A064539 All...are divisible by 3 but why?

[Ignacio Larrosa Cañestro]

----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Friday, June 24, 2005 8:28 AM
Subject: A064539 All...are divisible by 3 but why?


> Dear seqfans,
> A064539= (*Numbers n such that 2^n+n^2 is
> prime*){1,3,9,15,21,33,2007,2127,3759,29355,34653,57285,99069}All
> except first are divisible by 3 but why? zak

Because n must be odd. Then 2^n = 2 (mod 3). By the other side, if n =/= 0 
(mod 3), we have that n^2 = 1 (mod 3).

Then n =/= 0 (mod 3)  ===>  2^n + n^2 = 0 (mod 3), for all n odd >= 1.

Saludos,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa at mundo-r.com