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







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












$begingroup$


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

92539




92539












  • $begingroup$
    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

votes


















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043903%2fprove-x-y-in-bbb-f-q2-y2-qx-q-1-in-a-finite-field%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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












$begingroup$

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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043903%2fprove-x-y-in-bbb-f-q2-y2-qx-q-1-in-a-finite-field%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more