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# 1 million dollar prize to prove or disprove the Beal Conjecture

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, 9 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. · 3 years, 9 months ago

Comment deleted Jun 15, 2013

what if gcd(a,b) is 1? · 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 coprime · 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. · 3 years, 9 months ago

Ah. Yes · 3 years, 9 months ago

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

Maybe Fermat proved his last theorem using the Beal equation? (see if you can find out how) · 3 years, 9 months ago

thanks a lot for giving such a great problem i'll surely give it a try · 3 years, 9 months ago

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

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. · 3 years, 9 months ago

Comment deleted Jun 14, 2013

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. · 3 years, 9 months ago

Comment deleted Jun 14, 2013

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. · 3 years, 9 months ago

Ok then prove it :D · 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. · 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. · 3 years, 9 months ago

Comment deleted Jun 15, 2013

Yes, that is again a part of it which you proved. · 3 years, 9 months ago

too arrogant..

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