Number of solutions for x^2=a(mod m)

Question

I'm trying to understand how to know the numbers of solutions for x^2=a(mod m). I know that if a has roots they are solutions but how could I know the rest of them? Is there a need to use the Chinese reminder here? if so how?

The chinese reminder theorem allows you to reduce the problem when $m$ is a prime power, since a solution of $x^2=a\bmod m$ is in correspondence with the system of equations $x^2=a\bmod p_j^{k_j}$ where $m=p_1^{j_1}\dots {p_r}^{j_r}$ — Julian Mejia, May 25 '19 at 16:22
Thanks. I think I got how to do it with Chinese reminder. I posted an answer and will be glad of someone could confirm if it is the right way to do it. — Idan Daniel, May 25 '19 at 16:23
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For typesetting, please use MathJax. — dantopa, May 26 '19 at 02:20

Idan Daniel · Answer 1 · 2019-05-25T17:06:40.353

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I think I got it: I dismantle m to coprime numbers and check when the x^2=a(mod m) have solutions for each. lets say x^2=11^2(mod 1800) so it has solutions if and only if

x^2=11^2(mod 9)=4(mod 9)

x^2=11^2(mod 8)=1(mod 8)

x^2=11^2(mod 25)=21(mod 25)

Because 1800=9*8*25

x=+-2(mod 9)

x=+-1,+-3(mod 8)

x=+-11(mod 25)

so it has 2*4*2=16 solutions, is it right?

edited May 25 '19 at 17:06

answered May 25 '19 at 16:20

Idan Daniel

151

You are right. forgot I need to check numbers modulo. I edited it. is it right now? – Idan Daniel May 25 '19 at 16:47

Algebear · Answer 2 · 2019-05-25T16:34:14.003

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First, you want to know whether there are solutions: that's where the Legendre symbol comes in; there is a solution to $x^2\equiv a$ mod $m$ iff $\left(\frac{a}{m}\right)=1$. How to compute this Legendre symbol? Well, there are some fine tricks, one of them being the law of quadratic reciprocity.

After finding out it has a solution, one can find solutions the hard way by trying all numbers $1,2,3,\ldots,\lceil m\rceil$. Note that you don't have to checke the other half since if $a$ mod $m$ is a solution, then also $a-m$ mod $m$ is a solution.

There are more methods for finding the solutions but it all comes down to first checking the existence of a solution. For finding all solutions with less work, one can use Tonelli's algorithm, but you need to know about Legendre symbols and Jacobi symbols first.

You may find this post interesting: Number of solutions to $x^2\equiv a\pmod m$

edited May 25 '19 at 16:34

answered May 25 '19 at 16:23

Algebear

1,674

Thanks. I am aware of this but is works only if m is odd(at least I only learned Legendre symbol where m is odd). I asked this question because I got stuck on where m is even – Idan Daniel May 25 '19 at 16:27
@IdanDaniel Jacobi symbols are the extension of Legendre symbols and they deal with the cases where $m$ is not prime. – Algebear May 25 '19 at 16:28
I know. My professor only defined it for odd numbers though. – Idan Daniel May 25 '19 at 16:30
@IdanDaniel I should have looked better, but indeed it is defined for positive odd numbers only. – Algebear May 25 '19 at 16:32
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Note also that for Jacobi symbols, $\left(\frac{a}{b}\right)=-1$ tells you that $a$ is not a square modulo $b$, but if it is equal to $1$, it is still possible for $a$ to also not be a square modulo $b$. For example, that happens with $-1$ modulo $77$. – Arturo Magidin May 25 '19 at 17:15

Number of solutions for x^2=a(mod m)

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