2003-04-01, 14:35 | #1 |
Feb 2003
100000_{2} Posts |
if F(n) = B^(n+1) + 1, then F(a) | F(2ab+2b+a)
in trying to find primes of the form 101; 100001; 10000001; 100000001; etc., i've noticed that
if F(n) = B^(n+1) + 1, then F(a) divides F(2ab+2b+a), where B is any integer and a>=1 and b >=0. is there an existing theorem in number theory similar to this? |
2003-04-02, 01:28 | #2 |
Jan 2003
far from M40
5^{3} Posts |
Yes.
F(a) = B^(a + 1) + 1 F(2ab + 2b + a) = B^(2ab + 2b + a + 1) +1 = B^[(2b + 1)(a + 1)] + 1 = [B^(a + 1)]^(2b + 1) + 1 The general statements are: For x, y real and n a positive integer: 1) [x^(2n + 1) + y^(2n + 1)] / (x + y) ; x + y <> 0 = Sum(0 <= i <= 2n) (-1)^i * x^(2n - i) * y^i 2) (x^2n - y^2n) / (x + y) ; x + y <> 0 = Sum(0 <= i < 2n) (-1)^i * x^(2n - i - 1) * y^i 3) (x^n - y^n) / (x - y) ; x - y <> 0 = Sum(0 <= i < n) x^(n - i - 1) * y^i You can proof this by full induction over n. In your case statement 1) holds with x = B^(a + 1), y = 1 and n = b. Under the presumption that a, b and B are integers, you get the conjectured divisability. Benjamin |
2003-04-02, 10:43 | #3 |
Feb 2003
40_{8} Posts |
Thanks Benjamin.
Ray |