32. If $a$ and $b$ are positive integers such that $a^{2}+2 a b-3 b^{2}-41=0$, find the value of $a^{2}+b^{2}$.
32. Answer: 221
We have $a^{2}+2 a b-3 b^{2}=(a-b)(a+3 b)=41$. Since 41 is a prime number, and $a-b<\bullet+3 b$, we have $a-b=1$ and $a+3 b=41$. Solving the simultaneous equations gives $a=11$ and $b=10$. Hence $a^{2}+b^{2}=221$.