In diophantine $3b^2=a^2$ where $a$ and $b$ are coprime, does $3|a$? Integers $a$ and $b$ are co-prime and $3\cdot b^2=a^2$.
$3\cdot b^2=a^2$, implies $a^2$ is divisible by 3 since, $3b^2$ is divisible by 3.
Is $a$ divisible by 3?

Yes, $a$ must be divisible by $3$. If the integer $a=3k+1$ or $3k+2$ for arbitrary $k$, (which are the cases in which $a$ is not divisible by $3$) then the remainder is $9k^2+6k+1\equiv1\pmod3$ or $9k^2+12k+4\equiv1\pmod3$.
Therefore, $a^2|3$ cannot occur if $a$ is not divisible by $3$, and thus, $a$ must be divisible by $3$.
However, the equation $3a^2=b^2$ cannot be solved for coprime integers, so if you consider that, then it is impossible for $a|3$ since there are no $a$ that fit the equation.
