Number of Solutions to d(p)

[flouran]

What is the number of solutions d(p) of
N-n^2 \equiv 0 \pmod p

where p is a prime and n and N are positive and N => n?

[JustSam]

I'm not sure what you are really asking, but for each prime p, there are an infinite number of solutions (N,n) satisfying your criteria. Take n = 1, and N= k*p + 1, for k = 1, 2, 3, ...

[flouran]

I'm not sure what you are really asking, but for each prime p, there are an infinite number of solutions (N,n) satisfying your criteria. Take n = 1, and N= k*p + 1, for k = 1, 2, 3, ...
I have a trivial upper bound for d(p). That is, d(p) < p-1 for p\nmid N. I think that suffices for my usages of d(p) for now.

[flouran]

I am an idiot.
Hint: quadratic residue

[JustSam]

I still don't understand what you are asking.
What is the number of solutions d(p) of
N-n^2 \equiv 0 \pmod p

where p is a prime and n and N are positive and N => n?
Let p be a prime number. Define d(p) as the cardinality of

$
\{ (N,n) : N \ge 1, n \ge 1, N \ge n, N \equiv n^2 \pmod p \}


Clearly you intend something different.

[flouran]

I still don't understand what you are asking.

Let p be a prime number. Define d(p) as the cardinality of

$
\{ (N,n) : N \ge 1, n \ge 1, N \ge n, N \equiv n^2 \pmod p \}


Clearly you intend something different.

I can be clearer:
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?

[JustSam]

I can be clearer:
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?
Let p be a prime number. Define d(p) as:

$
\max_{N \ge 1} \left| \{ n \pmod p : N \equiv n^2 \pmod p \} \right|


Then d(2) = 1, d(p) = 2 for odd primes p.

[srijithju]

Let p be a prime number. Define d(p) as:

$
\max_{N \ge 1} \left| \{ n \pmod p : N \equiv n^2 \pmod p \} \right|


Then d(2) = 1, d(p) = 2 for odd primes p.

I dont think this is the right solution . Consider p = 3 and N = 2. We know that no square number is congruent to 2 modulo 3. So in this case d(3) = 0

[JustSam]

I dont think this is the right solution . Consider p = 3 and N = 2. We know that no square number is congruent to 2 modulo 3. So in this case d(3) = 0
Which is why it is important to have a precise definition of d(p). According to my second version, d(p) is the maximum number of solutions over all possible choices for N, so it makes sense that you can find particular values of N that have fewer than d(p) solutions.

Perhaps "d(p,N) = exact number of solutions" would be a more useful function.

[srijithju]

I think you mean that N is fixed , but

If the question is :
Find all N < p such that n^2 = N ( mod p) for some n ( By some n , I mean that corresponding to each N there will be one n) , then :

d(p) = greatest integer less than square root of p

[srijithju]

Which is why it is important to have a precise definition of d(p). According to my second version, d(p) is the maximum number of solutions over all possible choices for N, so it makes sense that you can find particular values of N that have fewer than d(p) solutions.

Perhaps "d(p,N) = exact number of solutions" would be a more useful function.

Considering d(p,N) :

take remainder of N divided by p . Let it be r.
Now if r is a perfect square , then d(p.N) will have infinite solutions of the form r + kp
else there is no solution

[flouran]

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.

[CRGreathouse]

Is m fixed or can it take any integer value? Are there any limits on the value of n or m?

[flouran]

Is m fixed or can it take any integer value? Are there any limits on the value of n or m?

There are no limits on the value of n or m. Nice to see you on the forums, by the way Charles!

[CRGreathouse]

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.

There are no limits on the value of n or m.

So there are, trivially, infinitely many solutions.

Nice to see you on the forums, by the way Charles!

You too.