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1 million dollar prize to prove or disprove the Beal Conjecture

Main post link -> http://newsfeed.time.com/2013/06/11/solve-this-math-problem-win-a-million-bucks/?hpt=hp_t3

We know many of you are into Fermat's last theorem and thought you might be interested in the Beal Conjecture as well. Expanding the frontier of number theory probably isn't the easiest way to make a million bucks, or even the best reason to work on that kind of problem. That is an impressive prize though, here is the prize page from the American Mathematical Society.

Note by Peter Taylor
3 years, 11 months ago

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There are 6(originally 7) unsolved conjectures known as the millennium prize problems and for solving each,the Clay Mathematical Institution offers a million bucks.The Poincare Conjecture is the only one that has been proved. Edward Elric · 3 years, 11 months ago

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Comment deleted Jun 15, 2013

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@Siddharth Kumar what if gcd(a,b) is 1? Aditya Parson · 3 years, 11 months ago

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@Brilliant Member Prove/ Disprove. The conjecture has to be proved first, to eliminate the possibility that \(gcd(A,B) \neq 1\). And if this equation also satisfies for \(gcd(A,B)=1\), then \(gcd(A,B,C)\) will of course be \(1\). So, you have to prove that the equation is not satisfied for any pair of CO-PRIME integers. Aditya Parson · 3 years, 11 months ago

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Wait maybe we can use fermat's little theorem to turn it into a more simpler problem.. Taehyung Kim · 3 years, 11 months ago

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Maybe Fermat proved his last theorem using the Beal equation? (see if you can find out how) Taehyung Kim · 3 years, 11 months ago

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thanks a lot for giving such a great problem i'll surely give it a try Tejas Kasetty · 3 years, 11 months ago

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Comment deleted Jun 14, 2013

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@Siddharth Kumar If \(A\) and \(B\) are primes\(\neq 2\), \(C\) will always be an even number \(>2\) and thus cannot be prime.

Similarly play around with the equation and it will be apparent that the equation will not hold for set of primes. Aditya Parson · 3 years, 11 months ago

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Comment deleted Jun 14, 2013

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@Siddharth Kumar I already said that it will not hold if you take prime numbers greater than 2. I just gave you the simplest way to prove it. Aditya Parson · 3 years, 11 months ago

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@Aditya Parson Ok then prove it :D Shivang Jindal · 3 years, 11 months ago

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@Shivang Jindal \(A,B,C\) can be co-prime as well. The equation will not hold if all \(3\) are primes. As such, we still need to prove that this equation is valid/not valid when if \(A,B,C\) are co-prime. Then we can conclude if there is a common factor of the three terms or not. Aditya Parson · 3 years, 11 months ago

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@Aditya Parson Also, there are cases when \(A,B,C\) are equal to \(2\) There is still a lot of work to be done to prove/disprove this equation. Aditya Parson · 3 years, 11 months ago

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Comment deleted Jun 15, 2013

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@Siddharth Kumar Yes, that is again a part of it which you proved. Aditya Parson · 3 years, 11 months ago

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@Aditya Parson too arrogant..

just read how technical the proof of fermats last theorem by andrew willes first before someone considers your boring proof!!! Mharfe Micaroz · 3 years, 11 months ago

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