Beal's Conjecture - a generalized Fermat's Last Theorem.
Banker Andrew Beal offers $1m for maths solution.
BEAL'S CONJECTURE:
If
$latex A^{x} + B^{y} = C^{z} $
where A, B, C, x, y and z are positive integers and
x, y and z are all greater than 2,
then A, B and C must have a common prime factor.
http://metro.co.uk/2013/06/05/texas-banker-offering-1m-for-answer-to-30-year-old-maths-problem-3829105/
This is an excellent post on Beal's Conjecture.
ReplyDeleteI have a really stupid question.
ReplyDeleteBased on the criteria:
If
A^{x} + B^{y} = C^{z}
A) where A, B, C, x, y and z are positive integers and
B) x, y and z are all greater than 2,
then A, B and C must have a common prime factor.
Why can't it just be disproved as such:
2^3 + 2^3 ≠ 2^3
(or any combination where A=B=C && x=y=z)
?
Clearly, it can't be that simple, but what criteria am I missing? Is it just implied that well duh, you can't have A=B=C AND x=y=z ?
In your examples, it's okay to have A=B=C,
and for x=y, and y=z
but not x=y=z.
Wait ...
that has to do with Fermat's Theorem being proven that
A^n + B^n ≠ C^n for integers greater than 2.
Never mind.
It's late. I haven't studied math since high school.
It is a million dollar question...
ReplyDelete