tag:blogger.com,1999:blog-72878725365270787812018-03-06T08:19:19.810-06:00Project PrimalityExploring prime numbers and their uses.John Y.http://www.blogger.com/profile/18231605600463483141noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-7287872536527078781.post-76210805512127313782011-08-17T17:00:00.010-05:002012-05-12T21:18:06.263-05:00Follow up: Simple Method for Fermat's Last Theorem and BeyondThis is a follow up to a previous blog entry. As previously conjectured:Where p is prime, 1 < a1 < a2 +...+ < am < p, if there exists {n, a1, a2, ..., am, p, x} that can satisfy (a1n + a2n +...+ amn) / pn-1 = x, then there exists, with the same number of terms (a) and exponent (n), {n, a1, a2, ... , am, b} that can satisfy a1n + a2n +...+ amn = bn.An example of a {n, a1, a2, a3, a4, p, x} that John Y.http://www.blogger.com/profile/18231605600463483141noreply@blogger.com0tag:blogger.com,1999:blog-7287872536527078781.post-23614340223955884752010-09-15T15:50:00.004-05:002010-09-16T21:48:23.853-05:00Collatz conjectureThe Collatz conjecture is also known as the 3n + 1 conjecture. Take any natural number n. If n is even, divide it by 2 to get n / 2, if n is odd multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1.For instance, starting with n = 19, generates the sequence 19, 58, 29, 88, 44, John Y.http://www.blogger.com/profile/18231605600463483141noreply@blogger.com0tag:blogger.com,1999:blog-7287872536527078781.post-11925306318510615682010-03-08T10:35:00.033-06:002011-08-18T03:10:40.768-05:00Simple Method for Fermat's Last Theorem and BeyondFermat's Last Theorem states that no three positive integers a, b, and c can satisfy the Diophantine equation an + bn = cn for any integer value of n greater than two.
All variables denote positive integers.
Where p is a prime number, n = 2, 1 < a < b < p,
(an + bn) can be evenly divided by pn-1.
Examples:
(22 + 42) / 51 = 4
(32 + 42) / 51 = 5
(22 + 102) / 131 = 8
Therefore, since (a2 + b2) John Y.http://www.blogger.com/profile/18231605600463483141noreply@blogger.com0