The 4^n Polynomial conjecture RESOLVED

[cino hilliard]

Hi, Seq Fans
Thanks all for your help on my conjecture.

>From: Richard Guy <rkg at cpsc.ucalgary.ca>

>If  n  is even, then  4^n + ... + 1 = (4^(n+1) - 1)/3

The expansion
(1) a^n - b^n = (a+b)(a^(n-1) - a^(n-2) + . . . - 1)  for odd values of n

The sum of a finite geometric series
(2)  S_n+1 = a + aq +a q^2 +a q^3 +a q^4  + . . . + a q^(n) = 
a(1-q^(n+1))/(1-q)
if q = 4,a=1, we have
S_n+1 = 1 + 4 + 4^2 + 4^3 + 4^4 + ... + 4^n = (1-4^(n+1))/(1-4) = 
(4^(n+1)-1)/(3)
if n+1  is even  then S_n+1 = (2^(n+1) -1)(2^(n+1)+1) /3 as stated below by 
R.
if n+1 is odd then S_n+1 = (4+1)(4^n - 4^(n-1) - . . . - 1)/3 from (1) above 
and 5 is a factor of S_n+1
Thus, the sum of the geometric series 4^n + ... + 1 is composit for all n>0 
as was desired to show.

It remains, of course to prove (1) and (2).


 >
>= (2^(n+1) - 1)(2^n+1) + 1)/3.   R.
>

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