Number of Solutions to d(p) - Page 2

CRGreathouse · 2009-08-30, 03:51

Quote:

Originally Posted by flouran

Let F(n) be a polynomial of degree g => 1 with integer coefficients. Let d(p) denote the number of solutions to the congruency $F(n) \equiv 0 \pmod p$ for all primes p (and suppose that d(p) < p for all p). We may take F(n) = N-n^2, where N is an integer greater than (or equal to) n. What is d(p) then in this case?

I still don't understand. Are you saying that d is a function of p and N, or that you can take N arbitrary, or take all N, or something else?

CRGreathouse · 2009-08-30, 03:54

Quote:

Originally Posted by flouran

I think d(2) = 1, and d(p) = 2 for all odd primes p.

Follwing JustSam?

flouran · 2009-08-30, 05:25

Quote:

Originally Posted by CRGreathouse

Follwing JustSam?

Yeah, apparently he understands the problem much better. I've already explained it as well as I can (and he gets it, and I get it; so it's all good).

Nice to see that you stalk my posts from other websites

S485122 · 2009-08-30, 06:25

Quote:

Originally Posted by flouran

Let F(n) be a polynomial of degree g => 1 with integer coefficients. Let d(p) denote the number of solutions to the congruency $F(n) \equiv 0 \pmod p$ for all primes p (and suppose that d(p) < p for all p). We may take F(n) = N-n^2, where N is an integer greater than (or equal to) n. What is d(p) then in this case?

If one takes a fixed N, for d(7) for instance we have the following solutions for N-n² $\equiv$ 0 mod 7 with n<=N depending on the value of N mod 7.
N=7*k => n=7*k' => d(7)=N/7+1
N=7*k+1 => n=7*k'+1 => d(7)=(N-1)/7+1
N=7*k+2 => n=7*k'+3 and n=7*k"+4 => d(7)=2*(N-2)/7
N=7*k+3 => d(p)=0
N=7*k+4 => n=7*k'+2 and n=7*k"+5 => d(7)=2*(N-4)/7+1
N=7*k+5 => d(p)=0
N=7*k+6 => d(p)=0
Another thing for any p if N = k*p and p-1 <= k [equivalent to p*(p-1)<=N], it is not true that d(p)<p. So this would imply that fixed N of arbitrary size is not part of your problem...

Quote:

Originally Posted by CRGreathouse

I still don't understand. Are you saying that d is a function of p and N, or that you can take N arbitrary, or take all N, or something else?

I concur, Flouran, you should better state the problem.

Jacob

flouran · 2009-08-30, 06:28

Quote:

Originally Posted by S485122

I concur, Flouran, you should better state the problem.

Maybe next time. I already got my answer

S485122 · 2009-08-30, 07:48

Quote:

Originally Posted by flouran

Maybe next time. I already got my answer

Can you explain your answer so that we can understand what the question was ?

Jacob

flouran · 2009-09-07, 03:56

Quote:

Originally Posted by S485122

Can you explain your answer so that we can understand what the question was ?

Jacob

d(p) is at most 2. For me, all that matters is an upper-bound (an equality would be great, but an upper-bound works fine as well).

CRGreathouse · 2009-09-07, 04:05

Quote:

Originally Posted by flouran

Nice to see that you stalk my posts from other websites

Yes, I signed up for an account, made thousands of posts, and was made a Science Advisor (and HH) just so I could stalk you.

Apart from the chronological impossibility, I'd buy that.

flouran · 2009-09-07, 23:04

Quote:

Originally Posted by flouran

d(p) is at most 2. For me, all that matters is an upper-bound (an equality would be great, but an upper-bound works fine as well).

Would d(p) be different if it was instead the number of solutions to:
$F(n) \equiv 0 \pmod p$ ? Here F(n) = n - q where q = m^2 is a perfect square where m is a natural number.

S485122 · 2009-09-08, 05:48

Flouran,

Once again you come up with the same kind of question. You do not give enough information for your question to be meaningful. As your new question is stated it makes no difference whether F(n)=N-n^2 or n-q^2.

You must plainly state the conditions for your problem because it is obvious that the way you explained your previous problem had no relation with the answer you gave. For instance : are the solutions modulo P as well ? What are constants, variables, what are the relations between them ?

When I asked an explanation of your answer, what I asked for was why and how your answer fitted the problem, not a restatement of the answer.

Jacob

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2009-09-08, 05:48	#21
S485122 "Jacob" Sep 2006 Brussels, Belgium 2·3²·5·19 Posts	Flouran, Once again you come up with the same kind of question. You do not give enough information for your question to be meaningful. As your new question is stated it makes no difference whether F(n)=N-n^2 or n-q^2. You must plainly state the conditions for your problem because it is obvious that the way you explained your previous problem had no relation with the answer you gave. For instance : are the solutions modulo P as well ? What are constants, variables, what are the relations between them ? When I asked an explanation of your answer, what I asked for was why and how your answer fitted the problem, not a restatement of the answer. Jacob