By simple induction, $a_{n+m}=\sqrt{a_n^2+m}$, so $a_m=\sqrt{m+a_0^2}$. For $1$, $a_{p^2-a_0^2}=p$ is integral while $a_{p-a_0^2}=\sqrt p$ is irrational for sufficiently large primes $p>a_0^2$ (of which there are infinitely many). For $2$, we need $a_{2010}^2-a_0^2=2010$, impossible by$\pmod4$ since: $$a_{2010}^2-a_0^2\equiv0,1,3\pmod4,2010\equiv2\pmod4.$$