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Topic: Is this well known?
Replies: 7   Last Post: Jul 26, 2011 11:05 AM

 Messages: [ Previous | Next ]
 Bill Dubuque Posts: 1,739 Registered: 12/6/04
Re: Is this well known?
Posted: Jul 26, 2011 11:05 AM

quasi <quasi@null.set> wrote:
>Danny73 <fasttrack2a@att.net> wrote:
>

>> 2^k +1, k = 16,+32,+32,+32,+32,+ --->oo
>>
>> After each accumulated sum before each
>> comma where k is an exponent of 2.
>>
>> Then (2^k +1) == 0 (mod (2^16 +1))

>
> Yes, it's a triviality.
> It follows from the fact that if n is odd,
>
> x^n + 1 is divisible by x + 1
>
> by virtue of the polynomial identity (assuming n is odd)
>
> x^n + 1 = (x + 1) (x^(n-1) - x^(n-2) + ... - x + 1)

which follows by specializing f(x) = x^n, k = -1 in the

FACTOR THEOREM x-k | f(x)-f(k) for k in Z, f in Z[x]

Or, mod x+1: x = -1 => x^n = (-1)^n = -1 by n odd.

--Bill Dubuque

Date Subject Author
7/23/11 dan73
7/23/11 Brian Q. Hutchings
7/23/11 dan73
7/23/11 dan73
7/23/11 quasi
7/24/11 dan73
7/24/11 quasi
7/26/11 Bill Dubuque