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.

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TopNewestThere 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, 9 months ago

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– Aditya Parson · 3 years, 9 months ago

what if gcd(a,b) is 1?Log in to reply

– Fiyi Adebekun · 3 years, 9 months ago

We have to remember. The conjecture is not asking for \(A,B,C\) are COPRIME. It is asking for the three numbers to have a common prime factor. gcd(a,b)=1 only determines is the two numbers are coprimeLog in to reply

– Aditya Parson · 3 years, 9 months ago

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.Log in to reply

– Fiyi Adebekun · 3 years, 9 months ago

Ah. YesLog in to reply

Wait maybe we can use fermat's little theorem to turn it into a more simpler problem.. – Taehyung Kim · 3 years, 9 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, 9 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, 9 months ago

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Also, the prime factorization of A, B, and C must contain a common number "n" – Fiyi Adebekun · 3 years, 9 months ago

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I'm not really sure what this problem takes. But I'm thinking that \(C^z\) must be an integer power of \(A^x+B^y\), We have

\(log_{A^x+B^y}C^z\) must be an integer. – Fiyi Adebekun · 3 years, 9 months ago

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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, 9 months ago

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– Aditya Parson · 3 years, 9 months ago

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.Log in to reply

– Shivang Jindal · 3 years, 9 months ago

Ok then prove it :DLog in to reply

– Aditya Parson · 3 years, 9 months ago

\(A,B,C\) 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.Log in to reply

– Aditya Parson · 3 years, 9 months ago

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.Log in to reply

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– Aditya Parson · 3 years, 9 months ago

Yes, that is again a part of it which you proved.Log in to reply

just read how technical the proof of fermats last theorem by andrew willes first before someone considers your boring proof!!! – Mharfe Micaroz · 3 years, 9 months ago

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– Fiyi Adebekun · 3 years, 9 months ago

Dude, chill. If you hate math, then leave. This is for people who appreciate the beauty of mathematics. I would highly suggest that you quit if you find math boring. And if it is boring, why would you look at this thread? Because it said one million dollars? Let us have our own fun. Go do something else. Math isn't for you.Log in to reply