.”[/quote]

I agree with you one hundred percent haha

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By: Ravage https://www.damninteresting.com/the-longest-standing-math-problem/#comment-17430 Thu, 27 Sep 2007 15:49:15 +0000 https://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.

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