We can define $a_n$ by $a_n=4a_{n-1}+a_{n-2}, a_0=2, a_1=4$. By induction, we know $$a_n={(2+\sqrt{5})^n+(2-\sqrt{5})^n}$$So $\lfloor (2+\sqrt5)^{2019} \rfloor=a_{2019}-1$. And by $mod \; 3$, $$a_0 \equiv -1, a_1 \equiv 1, a_2 \equiv 0, a_3 \equiv 1, a_4 \equiv 1, a_5 \equiv -1, a_6 \equiv 0, a_7 \equiv -1, a_8 \equiv -1, a_9 \equiv 1$$Therefore $a_n$ by $mod \; 3$ is periodic and its period is $8$. $a_{2019} \equiv a_3 \equiv 1 (mod \; 3)$ and $ 3\;| \; \lfloor (2+\sqrt5)^{2019} \rfloor=a_{2019}-1>3$

$a_n={(2+\sqrt{5})^n+(2-\sqrt{5})^n}\implies \lfloor (2+\sqrt5)^{2019} \rfloor=a_{2019}$ $a_{n+2}\equiv a_n \pmod 4$ $\implies a_{2019}\equiv a_1 \equiv 0 \pmod 4$

Let $(2+\sqrt5)^{2019}=R$ which has a fractional part $f$ and an integral part $I$, so we have $R=I+f$ where $0<f<1$. $$R=(2+\sqrt5)^{2019}=^{2019}C_0(\sqrt{5})^{2019}+^{2019}C_1(\sqrt{5})^{2018}\cdot 2+\dotsc+ ^{2019}C_{2019}(2)^{2019}$$Similarly we have $$G=(2-\sqrt5)^{2019}=^{2019}C_0(\sqrt{5})^{2019}-^{2019}C_1(\sqrt{5})^{2018}\cdot 2+\dotsc-^{2019}C_{2019}(2)^{2019}$$Note that $0<2-\sqrt{5}<1 \implies 0<(2-\sqrt5)^{2019}<1$. $$R-G=2\underbrace{(^{2019}C_1(\sqrt{5})^{2018}\cdot 2-^{2019}C_3(\sqrt{5})^{2016}\cdot 2^3+\dots)}_\text{let this be some integer k}$$So we have $R-G=I+f-G=k$. Note that $RHS \in \mathbb{Z} \implies LHS \in \mathbb{Z} \implies f-G \in \mathbb{Z} \implies f=G \implies I=2k$. So we have $\lfloor (2+\sqrt5)^{2019} \rfloor=2k$ and because an even number except 2 cannot be a prime we are done!