$b_n=2a_n$ $2a_n+2a_{n+1}=4a_{n+2}a_{n+3}+4032 \to b_n+b_{n+1}=b_{n+2}b_{n+3}+4032$ $b_{n+2}-b_n=b_{n+2}+b_{n+1}-(b_n+b_{n+1})=b_{n+3}b_{n+4}-b_{n+2}b_{n+3}=b_{n+3}(b_{n+4}-b_{n+2})$ $b_{n+2}-b_n=b_{n+3}(b_{n+4}-b_{n+2})=b_{n+3}b_{n+5}(b_{n+6}-b_{n+4})=....=b_{n+3}b_{n+5}...b_{n+2k+1}(b_{n+2k+2}-b_{n+2k})$ for every natural $n,k$ Let $b_3-b_1 =m \neq 0$ than $|m|=|b_3-b_1|=|b_4b_6....b_{2k+2}(b_{2k+3}-b_{2k+1})| \geq 2^{k+1}$ for every $k$. Obvious, that for every $m$ there is such $k$ that $|m|<2^{k+1}$ So $m=0$ Similar way we can prove, that $b_4-b_2=0$ (Also, we can prove, that $b_{2k}=b_2,b_{2k+1}=b_1$ for every $k$) $b_1=b_3=x,b_2=b_4=y$ $x+y=xy+4032 \to (x-1)(y-1)+4031=0 \to (x-1)(y-1)=-29*139$ Possible solutions : $(x,y)=(-28,140),(140,-28),(-138,30),(30,-138),(0,4032),(4032,0)$ Answer: possible solutions $(a_1,a_2)=(-14,70),(70,-14),(-69,15)(15,-69)(0,2016),(2016,0)$

Claim. $a_1=a_3,a_2=a_4$ Proof. Let $s_n=a_{2n+2}-a_{2n}$. Notice that $$a_{2n}+a_{2n+1}=2a_{2n+2}a_{2n+3}+2016$$$$a_{2n+1}+a_{2n+2}=2a_{2n+3}a_{2n+4}+2016$$Take the difference of them yields $$a_{2n+2}-a_{2n}=2a_{2n+3}(a_{2n+4}-a_{2n+2})$$$$s_n=2a_{2n+3}s_{n+1}$$Now suppose on the contrary that $s_n>0$ for all $n$ then $|a_{2n+3}|\geq 1$ for all $n$ hence $$|s_n|=2|a_{2n+3}||s_{n+1}|>|s_{n+1}| \hspace{20pt}(1)$$which implies $\{s_n\}$ is a strictly decreasing sequence of positive integers, contradiction. Therefore, $s_n=0$ for some $n\in\mathbb N$, by $(1)$ we have $s_{n+1}=0$ implies $s_n=0$, so by backward induction we have $s_1=0$. Therefore, $a_2=a_4$ and by symmetry $a_1=a_3$. $\blacksquare$ Therefore, $$a_1+a_2=2a_3a_4+2016=2a_1a_2+2016$$$$2a_1-1=\frac{4031}{1-2a_2}\in\mathbb Z$$We have $4031=29\times 139$ and hence $$(1-2a_2,2a_1-1)=(-4031,-1),(-139,-29),(-29,-139),(-1,-4031),(1,4031),(29,139),(139,29),(4031,1)$$so$$(a_1,a_2)=(0,2016),(-14,70),(-69,15),(-2015,1),(2016,0),(70,-14),(15,-69),(1,-2015)$$conversely, suppose $(a_1,a_2)$ are the values above then define $a_{2n}=a_2,a_{2n+1}=a_1$, we easily see that it satifies the requirements.