$x^2\not \equiv 0 \pmod{165}$ for $0

Question

So I have this question that goes like this:

Show that $x^2\not \equiv 0 \pmod {165}$ for $0<x<165$. I think I have a solution, but I'm not quite sure if it's correct:

if $x^2\equiv 0 \pmod{165}$ then $$x^2\equiv 0 \pmod{3},\ x^2\equiv 0 \pmod{5},\ x^2\equiv 0 \pmod{11},$$ but if this is true then $$x\equiv 0 \pmod{3},\ x\equiv 0 \pmod{5},\ x\equiv 0 \pmod{11}$$ and then CRT gives that $$x\equiv 0 \pmod{165},$$ but for $x\in \{ 1,...,164\}$ this cannot be true.

Is this correct? Thank you for your time.

Generally $, n,$ is squarefree $ \iff\ [,n\mid x^2\Rightarrow, n\mid x,]\ $ (or $\bmod n!:\ x\not\equiv 0,\Rightarrow, x^2\not\equiv 0,,$ equivalently), so this property characterizes squarefree integers. Follow the link for many more such characterizations. — Bill Dubuque, Dec 23 '19 at 21:43

score 2 · Answer 1 · edited Dec 23 '19 at 20:44

2

Notation questions aside, the math is good. A similar argument applies using any square-free natural number (such as $165=3×5×11$) as the modulus.

edited Dec 23 '19 at 20:44

J. W. Tanner

60,406

answered Dec 23 '19 at 20:37

Oscar Lanzi

39,403

$x^2\not \equiv 0 \pmod{165}$ for $0

1 Answers1