Re: Divisibility problem

In article
<88e82b00-bba5-40e9-996b-b983796dc739@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Xan <xancorreu@xxxxxxxxx> wrote:

On Jan 16, 12:14 am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article
<0b5f077a-1973-4261-812b-8e01a23a5...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,



Xan <xancor...@xxxxxxxxx> wrote:
On Jan 15, 3:39 am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article
<fc62f3bb-420e-496d-867e-476693271...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

Xan <xancor...@xxxxxxxxx> wrote:
Hi,

Let be p(x), q(x) in Z[x] (polynomials with integer coeficients)
And let be the following set:
X_k = X_k(p,q) = {(s, n) in Z^2 | p(s) divides q(n), p(s+1) divides
q(n
+1), ...., p(s+k) divides q(s+k)}

For example if p(x) = x^3 and q(x) = x, then we have that X_k = {(s,
n) in Z^2 | s^3 divides n, (s+1)^3 divides n+1, ... (s+k)^3 divides n
+k}

Well, I have the problem of study the cardinality of X_k, for p(x)
and
q(x) fixed polynomials.

I conjecture the following, but I don't know how to prove or
disprove:

Conjecture 1: There is k_0 in N such that X_k = emptyset
Conjecture 2: The k such that X_k is the emptyset are the majorty
Conjecture 3: There is k_0 in N such that for all k>=k_0, X_k is the
emptyset

Can anyone give some hint?

Thanks a lot in advance,
Xan.

You're going to want some conditions on p & q.
At the very least, you're going to want to assume
that q is not a (polynomial) multiple of p.

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

Yes, you can assume that p and q are coprime polynomials.

Here's another case you might want to rule out; suppose p is
the constant polynomial, 2, and q(x) = x (x + 1). Then
p(s) divides q(n) for all s and all n.

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

Okay, if p=2, q(x) = x (x+1), then X_k = Z^2
So we have that in this case, all the conjectures are false.

But what about p and q not constants and coprime polynomials?
It's in the case that I'm in trouble

Regards,
Xan.

Say p(x) = x and q(x) = (x - 1) (x - 2).

Then p(5) divides q(6), p(5 + 1) divides q(6 + 1), ...,
p(5 + k) divides q( 6 + k) for all k.

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

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