I need to prove that $ 7 \mid 3^{2n} - 2^n $ for every natural $n$. I have used induction on the modular expression $ 3^{2n} - 2^n \equiv 0 \bmod (7) $; the base case is trivial, and the thesis is
$$3^{2k+2}-2^{k+1} \equiv 0 \mod 7 \iff 9\cdot3^{2k}-2\cdot2^k \equiv 0 \mod 7,$$
but I can't really see how to use my inductive hypothesis in order to complete the proof. How should I proceed?
Let $3^{2n}-2^n \equiv 0 \pmod 7$ then there exist an integer $N$ satisfying $9^n - 2^n = 7N$. Then, $9^{n+1}- 2^{n+1} = 9 \cdot 9^n - 2 \cdot 2^n =9\cdot (7N + 2^n) - 2\cdot 2^n =63 N + 7\cdot 2^n\equiv 0 \pmod 7$.
Better way to use modular arithmetic is to proceed like \begin{align*} 9^n - 2^n \equiv 2^n - 2^n = 0 \pmod{7} \end{align*} without induction.