abstract algebra - Proof that the sqrt of a Natural Number is irrational outside of square numbers.

I'm reading a book on Abstract Algebra and I'm covering Rings at the moment. Now I'm fairly new to this so my terminology may be incorrect, but we are looking at examples in which the set of Integers can be composed into sets with different remainders under the integer division of a Natural Number $n$

For instance if we consider Integers under the division of $3$

Here let $[n]_{m}$ is the set of Integers with remainder $n$ under integer division $m$.

Take for instance $m = 3$, then $n = 0,1,2$

Where

$[0]_{3} = \{ 3k \: | \: k \in \mathbb{Z} \}$

$[1]_{3} = \{ 3k + 1 \: | \: k \in \mathbb{Z} \}$

$[2]_{3} = \{ 3k + 2 \: | \: k \in \mathbb{Z} \}$

Then if we consider the Set

$S = \{ [0]_{3}, [1]_{3}, [2]_{3} \}$

Under the definition of Addition and Multiplication as defined in the Commutative Field $\mathbb{Z}$ we see that Addition and Multiplication form a Commutative Ring on $S$.

What I noticed however is that for an Integer $x$ that $x^{2} \in [0]_{3} \iff x \in [0]_{3}$

This got me thinking about a result I saw that for a non-square integer $p$ that $\sqrt{p} \in \mathbb{R} - \mathbb{Q}$

Applying the $\sqrt{2}$ proof of being Irrational for a non-square $p$ we assume Rationality. In the following $a,b \in \mathbb{Z}$ where $\gcd(a,b) = 1$ and $b \neq 0$ then,

$\sqrt{p} = \frac{a}{b} \longrightarrow p = \frac{a^{2}}{b^{2}} \longrightarrow b^{2}p = a^{2}$

And so that $a^{2}$ is divisible by $p$

Now I believe that implies that $a$ must also be divisible by $p$, i.e.

$a^{2} \in [0]_{p} \iff a \in [0]_{p}$

Or alternatively that $0 \equiv a^{2} \pmod p \iff 0 \equiv a \pmod p$

I'm sure this have been proven but my limited knowledge of the field has left me not able to find the correct search terms!

Can someone please provide a link to the proof or some hints to get started.

Thanks