From the recurrent relation, we have: \begin{eqnarray*}x_{n+2}x_n&=&x_{n+1}^2+3\\x_{n}^2+3&=&x_{n+1}x_{n-1}\end{eqnarray*} Adding the equations and factoring yields \[ \frac{x_{n}+x_{n+2}}{x_{n+1}+x_{n-1}}=\frac{x_{n+1}}{x_n}\qquad(1)\] If we substitute $ n-1$ for $ n$ in $ (1)$, we find that \[ \frac{x_{n-1}+x_{n+1}}{x_{n}+x_{n-2}}=\frac{x_{n}}{x_{n-1}}\qquad(2)\] In fact, we can generate $ n-1$ of these equalities, which will all telescope when multiplied together. Thus:\begin{eqnarray*}\left(\frac{x_{n}+x_{n+2}}{x_{n+1}+x_{n-1}}\right)\times\left(\frac{x_{n-1}+x_{n+1}}{x_{n}+x_{n-2}}\right)\times\cdots\times\left(\frac{x_{2}+x_{4}}{x_{3}+x_{1}}\right)&=&\frac{x_{n+1}}{x_n}\times\frac{x_{n}}{x_{n-1}}\times\cdots\times\frac{x_3}{x_2}\\&\iff &x_{n}+x_{n+2}=\left(\frac{x_3+x_1}{x_2}\right)\cdot x_{n+1}\\&\iff&x_{n+2}-4x_{n+1}+x_n=0\qquad(3)\end{eqnarray*} Thus $ x_{2n+1}=2x_{n+1}^2-1$. The two recurrent equations established thus far are enough to establish these two properties of $ \{x_n\}$:\[ x_{2n-1}=2x_n^2-1\qquad \text{ and } \qquad x_{2n}=\left(x_{n+1}-x_n\right)^2+1\qquad(3)\] When $ n=1$, we check that $ x_1=2-1$ and $ x_2=(2-1)^2+1$ as claimed. Suppose the results in $ (4)$ hold up until $ n=k$. Using the equation $ (3)$, we compute \begin{eqnarray*}\frac{x_{2n+1}-(2x_{n+1}^2-1)}{2}&&\qquad\text{(trying to show this is exactly zero)}\\&=&\frac{(4x_{n+1}^2+4x_{n}^2-8x_nx_{n+1}+4) - (2x_n^2-1) - (2x_{n+1}^2-1)}{2}\\&=& x_n^2+x_{n+1}^2+3-4x_{n+1}x_n\qquad\text{(by the original recurrent equation)}\\&=&x_n^2+x_nx_{n+2}-4x_{n+1}x_n\\&=&x_n(x_{n+2}-4x_{n+1}+x_n)=0\end{eqnarray*} To complete the recursion, it remains to show the result for $ x_{2n+2}$: \begin{eqnarray*} x_{2n+2}+x_{2n}&=&4x_{2n+1}=8x_{n+1}^2-4\\ &=&8x_{n+1}^2+2(x_{n+1}^2-x_{n}x_{n+2})+2\qquad\text{(original recursion)}\\ &=&16x_{n+1}^2-2x_nx_{n+2}+2x_{n+1}^2-8x_{n+1}^2+2\\ &=&(x_n+x_{n+2})^2-2x_nx_{n+2}+2x_{n+1}^2-2x_{n+1}(x_n+x_{n+2})+2\qquad\text{(using derived linear recursion)}\\ &=&2x_{n+1}^2+x_n^2+x_{n+2}^2-2x_{n+1}(x_n+x_{n+2})+2\\ &=&(x_{n+1}-x_n)^2+1 +(x_{n+2}-x_{n+1})^2+1\\ &=&x_{2n}+(x_{n+2}-x_{n+1})^2+1\\ \end{eqnarray*} This completes the recursion. Hence, $ \boxed{x_{2008}=(x_{1005}-x_{1004})^2+1}$ is precisely the sum of two squares$ \square$

I will prove by induction that $ x_{n+2}=4x_{n+1}-x_{n}$ for all positive integers $ n$. Base Case: $ n=1$. Calculate $ x_3=\frac{x_2^2+3}{x_1}=\frac{2^2+3}{1}=7=4(2)-1=4x_2-x_1$. Inductive Step: Assume that $ x_{n+1}=4x_{n}-x_{n-1}$. Then \begin{align*} x_{n+2}x_{n}-x_{n+1}^2&=3\\ x_{n+1}x_{n-1}-x_{n}^2&=3\\ x_{n+2}x_{n}-x_{n+1}^2&=x_{n+1}x_{n-1}-x_{n}^2\\ x_{n+2}x_{n}-x_{n+1}(4x_{n}-x_{n-1})&=x_{n+1}x_{n-1}-x_{n}^2\\ x_{n+2}x_{n}-4x_{n+1}x_{n}&=-x_{n}^2\\ x_{n+2}-4x_{n+1}&=-x_{n}\\ x_{n+2}&=4x_{n+1}-x_{n}\\ \end{align*} (Dividing by $ x_n$ is legal, because it is not zero; if $ n=1$ or $ 2$, then $ x_n=1$ or $ 2$, and if $ n>2$, then $ x_{n}=\frac{x_{n-1}^2+3}{x_{n-2}}$ and $ x_{n-1}^2+3\ge 3$ cannot be zero.) This completes the induction. This shows that $ x_1,x_2,...$ is an integer sequence, since if $ x_{n}$ and $ x_{n+1}$ are integers, then $ x_{n+2}=4x_{n+1}-x_{n}$ is an integer. Now, I will now show by induction that $ x_n-1$ is a perfect square for even $ n$. Base Case: $ n=2$. We have $ x_2-1=2-1=1^2$ is a perfect square. Inductive Step: Assume that $ x_{n}-1$ is a perfect square; I will show that $ x_{n+2}-1$ is as well. \begin{align*} (x_{n}-1)(x_{n+2}-1)&=x_{n}x_{n+2}-(x_{n}+x_{n+2})+1\\ (x_{n}-1)(x_{n+2}-1)&=(x_{n+1}^2+3)-4x_{n+1}+1\\ (x_{n}-1)(x_{n+2}-1)&=x_{n+1}^2-4x_{n+1}+4\\ (x_{n}-1)(x_{n+2}-1)&=(x_{n+1}-2)^2\\ \end{align*} Now, since $ (x_{n+1}-2)^2$ and $ x_{n}-1$ are perfect squares, and $ x_{n+2}-1$ is an integer it must also be a perfect square. This completes the induction. Now, we have that $ x_{2008}-1$ is a perfect square, and since $ 1$ is a perfect square as well, $ x_{2008}$ is the sum of two squares.

$ x_{n + 2}x_n = x_{n + 1}^2 + 3$ ,$ x_{n + 1}x_{n - 1} = x_n^2 + 3$ $ \frac {x_{n + 2} + x_n}{x_{n} = \frac {x_{n + 1} + x_{n - 1}}{x_{n - 1} = .... = \frac {x_{3} + x_1}{x_{1} = 7$ $ x_{n + 2} = 6x_n$ ->$ x_2008 = 6^{1004}x_1$

We write the condition in the form $ x_nx_{n+2}=x_{n+1}^2+3$ and if we write it for $ n$ the $ n-1$ we get $ x_{n-1}x_{n+1}=x_n^2+3$ and substracting these two we have that $ \frac{x_n}{x_{n+1}+x_{n-1}}=\frac{x_{n+1}}{x_{n+2}+x_n}=\frac{x_2}{x_3+x_1}=\frac{1}{4}$ so $ x_{n+1}=4x_n-x_{n-1}$ and from here easily we obtain that $ x_{n+2}=14x_n-x_{n-2}$ so if $ y_n=x_{2n}$ then we have that $ y_{n}=14y_{n-1}-y_{n-2}$ Now we will prove by induction that there exist a sequence $ b_n$ with $ b_{n+1}=4b_n-b_{n-1}$ and $ b_1=1$ and $ b_2=5$ such that $ y_n=b_n^2+1$ which will be the desired . The result is obvious for $ n=1$ and $ n=2$ so by strong induction we suppose that it is true for all $ k\leq n$ and we will prove it for $ n+1$ . We have that $ y_{n+1}=15y_n-15y_{n-1}-y_{n-2}$ or $ y_{n+1}-1=15(y_n-1)-15(y_{n-1}-1)+y_{n-2}-1$ or $ y_{n+1}-1=15b_n^2-15b_{n-1}^2+b_{n-2}^2$ or $ y_{n+1}-1=15b_n^2-15b_{n-1}^2+(4b_{n-1}-b_n)^2$ which means that $ y_{n+1}-1=(4b_n-b_{n-1})^2=b_{n+1}^2$ which is what we want and we are done .