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Math Forum / Mathematics / Undergraduate Math / September 2005



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proving something is an integer

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cndc - 30 Sep 2005 02:35 GMT
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).

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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)

?

Signature

======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
   --- Calvin ("Calvin and Hobbes")
======================================================================

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).
 
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