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Banker Offers $1M To Solve Beal Conjecture
oxide7 writes "A Texas banker with a knack for numbers has offered $1 million for anyone who can solve a complex math equation that has stumped mathematicians since the 1980s. The Beal Conjecture states that the only solutions to the equation A^x + B^y = C^z, when A, B and C are positive integers, and x, y and z are positive integers greater than two, are those in which A, B and C have a common factor. Like most number theories, it's "easy to say but extremely difficult to prove.""
Only if x=y=z. For instance, somebody above suggested 3^3 + 6^3 = 3^5 (27+216=243). If we factor out the common 3, we get 3^2 + 2*(6^2) = 3^4 (9+72=81), which no longer has the right form because 72 is not a power of any number.
If x=y=z, and if A^x+B^x=C^x where A,B,C had the same greatest common factor n, then you could divide all three numbers by n^x and get a new formula (A/n)^x+(B/n)^y=(C/n)^z where A/n, B/n, and C/n had no common factor, and if Beal's conjecture is true then these numbers cannot exist if x>2. Therefore A^x+B^x=C^x has no nontrivial solution for x>2, which is Fermat's last theorem.