Indija TS

Consider points $P_j=(a_j,b_j)$ in the plane. The slope of the lines $P_n{P}_{n-1}$ and $P_n{P}_{n-2}$ are $$r=\frac{b_n-b_{n-1}}{a_n-a_{n-1}},s=\frac{b_n-b_{n-2}}{a_n-a_{n-2}},$$ respectively. The given condition implies that $rs=-1$. Hence it follows that the lines $P_n{P}_{n-1}$ and $P_n{P}_{n-2}$ are perpendicular to each other. This implies that $P_n$ lies on the circle $C_n$ with diameter $P_{n-1}{P}_{n-2}$. Let $f(n)=|P_{n-1}-P_{n-2}{|}^2$. Then $f(n)$ is an integer. The observation that $P_n$ lies on the circle $C_n$ shows that $f(n)\le f(n-1)$. Thus we get a non-increasing sequence of positive integers. This must be constant after certain stage. Thus $f(n)$ is constant for $n\ge N$, for some positive integer $N$. This implies that the diameter of $C_n$ is constant for all $n\ge N$. Hence $C_n$ are all equal circles for $n\ge N$. This implies that $P_n=P_{n+2}$ for all $n\ge N$. But then $a_n=a_{n+2}$ for all $n\ge N$. Hence $$a_{N+2011}=a_{N+(2011{)}^{2011}}.$$