False Proof that $\sqrt{4}$ is Irrational

Just because $a^2$ is divisible by $4$, that doesn't mean $a$ is.

Your error is stating that if $a^2$ is divisible by 4 so must $a$ be. The fundimental theorem states if a prime $p $ divides $ab $ then $p $ must divide $a $ or $p$ must divide $b $. That is true because $p $ is indivisable.

But if $p $ is composite it doesn't hold. $p$ could equal $jk $ and $j$ could divide $a $ and $k $ divide $b $. Example: 3 divides 4 times 9 so 3 either divides 4 or 3 divides 9 because 3 is prime. But 6 divides 4 times 9 but 6 neither divides 4 nor 9 but instead 3 divides 9 while 2 divides 4 so 6=2 times 3 divide 4 times 9.

So 4 divides $a^2$ means $2*2$ divides $a*a$ so 2 divides $a $ is all you can conclude with certainty. ... because 4 is not prime.

Actually because 4 is not square free. All of its prime factors must divide into $a$ but the square powers can be distributed among (and are) distributed among the square powers of $a$.