Prove: $|{(x,y)in Bbb F_q^2 : y^2 = Q(x) }|=q-1$ in a finite field












1












$begingroup$


Let $Bbb F_q$ be a finite field with q elements, with q odd. For a quadratic polynomial $Q(T)=T^2+a in Bbb F_q[T]$, show that if $a neq 0$, then:
$$|{(x,y)in Bbb F_q^2 : y^2 = Q(x) }|=q-1$$




If $a$ is a quadratic residue, then we have $q-1$ solutions of the form: $(0,b),(b,0)$ when $b$ is a quadratic residue. How do I show that those are the only solutions?
Also, what if $a$ is not a quadratic residue?






number-theory polynomials finite-fields







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$endgroup$












  • $begingroup$
    I didn't mention this in my answer, but I think you need to recheck your statement implying that, when $a$ is a quadratic residue, you always have solutions of the form $(0,b)$ and of the form $(b,0)$ and that this provides $q-1$ solutions. If $a=b^2$ is a quadratic residue then $(x,y)=(0,b)$ is indeed a solution to $y^2=x^2+a$, as is $(0,-b)$. If $-a$ is also a quadratic residue, which it will be when $qequiv1pmod{4}$, then $(c,0)$ and $(-c,0)$ will be solutions, where $c^2=-a$. This gives at most four solutions. If $-a$ is not a quadratic residue, then you'll only get two solutions.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:20
















1












$begingroup$


Let $Bbb F_q$ be a finite field with q elements, with q odd. For a quadratic polynomial $Q(T)=T^2+a in Bbb F_q[T]$, show that if $a neq 0$, then:
$$|{(x,y)in Bbb F_q^2 : y^2 = Q(x) }|=q-1$$




If $a$ is a quadratic residue, then we have $q-1$ solutions of the form: $(0,b),(b,0)$ when $b$ is a quadratic residue. How do I show that those are the only solutions?
Also, what if $a$ is not a quadratic residue?






number-theory polynomials finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    I didn't mention this in my answer, but I think you need to recheck your statement implying that, when $a$ is a quadratic residue, you always have solutions of the form $(0,b)$ and of the form $(b,0)$ and that this provides $q-1$ solutions. If $a=b^2$ is a quadratic residue then $(x,y)=(0,b)$ is indeed a solution to $y^2=x^2+a$, as is $(0,-b)$. If $-a$ is also a quadratic residue, which it will be when $qequiv1pmod{4}$, then $(c,0)$ and $(-c,0)$ will be solutions, where $c^2=-a$. This gives at most four solutions. If $-a$ is not a quadratic residue, then you'll only get two solutions.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:20














1












1








1





$begingroup$


Let $Bbb F_q$ be a finite field with q elements, with q odd. For a quadratic polynomial $Q(T)=T^2+a in Bbb F_q[T]$, show that if $a neq 0$, then:
$$|{(x,y)in Bbb F_q^2 : y^2 = Q(x) }|=q-1$$




If $a$ is a quadratic residue, then we have $q-1$ solutions of the form: $(0,b),(b,0)$ when $b$ is a quadratic residue. How do I show that those are the only solutions?
Also, what if $a$ is not a quadratic residue?






number-theory polynomials finite-fields







share|cite|improve this question









$endgroup$




Let $Bbb F_q$ be a finite field with q elements, with q odd. For a quadratic polynomial $Q(T)=T^2+a in Bbb F_q[T]$, show that if $a neq 0$, then:
$$|{(x,y)in Bbb F_q^2 : y^2 = Q(x) }|=q-1$$




If $a$ is a quadratic residue, then we have $q-1$ solutions of the form: $(0,b),(b,0)$ when $b$ is a quadratic residue. How do I show that those are the only solutions?
Also, what if $a$ is not a quadratic residue?






number-theory polynomials finite-fields




number-theory polynomials finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 12:57









user401516user401516

925310




925310












  • $begingroup$
    I didn't mention this in my answer, but I think you need to recheck your statement implying that, when $a$ is a quadratic residue, you always have solutions of the form $(0,b)$ and of the form $(b,0)$ and that this provides $q-1$ solutions. If $a=b^2$ is a quadratic residue then $(x,y)=(0,b)$ is indeed a solution to $y^2=x^2+a$, as is $(0,-b)$. If $-a$ is also a quadratic residue, which it will be when $qequiv1pmod{4}$, then $(c,0)$ and $(-c,0)$ will be solutions, where $c^2=-a$. This gives at most four solutions. If $-a$ is not a quadratic residue, then you'll only get two solutions.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:20


















  • $begingroup$
    I didn't mention this in my answer, but I think you need to recheck your statement implying that, when $a$ is a quadratic residue, you always have solutions of the form $(0,b)$ and of the form $(b,0)$ and that this provides $q-1$ solutions. If $a=b^2$ is a quadratic residue then $(x,y)=(0,b)$ is indeed a solution to $y^2=x^2+a$, as is $(0,-b)$. If $-a$ is also a quadratic residue, which it will be when $qequiv1pmod{4}$, then $(c,0)$ and $(-c,0)$ will be solutions, where $c^2=-a$. This gives at most four solutions. If $-a$ is not a quadratic residue, then you'll only get two solutions.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:20
















$begingroup$
I didn't mention this in my answer, but I think you need to recheck your statement implying that, when $a$ is a quadratic residue, you always have solutions of the form $(0,b)$ and of the form $(b,0)$ and that this provides $q-1$ solutions. If $a=b^2$ is a quadratic residue then $(x,y)=(0,b)$ is indeed a solution to $y^2=x^2+a$, as is $(0,-b)$. If $-a$ is also a quadratic residue, which it will be when $qequiv1pmod{4}$, then $(c,0)$ and $(-c,0)$ will be solutions, where $c^2=-a$. This gives at most four solutions. If $-a$ is not a quadratic residue, then you'll only get two solutions.
$endgroup$
– Will Orrick
Dec 17 '18 at 19:20




$begingroup$
I didn't mention this in my answer, but I think you need to recheck your statement implying that, when $a$ is a quadratic residue, you always have solutions of the form $(0,b)$ and of the form $(b,0)$ and that this provides $q-1$ solutions. If $a=b^2$ is a quadratic residue then $(x,y)=(0,b)$ is indeed a solution to $y^2=x^2+a$, as is $(0,-b)$. If $-a$ is also a quadratic residue, which it will be when $qequiv1pmod{4}$, then $(c,0)$ and $(-c,0)$ will be solutions, where $c^2=-a$. This gives at most four solutions. If $-a$ is not a quadratic residue, then you'll only get two solutions.
$endgroup$
– Will Orrick
Dec 17 '18 at 19:20










1 Answer
1






active

oldest

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2












$begingroup$

Let's look at an example, say $mathbf{F}_5$ and $Q(x)=x^2+2$. Now $2=1cdot2=2cdot1=3cdot4=4cdot3$. We can convert each of these products to a solution, and there are $5-1=4$ of them. For example, $2=4cdot3=(1+3)(1-3)=1^2-3^2$ so that $(3,1)$ is a solution. Similarly $2=1cdot2=(4+2)(4-2)=4^2-2^2$ so that $(2,4)$ is a solution.



This can be generalized to any finite field of odd characteristic and any $a$. Note that odd characteristic is required when writing each product as the difference of two squares.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'm not sure I understand the generalization; why are there $q-1$ ways to write $a$ as the difference of two squares?
    $endgroup$
    – user401516
    Dec 17 '18 at 19:02






  • 2




    $begingroup$
    There are $q-1$ ways to write $a$ as a product, $a=rs$ (because $s=a/r$ makes sense for any non-zero $r$). Now you want to let $r=y+x$ and $s=y-x$. This system of linear equations always has a unique solution when the field is of odd characteristic.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:23











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1 Answer
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$begingroup$

Let's look at an example, say $mathbf{F}_5$ and $Q(x)=x^2+2$. Now $2=1cdot2=2cdot1=3cdot4=4cdot3$. We can convert each of these products to a solution, and there are $5-1=4$ of them. For example, $2=4cdot3=(1+3)(1-3)=1^2-3^2$ so that $(3,1)$ is a solution. Similarly $2=1cdot2=(4+2)(4-2)=4^2-2^2$ so that $(2,4)$ is a solution.



This can be generalized to any finite field of odd characteristic and any $a$. Note that odd characteristic is required when writing each product as the difference of two squares.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'm not sure I understand the generalization; why are there $q-1$ ways to write $a$ as the difference of two squares?
    $endgroup$
    – user401516
    Dec 17 '18 at 19:02






  • 2




    $begingroup$
    There are $q-1$ ways to write $a$ as a product, $a=rs$ (because $s=a/r$ makes sense for any non-zero $r$). Now you want to let $r=y+x$ and $s=y-x$. This system of linear equations always has a unique solution when the field is of odd characteristic.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:23
















2












$begingroup$

Let's look at an example, say $mathbf{F}_5$ and $Q(x)=x^2+2$. Now $2=1cdot2=2cdot1=3cdot4=4cdot3$. We can convert each of these products to a solution, and there are $5-1=4$ of them. For example, $2=4cdot3=(1+3)(1-3)=1^2-3^2$ so that $(3,1)$ is a solution. Similarly $2=1cdot2=(4+2)(4-2)=4^2-2^2$ so that $(2,4)$ is a solution.



This can be generalized to any finite field of odd characteristic and any $a$. Note that odd characteristic is required when writing each product as the difference of two squares.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'm not sure I understand the generalization; why are there $q-1$ ways to write $a$ as the difference of two squares?
    $endgroup$
    – user401516
    Dec 17 '18 at 19:02






  • 2




    $begingroup$
    There are $q-1$ ways to write $a$ as a product, $a=rs$ (because $s=a/r$ makes sense for any non-zero $r$). Now you want to let $r=y+x$ and $s=y-x$. This system of linear equations always has a unique solution when the field is of odd characteristic.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:23














2












2








2





$begingroup$

Let's look at an example, say $mathbf{F}_5$ and $Q(x)=x^2+2$. Now $2=1cdot2=2cdot1=3cdot4=4cdot3$. We can convert each of these products to a solution, and there are $5-1=4$ of them. For example, $2=4cdot3=(1+3)(1-3)=1^2-3^2$ so that $(3,1)$ is a solution. Similarly $2=1cdot2=(4+2)(4-2)=4^2-2^2$ so that $(2,4)$ is a solution.



This can be generalized to any finite field of odd characteristic and any $a$. Note that odd characteristic is required when writing each product as the difference of two squares.






share|cite|improve this answer









$endgroup$



Let's look at an example, say $mathbf{F}_5$ and $Q(x)=x^2+2$. Now $2=1cdot2=2cdot1=3cdot4=4cdot3$. We can convert each of these products to a solution, and there are $5-1=4$ of them. For example, $2=4cdot3=(1+3)(1-3)=1^2-3^2$ so that $(3,1)$ is a solution. Similarly $2=1cdot2=(4+2)(4-2)=4^2-2^2$ so that $(2,4)$ is a solution.



This can be generalized to any finite field of odd characteristic and any $a$. Note that odd characteristic is required when writing each product as the difference of two squares.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 13:41









Will OrrickWill Orrick

13.7k13461




13.7k13461












  • $begingroup$
    I'm not sure I understand the generalization; why are there $q-1$ ways to write $a$ as the difference of two squares?
    $endgroup$
    – user401516
    Dec 17 '18 at 19:02






  • 2




    $begingroup$
    There are $q-1$ ways to write $a$ as a product, $a=rs$ (because $s=a/r$ makes sense for any non-zero $r$). Now you want to let $r=y+x$ and $s=y-x$. This system of linear equations always has a unique solution when the field is of odd characteristic.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:23


















  • $begingroup$
    I'm not sure I understand the generalization; why are there $q-1$ ways to write $a$ as the difference of two squares?
    $endgroup$
    – user401516
    Dec 17 '18 at 19:02






  • 2




    $begingroup$
    There are $q-1$ ways to write $a$ as a product, $a=rs$ (because $s=a/r$ makes sense for any non-zero $r$). Now you want to let $r=y+x$ and $s=y-x$. This system of linear equations always has a unique solution when the field is of odd characteristic.
    $endgroup$
    – Will Orrick
    Dec 17 '18 at 19:23
















$begingroup$
I'm not sure I understand the generalization; why are there $q-1$ ways to write $a$ as the difference of two squares?
$endgroup$
– user401516
Dec 17 '18 at 19:02




$begingroup$
I'm not sure I understand the generalization; why are there $q-1$ ways to write $a$ as the difference of two squares?
$endgroup$
– user401516
Dec 17 '18 at 19:02




2




2




$begingroup$
There are $q-1$ ways to write $a$ as a product, $a=rs$ (because $s=a/r$ makes sense for any non-zero $r$). Now you want to let $r=y+x$ and $s=y-x$. This system of linear equations always has a unique solution when the field is of odd characteristic.
$endgroup$
– Will Orrick
Dec 17 '18 at 19:23




$begingroup$
There are $q-1$ ways to write $a$ as a product, $a=rs$ (because $s=a/r$ makes sense for any non-zero $r$). Now you want to let $r=y+x$ and $s=y-x$. This system of linear equations always has a unique solution when the field is of odd characteristic.
$endgroup$
– Will Orrick
Dec 17 '18 at 19:23


















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