1. Problem: Prove that $5^n - 1$ is divisible by 4 for all $n \in \mathbb{N}$. Step 1: Base Case ($n=1$): Calculate $5^1 - 1 = 5 - 1 = 4$, which is divisible by 4. So, the base case holds. Step 2: Inductive Hypothesis: Assume $5^k - 1$ is divisible by 4, i.e., $5^k - 1 = 4m$ for some integer $m$. Step 3: Inductive Step: Show $5^{k+1} - 1$ is divisible by 4. $$5^{k+1} - 1 = 5 \cdot 5^k - 1 = 5(5^k) - 1 = 5(4m + 1) - 1 = 20m + 5 - 1 = 20m + 4 = 4(5m + 1)$$ which is divisible by 4. Conclusion: The statement holds for all $n \in \mathbb{N}$. 2. Problem: Prove that $2^n > n^2$ for all integers $n \geq 5$. Step 1: Base Case ($n=5$): Calculate $2^5 = 32$ and $5^2 = 25$. Since $32 > 25$, base case holds. Step 2: Inductive Hypothesis: Assume $2^k > k^2$ for some integer $k \geq 5$. Step 3: Inductive Step: Show $2^{k+1} > (k+1)^2$. $$2^{k+1} = 2 \cdot 2^k > 2 \cdot k^2$$ We want to show: $$2 k^2 > (k+1)^2 = k^2 + 2k +1$$ This is equivalent to: $$2 k^2 - k^2 - 2k -1 = k^2 - 2k -1 > 0$$ At $k=5$, $25 - 10 - 1 =14 > 0$, and for $k > 5$, since $k^2$ grows faster than $2k + 1$, this inequality holds. Conclusion: The statement holds for all integers $n \geq 5$. 3. Problem: Prove that $7^n - 1$ is divisible by 6 for all $n \in \mathbb{N}$. Step 1: Base Case ($n=1$): $7^1 - 1 = 6$ which is divisible by 6. Base case holds. Step 2: Inductive Hypothesis: Assume $7^k - 1 = 6m$ for some integer $m$. Step 3: Inductive Step: Show $7^{k+1} - 1$ divisible by 6. $$7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(6m + 1) - 1 = 42m + 7 - 1 = 42m + 6 = 6(7m + 1)$$ which is divisible by 6. Conclusion: The statement holds for all $n \in \mathbb{N}$. 4. Problem: Prove that $n^3 - n$ is divisible by 6 for all $n \in \mathbb{N}$. Step 1: Base Case ($n=1$): $1^3 - 1 = 0$ is divisible by 6. Base case holds. Step 2: Inductive Hypothesis: Assume $k^3 - k$ divisible by 6 for some $k$. Step 3: Inductive Step: Show $(k+1)^3 - (k+1)$ divisible by 6. Expand: $$(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k -1 = k^3 + 3k^2 + 2k$$ Group terms: $$(k^3 - k) + 3k(k+1)$$ By hypothesis, $k^3 - k$ is divisible by 6. Also, among any two consecutive integers $k$ and $k+1$, one is even, so $k(k+1)$ is even. Multiplying by 3 gives a multiple of 6. Thus, the sum is divisible by 6. Conclusion: The statement holds for all $n \in \mathbb{N}$.