1. **State the problem:** We are given positive integers $a$ and $b$ such that $$a^2 + 2ab - 3b^2 - 41 = 0$$ and we need to find the value of $$a^2 + b^2$$. 2. **Rewrite the given equation:** $$a^2 + 2ab - 3b^2 = 41$$ 3. **Try to factor the left-hand side:** Consider it as a quadratic in $a$: $$a^2 + 2ab - 3b^2 = (a + 3b)(a - b)$$ Check: $$(a + 3b)(a - b) = a^2 - ab + 3ab - 3b^2 = a^2 + 2ab - 3b^2$$ which is correct. 4. So the equation becomes: $$(a + 3b)(a - b) = 41$$ 5. Since $a$ and $b$ are positive integers and 41 is a prime number, the possible positive factor pairs of 41 are only (1, 41) and (41, 1). So either: - $a + 3b = 41$ and $a - b = 1$, or - $a + 3b = 1$ and $a - b = 41$ (not possible since $a+3b ot= 1$ for positive $a,b$) 6. From the first system: $$a + 3b = 41$$ $$a - b = 1$$ 7. Subtract the second equation from the first: $$(a + 3b) - (a - b) = 41 - 1 \\ 4b = 40 \\ b = 10$$ 8. Substitute back to find $a$: $$a - 10 = 1 \\ a = 11$$ 9. Compute $a^2 + b^2$: $$a^2 + b^2 = 11^2 + 10^2 = 121 + 100 = 221$$ **Final answer:** $$\boxed{221}$$