Assume $f(k)$ $=$ $8^k$ $-$ $3^k$ is divisible by 5
I'm trying to prove $f(k+1)$ is divisible by 5. If I use $f(k+1) - f(k)$, and then add $f(k)$ to both sides, I get $f(k+1) = f(k) + 5(8^k)$ which is divisible by 5.
However, I wan't to directly prove $f(k+1)$ is divisible by 5 without adding an $f(k)$ term:
$f(k+1) = 8(8^k) - 3(3^k)$
How can I factorise this to get the same result?
note that $$8\cdot 8^k-3\cdot 3^k=5\cdot 8^k+3(8^k-3^k)=f(k+1)$$