1. **Problem Statement:** Show that if $\alpha$ and $\beta$ are roots of the quadratic equation $$x^2 - 14x + 36 = 0,$$ then for all positive integers $n$, the expression $$\alpha^n + \beta^n$$ is divisible by $$2^n$$. 2. **Step 1: Identify the roots and their properties.** From the quadratic equation $$x^2 - 14x + 36 = 0,$$ by Vieta's formulas: - Sum of roots: $$\alpha + \beta = 14$$ - Product of roots: $$\alpha \beta = 36$$ 3. **Step 2: Define the sequence and find a recurrence relation.** Let $$S_n = \alpha^n + \beta^n$$. Using the characteristic equation, the sequence $$S_n$$ satisfies the recurrence: $$S_n = (\alpha + \beta) S_{n-1} - (\alpha \beta) S_{n-2}$$ which becomes $$S_n = 14 S_{n-1} - 36 S_{n-2}$$ with initial values: $$S_0 = \alpha^0 + \beta^0 = 1 + 1 = 2$$ $$S_1 = \alpha + \beta = 14$$ 4. **Step 3: Prove divisibility by induction.** We want to prove $$2^n \mid S_n$$ for all $$n \geq 1$$. **Base cases:** - For $$n=1$$, $$S_1 = 14 = 2 \times 7$$, divisible by $$2^1$$. - For $$n=2$$, $$S_2 = 14 S_1 - 36 S_0 = 14 \times 14 - 36 \times 2 = 196 - 72 = 124$$. Check divisibility by $$2^2 = 4$$: $$124 \div 4 = 31$$, an integer, so divisible. 5. **Step 4: Inductive hypothesis.** Assume for some $$k \geq 2$$, $$S_k$$ is divisible by $$2^k$$ and $$S_{k-1}$$ is divisible by $$2^{k-1}$$. 6. **Step 5: Inductive step.** Using the recurrence: $$S_{k+1} = 14 S_k - 36 S_{k-1}$$ By the inductive hypothesis: - $$S_k = 2^k m$$ for some integer $$m$$ - $$S_{k-1} = 2^{k-1} n$$ for some integer $$n$$ Substitute: $$S_{k+1} = 14 (2^k m) - 36 (2^{k-1} n) = 2^k (14 m) - 2^k (18 n) = 2^k (14 m - 18 n)$$ Factor out $$2^k$$: $$S_{k+1} = 2^k (14 m - 18 n)$$ Since $$14 m - 18 n$$ is an integer, $$S_{k+1}$$ is divisible by $$2^k$$. 7. **Step 6: Show divisibility by $$2^{k+1}$$.** Rewrite: $$S_{k+1} = 2^k (14 m - 18 n) = 2^{k+1} \times \frac{14 m - 18 n}{2}$$ We need to show $$\frac{14 m - 18 n}{2}$$ is an integer. Since $$m$$ and $$n$$ are integers, $$14 m$$ and $$18 n$$ are even numbers (both divisible by 2), so their difference is even. Therefore, $$\frac{14 m - 18 n}{2}$$ is an integer. Hence, $$S_{k+1}$$ is divisible by $$2^{k+1}$$. 8. **Step 7: Conclusion.** By mathematical induction, $$S_n = \alpha^n + \beta^n$$ is divisible by $$2^n$$ for all positive integers $$n$$. **Final answer:** $$\boxed{2^n \mid \alpha^n + \beta^n \text{ for all } n \geq 1}$$