$\sqrt {3} $ is irrational (proof verification)

Question

$\sqrt {3} \in Q$. Then, $\sqrt{3} = \frac ab$ with the lowest term for $a,b \in Z$.

Then, $3b^2=a^2$, which implies that $a^2$ is divisible by 3.

That is, $a$ is also divisible by 3 (by fundamental theorem of arithmetic).

I don't understand here $a^2$ divisible by 3 implies $a$ divisible by 3.

Could you explain it?

Answer 1

Working with integers, recall that $cd$ divisible by a prime $k$ implies $c$ is divisible by $k$ or $d$ is divisible by $k$. $a^2$ divisible by 3 implies $a \cdot a$ divisible by 3. So, $a$ is divisible by $3$ or $a$ is divisible by $3$ (This statement occurs by taking the left $a$ and the right $a$; if you don't understand this, in the example mentioned let $c = a$ and $d=a$).

Answer 2

Consider the prime factorization of $a$.

Suppose on the contrary that $3$ is not a factor of $a$. Then squaring $a$ will also not make $3$ appears in the prime factorization of $a^2$. Hence $3$ will not be a factor of $a^2$.