Proof that $\sqrt{4}\notin\mathbb{Q}$ of course wrong but where is the flaw?

Question

Assume $$\eqalign{ \sqrt{4}\in\mathbb{Q}&\Longrightarrow(\exists a,b\in\mathbb{Z})\sqrt{4}=\frac{a}{b}\text{ and }\gcd(a,b)=1\\ &\Longrightarrow 4b^2=a^2\Longrightarrow a\text{ is even}\\ &\Longrightarrow a=2k\Longrightarrow b^2=4k^2\Longrightarrow b\text{ is even}\\ &\Longrightarrow \gcd(a,b)\ne1 \text{ contradiction}}$$ Where's the flaw?

Answer 1

You get $4b^2=a^2=(2k)^2=4k^2$, so $b^2=k^2$.

Answer 2

You simplify by 4 without throwing the 4 on the RHS away.