33. Find the largest positive integer $n$ such that $\sqrt{n-100}+\sqrt{n+100}$ is a rational number.
33. Answer: 2501
First we note that if $\sqrt{n-100}+\sqrt{n+100}$ is a rational number, then both $\sqrt{n-100}$ and $\sqrt{n+100}$ are rational. Since $n$ is a positive integer, this implies that $\sqrt{n-100}$ and $\sqrt{n+100}$ are integers. Let $\sqrt{n-100}=k$ and $\sqrt{n+100}=\ell$. Squaring both sides of the equations, we obtain $n-100=k^{2}$ and $n+100=\ell^{2}$. This gives
$$
200=\ell^{2}-k^{2}=(\ell-k)(\ell+k) .
$$

Now for $n$ to be the largest positive integer for which $\sqrt{n-100}+\sqrt{n+100}$ is rational, $\ell-k=\sqrt{n+100}-\sqrt{n-100}$ should be as small as possible. We rule out $\ell-k=1$ since it leads to $\ell$ and $k$ being non-integers from equation (1). Thus we have $\ell-k=2$, and it follows from equation (1) that $\ell+k=100$. Hence $\ell=51$ and $k=49$, and $n=k^{2}+100=49^{2}+100=2501$.