Re: all prime factors of f(n) are mutually congruent mod m

On Fri, 29 Jun 2007 18:14:14 EDT, tommy1729 <tommy1729@xxxxxxxxx>
wrote:

quasi wrote:

Conjecture (1):

If for some integer m>1, there exists a univariate,
nonconstant polynomial f with integer coefficients
such that, for all integers n, all prime factors of f(n)
are mutually congruent mod m, then they are
all congruent to 1 mod m.

if q and m are relatively prime , there exist infinite
primes of q mod m.

Right.

(can be proven in at least 3 ways but not easy proofs)

Ok.

not all polynomials are solvable in roots and diophantic
forms above deg 4 can be unsolvable ( as in solving the
roots , how nice coincidence at first sight ).

The above does not seem relevant, but maybe I'm missing your point.

so why stop q at 1 ??

I think this is your real point, right?

You are wondering why I focused on residues of 1, rather than
arbitrary residues.

As a test case, try this ...

Can you find a nonconstant univariate polynomial f with integer
coefficients such that, for all integers n, all prime factors of f(n)
are congruent to 2 mod 3?

If my conjecture (1) is true, then the answer is no.

i admit i havent entirely investigated this ,
but some statistical data implies otherwise...

counterconjecture :

some septic polynomial has primefactors
only of the form q mod m where q and m
are fixed and prime.

(yes i assume prime not just relatively prime ,
that seems to weak for a condition )

Well, go for it.

But I'll bet against.

quasi
.

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