Not enough room in comment
I have discovered an ingenious proof for this, but alas, there is insufficient room in this comment box to elaborate.
200TB proof cracks puzzler
Three computer scientists from the University of Texas have broken the record for the largest ever mathematical proof. The Boolean Pythagorean triples problem - first proposed in the 1980s - was cracked by supercomputers and required 200TB of information to solve. This is estimated to be more than 16 million times more data …
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Friday 27th May 2016 12:13 GMT Anonymous Coward
Re: Not enough room in comment
From a quick google then looks like the Boolean Pythagorean property was a "conjecture" and the "proof" looks to me more of a case fo showing that it doesn't hold by finding a counter example by what looks liek a brute force approach of trying all colourings of truples < a given N and checking if for any the conjecture holds ... and when N got to 7825 they showed that no colouring where the conjecture held was possible and hence they had "proved" it false.
So interesting in terms of running huge data problems but from a mathematical point of view its probably less interesting - its much more interesting to show that somethine must always holds.
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Friday 27th May 2016 13:06 GMT Len Goddard
It is a proof
Proof of impossibility by exhaustive examination of all possible cases is about as strong a proof as you can get since it cannot be subject to logical flaws.
It may be intellectually unsatisfying and lacking general applicability but it is still a valid proof. Hopefully at some stage in the future someone will produce a formal logic proof which casts light on the underlying structures. The fact that the validity or otherwise of the conjecture is now established may actually help this process in that future workers will know what they need to (dis)prove. This has happened before with computer proofs.
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Friday 27th May 2016 14:58 GMT Anonymous Coward
Re: It is a proof
Proof of impossibility by exhaustive examination of all possible cases is about as strong a proof as you can get since it cannot be subject to logical flaws.
Sorry, ALL?
That's not what they did here. I don't recommend trying to exhaustively examine all cases when you're talking about and unconstrained problem with integers (no maximum given)
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Friday 27th May 2016 18:00 GMT Len Goddard
Re: It is a proof
Lazy English. They tested all possible mappings for integers < an upper bound N to see if a mapping existed where there was a unicolour triplet. They then increased N until they found a value where, despite testing all possible mappings, there was no mapping with such a triplet. Thats the ALL is was referring to.
Actually, everywhere I have looked this thing is specified as a question, not an assertion, so this team has not actually proven a theorum so much as answered a question. Now that the question has been answered we have a theorum which states that you can't do it and that is what now needs to be proven from the tenets of number theory.
I accept that you could not do an exhaustive analysis of all possible mappings for all possible numbers. The approach taken by this team could only produce a result if the assertion that you cannot partition the integers and find a triplet in one partition was false for a "reasonably small" upper bound (where reasonably small increases over time along the lines of Moore's Law).
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Friday 27th May 2016 14:01 GMT Lee D
Re: Not enough room in comment
It's actually a proof by contradiction.
The conjecture asserts "we can always colour a, b, c such that a^2 + b^2 = c^2 and a, b, c are never the same colour". The proof was finding an instance where it WASN'T true. Thus the "proof" is provably incorrect or - to put it another way - you're proved the opposite (there there ISN'T a way to colour these things differently like that in all instances).
This isn't like the four-colour theorem where people narrowed it down to a special set of cases and then exhaustively eliminated them all by computer search. It's a case where - after 7000-something numbers - brute-force cannot find a single solution thus "proving" that you can't always colour in different colours.
If the number hadn't been 7000 but 2 million, it would still be calculating and the "conjecture" would still look as if it were true.
P.S. In case you hadn't guessed, I'm a mathematician. This isn't a mathematical proof. This is finding an opposite-proof-by-counterexample by brute force. They conjecture that the same is true for any amount of colours. To prove that, you will need "real" maths. And that's not been done and probably won't for a long time yet.
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Friday 27th May 2016 16:52 GMT Mpeler
Re: Not enough room in comment
I have also discovered a proof, indeed 151 proof. Sadly, the flask will not fit in this comment box.
I have been rum-inating on possible means of fitting it, but they cola-psed (hic, er, sic)...
(My apologies for the beer icon - Bacardi and beer - just NO. Cola, ahh, for college days, room spinning...).
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Biting the hand that feeds IT
