Primes of the form (3^n + 5^n)/2.

[cino hilliard]

>From: zak seidov <zakseidov at yahoo.com>
>To: seqfan at ext.jussieu.fr
>Subject: RE: Primes of the form (3^n + 5^n)/2.
>Date: Sun, 10 Sep 2006 13:22:14 -0700 (PDT)
>
>Thanks to all responding!!!
>
>Zak
>
>Just submited:
>
>
>%I A000001
>%S A000001 19, 421, 10039, 95383574161, 2384331073699
>%N A000001 Primes of the form (3^n+5^n)/2^3
>%C A000001 Corresponding values of n's are 3, 5, 7,
>17, 19.
>There is only one prime of the form (3^n+5^n)/2^2
>(n=1, p=2),
We all know that  a^n+b^n is div by a+b if n is odd. Call this theorem 1

then a=3,b=5 => 8 divides 3^n + 5^n for odd n and (3^n + 5^n)/4 is div by 2. 
if n=1
this is true.
Except for this case n is not odd since n odd => 3^n+5^n is div by 8 
(theorem 1) =>
(3^n+5^n)/4 =2m not prime.

So n is even
Rewrite 3 = 2+1 and 5=4+1. Expanding, (2+1)^(n) + (4+1)^n we get 4H + 4K + 2 
which
is not divisible by 4.
Therefore the statement
There is only one prime of the form (3^n+5^n)/2^2
is true


>and there are only three known primes of the form
>(3^n+5^n)/2^1:
>17, 353, 198593 (for n=2,4,8). Cf. A074606 3^n + 5^n

n must be of the form 2^k
so n is even. If n is of the form 2k(2m+1) then
3^(2k)(2m+1) + 5^(2k)(2m+1) is div by (3^2^k + 5^2k) by theorem 1.

Thus,
n is of the form 2^k and that is all we need to test for.
Cino