Prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ Is it possible to prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ for $n \geq 1 $ . I know that it is true because $ \frac{3^n - 4(2^n) + (-1)^n + 6}{24}$ represents the number of ways to uniquely $4$-colour an n-cycle , excluding permutations of colours.   

Using weak induction,
if $f(n)=3^n - 4(2^n) + (-1)^n + 6,$
$$f(m+2)-f(m)=3^m(3^2-1)-4\cdot2^m(2^2-1)$$ which is clearly divisible by $3\cdot8$ for $m\ge1$
So, $24\mid f(m)\iff24\mid f(m+2)$
Now establish the base cases $f(1),f(2)$
