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flouran
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What is the number of solutions d(p) of
[tex]N-n^2 \equiv 0 \pmod p[/tex]
where p is a prime and n and N are positive and N => n?
[tex]N-n^2 \equiv 0 \pmod p[/tex]
where p is a prime and n and N are positive and N => n?
I have a trivial upper bound for d(p). That is, d(p) < p-1 for [tex]p\nmid N[/tex]. I think that suffices for my usages of d(p) for now.JustSam said: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, ...
Let p be a prime number. Define d(p) as the cardinality offlouran said:What is the number of solutions d(p) of
[tex]N-n^2 \equiv 0 \pmod p[/tex]
where p is a prime and n and N are positive and N => n?
JustSam said:I still don't understand what you are asking.
Let p be a prime number. Define d(p) as the cardinality of
[tex]$
\{ (N,n) : N \ge 1, n \ge 1, N \ge n, N \equiv n^2 \pmod p \}
[/tex]
Clearly you intend something different.
Let p be a prime number. Define d(p) as:flouran said: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 [tex]F(n) \equiv 0 \pmod p[/tex] 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 said:Let p be a prime number. Define d(p) as:
[tex]$
\max_{N \ge 1} \left| \{ n \pmod p : N \equiv n^2 \pmod p \} \right|
[/tex]
Then d(2) = 1, d(p) = 2 for odd primes p.
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.srijithju said:I don't 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 said: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.
CRGreathouse said:Is m fixed or can it take any integer value? Are there any limits on the value of n or m?
flouran said:Would d(p) be different if it was instead the number of solutions to:
[tex]F(n) \equiv 0 \pmod p[/tex]? Here F(n) = n - q where q = m^2 is a perfect square where m is a natural number.
flouran said:There are no limits on the value of n or m.
flouran said:Nice to see you on the forums, by the way Charles!
"Number of Solutions to d(p)" refers to the number of positive integer solutions to the equation d(p) = n, where d(p) represents the number of positive divisors of a given positive integer p and n is a positive integer.
Yes, there is a formula known as the divisor function that can be used to calculate the number of solutions to d(p). The formula is d(p) = (a+1)(b+1)(c+1)... where p = a^x * b^y * c^z * ..., where a, b, c, etc. are prime numbers and x, y, z, etc. are their respective powers.
No, the number of solutions to d(p) can never be greater than p. This is because the maximum number of divisors a positive integer p can have is p itself, and therefore, the maximum number of solutions to d(p) is also p.
The number of solutions to d(p) is directly related to prime numbers. For any prime number p, the only positive integer solution to d(p) = n is when n = 2, as a prime number only has two divisors (1 and itself). Therefore, the number of solutions to d(p) for any prime number p is always 1.
Yes, the number of solutions to d(p) has applications in number theory and cryptography. It is also used in understanding the distribution of prime numbers and in factoring large numbers, which is important for many encryption algorithms.