33. It is known that there is only one pair of positive integers $a$ and $b$ such that $a \leq b$ and $a^{2}+b^{2}+8 a b=2010$. Find the value of $a+b$.
33. Ans: 42 .
Since $a \geq 1,2010=a^{2}+b^{2}+8 a b \geq 1+b^{2}+8 b . b^{2}+8 b-2009 \leq 0$. However $b^{2}+8 b-2009=0$ has an integer solution 41. So $a=1$ and $b=41$. The result follows.