Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdot3^n$ is divisible by $4$.
After that I am not sure what to do.

If $a$ is divisible by $4$, then $b$ is divisible by $4$ if and only if $b-a$ is divisible by $4$.
So, you can assume $6\cdot 7^n-2\cdot 3^n$ is divisible by $4$ and try for
\begin{align}
(6\cdot 7^{n+1}-2\cdot 3^{n+1})-(6\cdot 7^n-2\cdot 3^n)
&=
(6\cdot 7^{n+1}-6\cdot 7^{n})-(2\cdot 3^{n+1}-2\cdot 3^{n})
\\
&=
(6\cdot 7\cdot 7^{n}-6\cdot 7^{n})-(2\cdot 3\cdot 3^{n}-2\cdot 3^{n})\\
&=
\dots
\end{align}

Of course, using congruences modulo $4$ is easier:
$$
6\cdot 7^n-2\cdot 3^n\equiv 2\cdot 3^n-2\cdot 3^n\equiv 0 \pmod{4}
$$
