1. **State the problem:** We are given positive integers $a$ and $b$ such that $$a^2 + 2ab - 3b^2 - 41 = 0$$ We need to find the value of $a^2 + b^2$. 2. **Rewrite the equation:** $$a^2 + 2ab - 3b^2 = 41$$ 3. **Attempt to factor or rewrite the left side:** Note that $$a^2 + 2ab - 3b^2 = (a + 3b)(a - b)$$ Verify: $$(a+3b)(a - b) = a^2 - ab + 3ab - 3b^2 = a^2 + 2ab - 3b^2$$ 4. **Set the factored form equal to 41:** $$(a + 3b)(a - b) = 41$$ Since $a$ and $b$ are positive integers, both factors are integers. 5. **Prime factorization of 41:** 41 is prime, so its positive factor pairs are only (1, 41) or (41, 1). 6. **Check possible values:** Case 1: $$a + 3b = 41, \, a - b = 1$$ Add the two equations: $$2a + 2b = 42 \, \Rightarrow \, a + b = 21$$ From second equation: $$a - b = 1$$ Add these two: $$2a = 22 \, \Rightarrow \, a = 11$$ Then, $$b = 21 - a = 21 - 11 = 10$$ Case 2: $$a + 3b = 1, \, a - b = 41$$ Since $a$ and $b$ are positive integers, this case is impossible because $a + 3b$ would be at least 1 but $a - b = 41$ suggests $a$ is much larger than $b$; however $a + 3b=1$ is too small to accommodate positive integers. 7. **Thus, $a=11$, $b=10$.** 8. **Compute $a^2 + b^2$:** $$a^2 + b^2 = 11^2 + 10^2 = 121 + 100 = 221$$ **Answer:** $\boxed{221}$