Mathematical Induction for divisibility by $7$

Question

Not entirely sure if this is where I should post, but I need help.

I need to prove $7\mid (9^n - 2^n)$ for all $n\ge 1$. I have the parts for $n = 1$. But when it comes to solving $k \implies k+1$, I run into issues.

I get that $(9^k - 2^k) = 7a$. But here $(9\cdot 9^k - 2\cdot 2^k)$ is where I have trouble factoring out a $7$.

Any help would be appreciated.

Answer 1

Inductive step:

$$9^{k+1}-2^{k+1}=9(9^k-2^k)+9\times2^k-2\times2^k=9(9^k-2^k)+7\times2^k$$

Answer 2

Clearly holds for $n=1$.

Assume that it holds for $n=k$, therefore 7 divides $9^k - 2^k$.

Now prove for $n=k+1$: $9^{k+1} - 2^{k+1}= 9(9^k - 2^k) + 9(2^k)- 2(2^k)=9(9^k - 2^k)+7(2^k).$ Now clearly, 7 divides $7(2^k)$, and 7 divides $9^k - 2^k$ by our inductive step. This completes the proof.