all prime factors of f(n) are congruent to 1 mod 3

In article <20070627.164954@xxxxxxxx>, rob@xxxxxxxxxxxxxx (Rob Johnson)
wrote:

In article <4bp583tllhfa1d4ib4lajrid87p2oqgpqu@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:

What about m=3? Does there exist a univariate polynomial f with
integer coefficients such that, for every integer n, all prime factors
of f(n) are congruent to 1 mod 3?

Say p = 2 mod 3 is prime, consider the quadratic character of -3 mod p.

If p = 1 mod 4 then (-3|p) = (3|p) = (p|3) = (2|3) = -1,
where (a|p) is the Legendre symbol.

If p = 3 mod 4 then (-3|p) = - (3|p) = (p|3) = -1.

So n^2 + 3 can't be divisible by any such prime p.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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