How can I solve the following congruence $x^2 \equiv 9 \pmod {2^3 . 3 . 5^2}$?

Question

The problem is that I do not know the number of solutions of $x^2 \equiv 9 \pmod { 3}$, it seems like either it is zero only or any multiple of 3 other than 0, could anyone explain for me why it is not any multiple of 3 other than 0?

Any $x\equiv 0\pmod{3}$ satisfies $x^2\equiv 9 (\equiv 0) \pmod{3}$. That is, $x$ could be $0$, or $3$ or $6$, or $-3$, or,.... — vadim123, Jun 14 '19 at 03:39
are you asking why the number of solutions is not any multiple of $3$, or why not any multiple of $3$ is a solution? — J. W. Tanner, Jun 14 '19 at 03:40
why not any multiple of 3 is a solution ? only the zero multiple is a solution? .... by the way how many solutions are there?@J.W.Tanner — Intuition, Jun 14 '19 at 03:44
$3^2\equiv9\pmod{2^3\cdot3\cdot5^2}$ but $6^2\equiv36\not\equiv9\pmod{2^3 \cdot3 \cdot5^2}$ — J. W. Tanner, Jun 14 '19 at 03:45
so we have only 0 and 3 for this congruence ..... correct ? @J.W.Tanner — Intuition, Jun 14 '19 at 03:50
I will make a comment of what I think your confusion is. Usually when you want to solve a modular equation like $x^2=9\bmod 3$. The solutions are not just $x$, but $x\bmod 3$. As you said, $x$ can be any multiple of $3$, but in that case $x\bmod 3=0\bmod 3$. So, the only solution of the equation is $0\bmod 3$, and that may be the reason that you have heard that $0$ was the only solution. — Julian Mejia, Jun 14 '19 at 03:51

score 2 · Accepted Answer · edited Jun 14 '19 at 05:38

2

$$\begin{align} x^2\equiv 9\bmod 8&\iff x\equiv 1\bmod 2\\ x^2\equiv 9\bmod 3&\iff x\equiv 0\bmod 3\\ x^2\equiv 9\bmod 25&\iff x\equiv \pm 3\bmod 25 \end{align}$$

So, $x^2\equiv 9\mod 600\iff x\equiv \pm3\bmod 150$.

So, the solutions $\pmod{600}$ are going to be $8$, namely $\pm 3\bmod{600}$, $150\pm 3\bmod{600}$, $300\pm3\bmod{600}$, and $450\pm 3\bmod{600}$.

edited Jun 14 '19 at 05:38

Arturo Magidin

417,286

answered Jun 14 '19 at 04:02

Julian Mejia

4,458

https://math.stackexchange.com/questions/3261661/without-actually-finding-then-determine-the-number-of-solutions-of-the-congruen – Intuition Jun 14 '19 at 04:03
The solution at the back of the book contains 8 solutions? – Intuition Jun 14 '19 at 04:05
@hopefully makes sense they give 8 solutions because the solutions actually should be given in the original modulus $\bmod(2^3\cdot 3\cdot 5^2)=\bmod(600)$. So, the actual 8 solutions will be $\pm 3,150\pm3,300\pm 3$ and $450\pm 3$ ($\bmod 600$) – Julian Mejia Jun 14 '19 at 04:15
from where you got the first equivalence? – Intuition Jun 14 '19 at 04:28
how do you remove the power from the prime in the mod ? – Intuition Jun 14 '19 at 04:33
The third congruence has solutions 18 and 7 – Intuition Jun 14 '19 at 05:36
what are 150 , 300 and 450? – Intuition Jun 14 '19 at 05:40
where are the 8 systems of linear congruences? – Intuition Jun 14 '19 at 05:41
@hopefully $odd^2\equiv 1\bmod 8$ that's how I got first equivalence. $18,7$ are not solutions of the third equation. $150,300,450$ are the multiples of $150$ smaller than $600$. There are $8$ solutions $\bmod 600$ count it.( When I use $\pm$ I am actually given 2 solutions at once) – Julian Mejia Jun 14 '19 at 19:57

How can I solve the following congruence $x^2 \equiv 9 \pmod {2^3 . 3 . 5^2}$?

1 Answers1

Linked