Find all $x \in \mathbb Z_{360}$ such that $x^2 ≡ 0 \pmod{360}$

Question

Find all $x \in \mathbb Z_{360}$ such that $$x^2 ≡ 0 \pmod{360}.$$

I know that this means to find all $x$ such that the result divides into $360$ evenly. I also know the prime factorization of $$360 = 2^3 \cdot 3^2 \cdot 5 .$$

Is there a technique to use this prime factorization to easily come up with all the numbers that would satisfy this equation?

Thank you.

Answer 1

Take any $x$ which is less than $360$.

Now, note that if $x = 2^a3^b5^c \times d$, where $d$ is some number coprime to $2,3,5$ (this can be done by unique factorization) then $x^2 = 2^{2a}3^{2b}5^{2c} d^2$.

We want $360$ to divide this number. That is, $360$ divides $2^{2a} 3^{2b} 5^{2c}d^2$. But, $360$ is coprime to $d^2$, because $d^2$ cannot share any prime factors with $360$ by the way we defined $d$.

Therefore, by a result we know :

If $a$ divides $bc$ and $a$ is coprime to $c$, then $a$ divides $b$.

We get that $360$ divides $2^{2a} 3^{2b}5^{2c}$. Now, what is $360$? It is $2^3 \times 3^2 \times 5$. For it to divide the above number, it follows that: $$ 2a \geq 3, 2b \geq 2, 2c \geq 1 \iff a \geq 2,b \geq 1, c \geq 1 \tag{*} $$

So, what we do know, is this : $360$ divides $2^{2a} 3^{2b} 5^{2c}$, if and only if the right hand side of $(*)$ hold true.

Let $n$ be any natural number. Note that if $360$ divides $2^{2a} 3^{2b} 5^{2c}$, then it also divides $2^{2a} 3^{2b}5^{2c} \times n$.

Therefore, put $a = 2,b=1,c=1$ : since the right hand side of $(*)$ is satisfied, so 360 divides the square of $2^2 \times 3 \times 5 = 60$, and therefore the square of every multiple of $60$ as well.

This tells us that the answer to the given question have to be multiples of $60$. Alternately, it is not difficult to see that every multiple of $60$ does satisfy the conditions. So the answer is precisely multiples of $60$ in $Z_{360}$.

Answer 2

Hint: $x^2 \equiv 0 \mod 360$ iff $x^2 \equiv 0 \mod 2^3, \mod 3^2 \ \text{and} \mod 5$.