Dan, A prime p will divide n^2+n+1 for some n if and only if n^2+n+1 = 0 has a solution modulo p if and only if GF(p) contains a primitive third root of unity (p=3 excluded) if and only if p = 1 modulo 3. This observation, by the way, can be used to give an elementary proof (a la Euclid) that there are infinitely many prime = 1 mod 3. Victor A similar argument works for other cyclotomic polynomials. On Fri, Apr 4, 2008 at 5:24 PM, Dan Asimov <dasimov@earthlink.net> wrote:

Playing around with the cyclotomic polynomial

P(n) = n^2 + n + 1

I noticed that there are many primes that cannot divide this quantity for any n (e.g., 5, 11). I suppose this is probably related to quadratic reciprocity.

Can it be determined just which primes can't divide n^2 + n + 1 for any n ? What about other polynomials, cyclotomic and otherwise?

--Dan

E.g.:

For 5, the residues of P(n) are 1,-2 (twice each) and 2 (once).

For 11, the residues of P(n) are 1,2,3,-1,-4 (twice each) and -2 (once).

Why these funny patterns?

_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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