I'll just post my solution in short notes . Let $ x_i=a_{i+1}-a_i$ and $ y_i=b_{i+1}-b_i$. Then the condition from the task is equivalent to: $ (x_n+x_{n-1})^2+(y_n+y_{n-1})^2+x_n^2+y_n^2=x_{n-1}^2+y_{n-1}^2$ From this we obtain $ x_n^2+y_n^2 \leq x_{n-1}^2+y_{n-1}^2$ which means the some of squares of the differences is decreasing or staying equal(at start it is equal to $ x_1^2+y_1^2$). If $ x_{n-1}^2+y_{n-1}^2=x_n^2+y_n^2$ then by combining it with the original condition we easily obtain $ x_n=-x_{n-1}$ and $ y_n=-y_{n-1}$. From which follows $ a_{n-1}=a_{n+1}$. So we take first 2008 $ a_i$s and we conclude: if the $ x_i^2+y_i^2$ is not decreasing then we have $ a_1=a_{2009}$; if not then $ x_i^2+y_i^2$ decreased atleast by 1($ x$ and $ y$ are integer rays). So we continue by taking next 2008,,,, etc. By final amount of steps($ x_1^2+y_1^2$ is final) we will come to a moment where $ x_i=y_i=0$, from that point on, all members of $ x$ and $ y$ are $ 0$, so all members of $ a$ and $ b$ are equal and we can get what we wanted. The end

In coordinate plane let point $ P_n=(a_n,b_n)$ and the given condition implies that $ P_n$ is on the circumcircle whose diameter is segment $ P_{n-1}P_{n-2}$. Denote by $ d_n$ the square of the distance of $ P_n$ and $ P_{n+1}$, namely $ d_n=|P_{n}P_{n+1}|^2$. Clearly $ d_n$ is non-increasing and $ d_n$ is an non-negetive integer. So take $ n$ to infinite we get that there exists $ n$ such that either $ 0=d_n=d_{n+1}=\ldots$ or $ 0<d_n=d_{n+1}=\ldots$, that is, $ P_{n}=P_{n+1}$ or $ P_{n}=P_{n+2}$. Either implies $ P_k=P_{k+2008}$.

Slightly more detailed version of the above.

Let $s_n=a_n-a_{n-1}$ and $t_n=b_n-b_{n-1}$, so the initial condition becomes \[ s_n^2+s_ns_{n-1}+t_n^2+t_nt_{n-1}=0 \]Then by Cauchy-Schwarz \[ (s_n^2+t_n^2)(s_{n-1}^2+t_{n-1}^2)\ge (s_ns_{n-1}+t_nt_{n-1})^2=(s_n^2+t_n^2)^2 \]Thus either $s_n^2+t_n^2\le s_{n-1}^2+t_{n-1}^2$ or $s_n^2+t_n^2=0$, in which case the first inequality holds anyway. As $s_n^2+t_n^2$ is now a monovariant, it must eventually converge to some value. If $s_n^2+t_n^2=0$ for all sufficiently large $n$, then $s_n=0$ for sufficiently large $n$, easily implying the desired results. So now assume that this is not the case; as $s_n^2+t_n^2$ converges to some positive value, equality must hold in each Cauchy-Schwarz, and $s_n=ks_{n-1}$, $t_n=kt_{n-1}$ for some $k$ and each $n$. But then $s_n^2+t_n^2=s_{n-1}^2+t_{n-1}^2$ implies $k^2=1$, and the initial condition implies $k^2+k=0$, so $k=-1$. Thus $s_n=-s_{n-1}$ for all sufficiently large $n$, yielding that $a_k$ eventually periodic with period $2$.

The key observation is that Lemma: For $A(x_1, y_1)$ and $B(x_2, y_2)$ in the plane, the equation of a circle with diameter $AB$ is given by $$C(x, y) \overset{\text{def}}{:=} (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.$$ (Proof) Follows from the fact that $AC \perp CB$ so $m_{AC}\cdot m_{BD}=-1.$ Define the sequence of points $\left(P_n(a_n, b_n)\right)_{n \ge 1}$ and observe that $P_{n+2}$ lies on the circle with diameter $P_nP_{n+1}$. It follows that the sequence $d_n=|P_nP_{n+1}|^2$ of positive integers is non-increasing, so for all large $n$ we have $P_{n+2} \in \{P_n, P_{n+1}\} \Longrightarrow P_k=P_{k+2008}$ for all large $k$ since the sequence is either eventually constant, or periodic with period $2$. Remark: Studying for boards isn't all that bad after all

define $x_n=a_n-a_{n-1} $ and $y_n=b_n-b_{n-1}$ then the given condition equivalent to $x_n^2+y_n^2+x_n.x_{n-1}+y_n.y_{n-1}=0 $ which is equivalent to $(2x_n+1)^2+(2y_n+1)^2=2 \implies |2x_n+1|=1$ then $x_n=0$ or $x_n=-1$ then $a_n=a_{n-1}$ or $a_n=a_{n-1}-1$ which implies $a_n=a_{n+2008} $ for a large enough $n$ and we win