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

49.1k870156




49.1k870156










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













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1 Answer
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active

oldest

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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

190k14216348




190k14216348












  • $begingroup$
    $9x^2+9x$ 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




















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