[seqfan] Re: p^2+6p+14

[Andrew Weimholt]

On 1/26/10, vincenzo.librandi at tin.it <vincenzo.librandi at tin.it> wrote:
>
>  because
>
>  p^2+6p+14 (with p prime odd)
>
>  is prime for p=3
>
>  and for all prime is divisible for 3 ?
>

There's nothing deep here. It is trivial to show
that 3|(p^2+6p+14) when p is a prime other than 3.
(And it has nothing to do with p^2+6p+14 being prime when
p is 3)

p^2 + 6p + 14 == p^2 + 2 mod 3

If p =/= 3 and p is prime, then p == +/- 1 mod 3,
so p^2 == 1 mod 3
therefore p^2 + 2 == 0 mod 3

Again, there is nothing special about p^2 + 6p + 14.
An infinite number of similar second order polynomials
can be constructed to produce the same result.

Andrew