1. **State the problem:** Prove by induction that $$\frac{2^{3^n} + 1}{3^{n+1}}$$ is an integer for all integers $n \geq 0$. 2. **Base case ($n=0$):** Calculate $$\frac{2^{3^0} + 1}{3^{0+1}} = \frac{2^1 + 1}{3^1} = \frac{2 + 1}{3} = \frac{3}{3} = 1$$ which is an integer. 3. **Inductive hypothesis:** Assume for some $k \geq 0$, $$\frac{2^{3^k} + 1}{3^{k+1}} = m$$ where $m$ is an integer. 4. **Inductive step:** We need to prove $$\frac{2^{3^{k+1}} + 1}{3^{k+2}}$$ is an integer. Note that $$3^{k+1} = 3 \cdot 3^k$$, so rewrite the numerator: $$2^{3^{k+1}} + 1 = 2^{3 \cdot 3^k} + 1 = \left(2^{3^k}\right)^3 + 1$$ Use the sum of cubes factorization: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ with $$a = 2^{3^k}$$ and $$b = 1$$: $$\left(2^{3^k}\right)^3 + 1^3 = \left(2^{3^k} + 1\right)\left(\left(2^{3^k}\right)^2 - 2^{3^k} \cdot 1 + 1^2\right)$$ So, $$2^{3^{k+1}} + 1 = \left(2^{3^k} + 1\right)\left(2^{2 \cdot 3^k} - 2^{3^k} + 1\right)$$ Divide by $$3^{k+2} = 3 \cdot 3^{k+1}$$: $$\frac{2^{3^{k+1}} + 1}{3^{k+2}} = \frac{\left(2^{3^k} + 1\right)\left(2^{2 \cdot 3^k} - 2^{3^k} + 1\right)}{3 \cdot 3^{k+1}} = \frac{2^{3^k} + 1}{3^{k+1}} \cdot \frac{2^{2 \cdot 3^k} - 2^{3^k} + 1}{3}$$ By the inductive hypothesis, $$\frac{2^{3^k} + 1}{3^{k+1}} = m$$ is an integer. It remains to show $$\frac{2^{2 \cdot 3^k} - 2^{3^k} + 1}{3}$$ is an integer. 5. **Show divisibility by 3:** Consider $$x = 2^{3^k}$$. Calculate $$x^2 - x + 1 \pmod{3}$$: Since $$2 \equiv -1 \pmod{3}$$, then $$2^{3^k} \equiv (-1)^{3^k} \pmod{3}$$. Because $$3^k$$ is odd (3 to any power is odd), $$(-1)^{3^k} = -1 \equiv 2 \pmod{3}$$. So, $$x \equiv 2 \pmod{3}$$ Then, $$x^2 - x + 1 \equiv 2^2 - 2 + 1 = 4 - 2 + 1 = 3 \equiv 0 \pmod{3}$$ Thus, $$3$$ divides $$x^2 - x + 1$$. 6. **Conclusion:** Since both factors are integers and the second factor is divisible by 3, the entire expression $$\frac{2^{3^{k+1}} + 1}{3^{k+2}}$$ is an integer. By induction, the statement holds for all $n \geq 0$. **Final answer:** The expression $$\frac{2^{3^n} + 1}{3^{n+1}}$$ is an integer for all integers $n \geq 0$.