Hi,
Let p be a prime number. I need to show that
2n+1
p + 1
K = ---------------
p + 1
where n >= 0 is an integer. This seems to work fine with examples
(i.e. (5^3 + 1)/(5+1) = 21)
but I'm not sure how to show this. Induction on n:
Case 0:
0 + 1
p + 1
----------------- = 1
p + 1
inductive step:
suppose it is true for k,
2k+1
p + 1
---------------
p + 1
we need to show it is true for k + 1:
2(k + 1) + 1 2k + 3
p + 1 p + 1
------------------------ = ------------------
p + 1 p + 1
but I don't see how to work the induction hypothesis in there.
The formula seems reminiscent of the Fundamental Theorem of Algebra
but I don't see how to use it.
Thank you,
-e
Paul Sperry - 30 Sep 2005 04:04 GMT
> Hi,
>
[quoted text clipped - 6 lines]
>
> where n >= 0 is an integer.
> [...]
No matter what, -1 is a root of x^(2n + 1) + 1 so x + 1 is a factor.
Let x = p (or any other integer).

Signature
Paul Sperry
Columbia, SC (USA)
Arturo Magidin - 30 Sep 2005 04:04 GMT
>Hi,
>
[quoted text clipped - 6 lines]
>
>where n >= 0 is an integer.
Hint: How much is
(x+1)(x^{2n} - x^{2n-1} + x^{2n-2} -...-x + 1)
?

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======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
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======================================================================
Arturo Magidin
magidin@math.berkeley.edu
cndc - 30 Sep 2005 04:07 GMT
Thank you Paul and Arturo! The root method seems easiest so I shall
use it.
-e
Paul Sperry - 30 Sep 2005 05:04 GMT
> Hi,
>
[quoted text clipped - 34 lines]
>
> but I don't see how to work the induction hypothesis in there.
(p^(2k + 3) + 1)/(p + 1) =
(p^(2k + 1) + 1)/(p + 1) + p^(2k + 1)((p^2 - 1)/(p + 1)). And, of
course, p^2 - 1 = (p + 1)(p - 1).

Signature
Paul Sperry
Columbia, SC (USA)
Virgil - 30 Sep 2005 06:06 GMT
> Hi,
>
[quoted text clipped - 40 lines]
> Thank you,
> -e
Let f(x) = x^(2*n+1) + 1, for anypositive integer n.
Then f(-1) = 0, so that f(x) is divisible by x - (-1).