How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$?
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Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?
Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?
(Please let me know if I should add some details to this problem)
calculus combinatorics number-theory elementary-number-theory discrete-mathematics
asked Nov 26 at 18:32
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163
add a comment |
up vote
3
down vote
favorite
Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?
Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?
(Please let me know if I should add some details to this problem)
calculus combinatorics number-theory elementary-number-theory discrete-mathematics
asked Nov 26 at 18:32
color
163
2
This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38
@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43
1
Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?
Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?
(Please let me know if I should add some details to this problem)
calculus combinatorics number-theory elementary-number-theory discrete-mathematics
asked Nov 26 at 18:32
color
163
Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?
Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?
(Please let me know if I should add some details to this problem)
calculus combinatorics number-theory elementary-number-theory discrete-mathematics
calculus combinatorics number-theory elementary-number-theory discrete-mathematics
asked Nov 26 at 18:32
color
163
asked Nov 26 at 18:32
color
163
edited Nov 27 at 0:44
asked Nov 26 at 18:32
color
163
asked Nov 26 at 18:32
color
163
asked Nov 26 at 18:32
color
163
163
2
This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38
@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43
1
Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56
add a comment |
2
This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38
@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43
1
Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56
2
2
This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38
This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38
@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43
@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43
1
1
Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56
Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56
add a comment |
1 Answer
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Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):
$d^{p-1}-1 ≡0mod p$
$ax^2+by^2=d^{p-1}$
If $y=0$ then:
$$x=frac{d^{(p-1)/2}}{a^{1/2}}$$
So
$x=<frac{d^{(p-1)/2}}{a^{1/2}}$
That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:
$$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$
That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:
$$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$
You can exclude 1 from solutions and find better formula.
answered Nov 29 at 17:26
sirous
1,5741513
add a comment |
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Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):
$d^{p-1}-1 ≡0mod p$
$ax^2+by^2=d^{p-1}$
If $y=0$ then:
$$x=frac{d^{(p-1)/2}}{a^{1/2}}$$
So
$x=<frac{d^{(p-1)/2}}{a^{1/2}}$
That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:
$$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$
That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:
$$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$
You can exclude 1 from solutions and find better formula.
answered Nov 29 at 17:26
sirous
1,5741513
add a comment |
up vote
0
down vote
Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):
$d^{p-1}-1 ≡0mod p$
$ax^2+by^2=d^{p-1}$
If $y=0$ then:
$$x=frac{d^{(p-1)/2}}{a^{1/2}}$$
So
$x=<frac{d^{(p-1)/2}}{a^{1/2}}$
That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:
$$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$
That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:
$$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$
You can exclude 1 from solutions and find better formula.
answered Nov 29 at 17:26
sirous
1,5741513
add a comment |
up vote
0
down vote
up vote
0
down vote
Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):
$d^{p-1}-1 ≡0mod p$
$ax^2+by^2=d^{p-1}$
If $y=0$ then:
$$x=frac{d^{(p-1)/2}}{a^{1/2}}$$
So
$x=<frac{d^{(p-1)/2}}{a^{1/2}}$
That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:
$$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$
That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:
$$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$
You can exclude 1 from solutions and find better formula.
answered Nov 29 at 17:26
sirous
1,5741513
Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):
$d^{p-1}-1 ≡0mod p$
$ax^2+by^2=d^{p-1}$
If $y=0$ then:
$$x=frac{d^{(p-1)/2}}{a^{1/2}}$$
So
$x=<frac{d^{(p-1)/2}}{a^{1/2}}$
That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:
$$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$
That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:
$$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$
You can exclude 1 from solutions and find better formula.
answered Nov 29 at 17:26
sirous
1,5741513
edited Nov 29 at 17:31
answered Nov 29 at 17:26
sirous
1,5741513
answered Nov 29 at 17:26
sirous
1,5741513
answered Nov 29 at 17:26
sirous
1,5741513
1,5741513
add a comment |
add a comment |
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2
This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38
@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43
1
Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56