American Bank Offers $1M To Anyone Able To Solve This Mathematical Equation (Page 3)

i rather stay normal than loose my sanity of $1 million, cos i dont understand a jerk in that equation..............

Posted: at 6-06-2013 10:04 AM (10 years ago) | Gistmaniac

ok

Posted: at 6-06-2013 10:29 AM (10 years ago) | Gistmaniac

ok

Posted: at 6-06-2013 01:47 PM (10 years ago) | Gistmaniac

Maths di n abia

Posted: at 6-06-2013 04:16 PM (10 years ago) | Hero

relect the maths to profesor  chika obi  chikina................self thought

Posted: at 6-06-2013 07:20 PM (10 years ago) | Gistmaniac

let get my pen and paper

Posted: at 6-06-2013 07:59 PM (10 years ago) | Upcoming

chike obi

Posted: at 6-06-2013 09:51 PM (10 years ago) | Hero

1 mil day small, I go manage 3 mil, so if them day ready make una let know.

Posted: at 7-06-2013 03:47 AM (10 years ago) | Upcoming

Let there be positive integers n > 2 and A, B, C such that A^n+B^n=C^n, where C is the smallest possible. Then, by the Beal's conjecture with x = y = z = n there exists a base p > 1 dividing each of A, B, and C. However, then (A/p)^n+(B/p)^n=(C/p)^n, which contradicts C being the smallest possible. This proof by infinite descent shows the positive integers A, B, C, and n > 2 cannot exist, thus proving Fermat's last theorem.

A variant of this proof is as follows: If A^n+B^n=C^n, then by Beal's conjecture (if proven), A, B, and C have at least one prime factor in common. Let q be the product of all their common prime factors (that is, the greatest common divisor of A, B, and C). Then we would also have (A/q)^n + (B/q)^n = (C/q)^n in which (A/q), (B/q), and (C/q) have no common prime factor, which would be impossible by Beal's conjecture.

Here is ma phone number +221 77 810 79 49 for informations on how to make the payment, please make una hurry coz i was hospitalized after going this, unless say una wan make i die, and ma spirit go kill who ever that will chope money.

Posted: at 7-06-2013 03:54 AM (10 years ago) | Newbie

Hi,
The mathematical is not clear. Can you send me the equation more clearly.

Best Regards,

Steve

Posted: at 7-06-2013 07:46 AM (10 years ago) | Newbie