Number of roots of a quadratic polynomial with coefficients in ring $mathbb{Z_{18}}$












3












$begingroup$


I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17
















3












$begingroup$


I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17














3












3








3





$begingroup$


I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups







share|cite|improve this question











$endgroup$




I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups




abstract-algebra group-theory polynomials ring-theory cyclic-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 22:16









Blue

48.6k870154




48.6k870154










asked Dec 17 '18 at 21:18









CoupeauCoupeau

1296




1296








  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17














  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17








1




1




$begingroup$
In the second and third line you don't get that these expressions are equal to zero.
$endgroup$
– Lukas Geyer
Dec 17 '18 at 21:51




$begingroup$
In the second and third line you don't get that these expressions are equal to zero.
$endgroup$
– Lukas Geyer
Dec 17 '18 at 21:51












$begingroup$
Oh, of course, thanks. I will correct it.
$endgroup$
– Coupeau
Dec 17 '18 at 22:07




$begingroup$
Oh, of course, thanks. I will correct it.
$endgroup$
– Coupeau
Dec 17 '18 at 22:07












$begingroup$
You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
$endgroup$
– Berci
Dec 17 '18 at 22:37






$begingroup$
You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
$endgroup$
– Berci
Dec 17 '18 at 22:37














$begingroup$
It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
$endgroup$
– Daniel Schepler
Dec 17 '18 at 23:17




$begingroup$
It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
$endgroup$
– Daniel Schepler
Dec 17 '18 at 23:17










1 Answer
1






active

oldest

votes


















3












$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25













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%2f3044451%2fnumber-of-roots-of-a-quadratic-polynomial-with-coefficients-in-ring-mathbbz%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









3












$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25


















3












$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25
















3












3








3





$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$



[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 23:07

























answered Dec 17 '18 at 23:02









Eric WofseyEric Wofsey

187k14215344




187k14215344












  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25




















  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25


















$begingroup$
$9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
$endgroup$
– lhf
Dec 17 '18 at 23:10




$begingroup$
$9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
$endgroup$
– lhf
Dec 17 '18 at 23:10












$begingroup$
Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
$endgroup$
– MathematicsStudent1122
Dec 18 '18 at 3:25






$begingroup$
Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
$endgroup$
– MathematicsStudent1122
Dec 18 '18 at 3:25




















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%2f3044451%2fnumber-of-roots-of-a-quadratic-polynomial-with-coefficients-in-ring-mathbbz%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