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Old 2003-04-01, 14:35   #1
1260
 
Feb 2003

1000002 Posts
Default 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?
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Old 2003-04-02, 01:28   #2
S80780
 
Jan 2003
far from M40

53 Posts
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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
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Old 2003-04-02, 10:43   #3
1260
 
Feb 2003

408 Posts
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Thanks Benjamin.


Ray
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