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

On Fri, 29 Jun 2007 22:53:43 -0500, quasi <quasi@xxxxxxxx> wrote:

On Fri, 29 Jun 2007 22:49:24 -0500, quasi <quasi@xxxxxxxx> wrote:

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

quasi wrote:

Conjecture (1a):

There does not exist an integer m>2 and a univariate,
degree 1 polynomial f with integer coefficients such
that, for all integers n, all prime factors of f(n) are
congruent to 1 mod m.

agreed , and relatively trivial i guess.

Actually, I initially assumed this would be easy to prove but I'm no
longer so sure about that. I haven't really played with it, but I
don't see an immediate proof. Also, except for your attempted proof
below, no one has yet supplied either a proof, a disproof, or a
reference.

since there are an infinite set of primes q mod m these already
do not factor like 1 mod m. and all lineairy forms devidable bye
q x + m at least for one value n , therefore do not completely
factor as 1 mod m.

this seems to me as a proof or as good as.
(at first sight)

While I can't quite follow the logic of the proof you sketched above,
it doesn't look like a valid proof to me.

The most puzzling is the last line, where it appears you conclude the
opposite of what you're trying to prove.

Ok, forget the above remark -- I understand the last line now.
Unfortunately, I don't understand the logic that precedes it.

Ok, I now understand it.

While it needs a little work, it's basically the right strategy.

My proof of conjecture (1a), which I'll try to post tomorrow, works
essentially along the lines of the proof you outlined.

quasi
.

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