Hypothesis

[Jonathan Post]

Exception 38.

After that, three consecutive primes are all odd.  Hence each is 2n+1
for some n.

(2a+1)^2 + (2b+1)^2 + (2c+1)^2 =
(4a^2 + 4a + 1) + (4b^2 + 4b + 1) + (4c^2 + 4c + 1)
= 4(a^2 + b^2 + c^2 + a + b + c) + 3.

Now do the same elementary algebra based on the fact that every prime
after 2 is of the form 4n+1 or 4n+3.

(4n+1)^2 = 16n^2 + 8n + 1 = 4(4n^2 + 2n) + 1
(4n+3)^2 = 16n^2 + 24n + 9 = 4(4n^2 + 6n + 2) + 1.

Some cleaning up to do.  Am I missing something?

On 9/28/07, Artur <grafix at csl.pl> wrote:
> Dear Seqfans.
> Is somebody able to proof hypothesis that sum of squares of three
> [consecutive] primes A133529
> <http://www.research.att.com/%7Enjas/sequences/A133529> is divisable by
> 3 (with exception 83)
> or countersample that isn't other than 83 ?
> BEST WISHES
> ARTUR
>