Proposition $a^{2}_{n+1} +3=a_{n} a_{n+2} \cdots ( \heartsuit )$ holds for all non-negative integers $n$. Proof of this proposition$a_{2} =5a_{1} -a_{0} =20-1=19$ $a_{3} =5a_{2} -a_{1} =95-4=91$ We can show the above equality by induction on $n\in \mathbb{Z}_{\geq 0}$. For $n=0$, $a^{2}_{1} +3=19=1\cdotp 19=a_{0} \cdotp a_{2}$. For $n=1$, $a^{2}_{2} +3=364=4\cdotp 91=a_{1} \cdotp a_{3}$. Suppose that $( \heartsuit )$ holds for $n=k$ and $n=k+1$ for some $k\in \mathbb{Z}_{\geq 0}$. ${a^{2}_{k+3} +3} =( 5a_{k+2} -a_{k+1})^{2} +3=25a^{2}_{k+2} -10a_{k+2} a_{k+1} +a^{2}_{k+1} +3$ $=$$25a^{2}_{k+2} -10a_{k+2} a_{k+1} +a_{k} \cdotp a_{k+2}$ $=a_{k+2}( 25a_{k+2} -10a_{k+1} +a_{k})$ It is sufficient to show that $a_{k+4} =25a_{k+2} -10a_{k+1} +a_{k}$ : $a_{k+4} =5a_{k+3} -a_{k+2} =5( 5a_{k+2} -a_{k+1}) -( 5a_{k+1} -a_{k})$ $=25a_{k+2} -10a_{k+1} +a_{k}$ Therefore, we have $a^{2}_{k+3} +3 =a_{k+2} \cdotp a_{k+4}$ , which implise $( \heartsuit )$ holds for $n=k+2$. Hence $( \heartsuit )$ holds for all non-negative integers $n$. $\blacksquare $ Proof of $a_{n}=x^2+3y^2$ for some $(x,y)$ $1=1^{2} +3\cdotp 0^{2}$ $4=2^{2} +3\cdotp 0^{2}$ By the proposition, we have $a^{2}_{n+1} +3=a_{n} a_{n+2}$. For simplicity, let $N=a^{2}_{n+1} +3$. Note that $a_{n} \mid N$ and $N\equiv 4\pmod 8$ or $N\equiv 1\pmod 2$ Hence we have $a_{n} =x^{2} +3y^{2}$ for some $( x,y) \in \mathbb{Z}^{2}$ due to the following two lemmas. $\blacksquare $ Lemma1 Let $n$ be an integer. If $p$ is an odd prime divisor of $n^{2} +3$, then $p=x^{2} +3y^{2}$ for some $( x,y) \in \mathbb{Z}^{2}$. proof of this lemmaFor $p=3$, $p=0^{2} +3\cdotp 1^{2}$. Suppose that $p >3$. There exist $( s,t) ,( s',t') \in \mathbb{Z}^{2}$ such that $s-nt\equiv s'-nt\pmod p$ and $s,s',t,t'\in \left[ 0,\sqrt{p}\right)$. Let $x=s-s'$ and $y=t-t'$. Then $x^{2} =( s-s')^{2} \equiv ( n( t-t')^{2} =n^{2} y^{2} \equiv -3y^{2}\pmod p$. Furthermore, we have $x^{2} +3y^{2} < p+3p=4p$. So we have $x^{2} +3y^{2} =kp$ for some $k\in \mathbb{N}$ with $k\leq 3$. Since $x^{2} +3y^{2} \not \equiv 2\pmod 4$, we have $x^{2} +3y^{2} \neq 2p$. If $k=3$, then we have $3( x/3)^{2} +y^{2} =p$. If $k=1$, then we have $x^{2} +3y^{2} =p$, of course. $\blacksquare $ Lemma2 $\left( x^{2} +3y^{2}\right)\left( z^{2} +3w^{2}\right) =( xz+3yw)^{2} +3( xw-yz)^{2}$ proof of this lemmaJust expand the LHS and RHS.

\begin{align*} a_0=1&=1^2+3\cdot 0^2 \\ a_1=4&=1^2+3\cdot 1^2 \\ a_2=19&=4^2+3\cdot 1^2 \\ a_3=91&=4^2+3\cdot 5^2 \\ a_4=436&=19^2+3\cdot 5^2 \\ a_5=2089&=19^2+3\cdot 24^2 \\ a_6=10009&=91^2+3\cdot 24^2 \\ a_7=47956&=91^2+3\cdot 115^2 \\ \end{align*}In general, $a_{2n}=a^2_n + 3\cdot \left(\sum^{n-1}_{i=0} a_i\right)^2$ and $a_{2n+1}=a^2_n + 3\cdot \left(\sum^{n}_{i=0} a_i\right)^2$ for all $n\geq 1$. This can be proved using induction based on the difference $a_n-a_{n-1}$ for odd and even $n$.