Prove by induction that $21^n-21$ is divisible by $5$ for all natural numbers.
I have tried this question several times but am always having to use that $0$ is divisible by $5$ in step one
So far I have done;
$21^1-21=0$ which is divisible by 5
then assumed true for n=k
I cannot think of a way to use these steps for n+1
Best Answer
Base Case ($n = 1$): $21^1 = 21 = 0$, and $0$ is of course divisible by $5$.
Induction Hypothesis: Assume $21^k - 21$ is divisible by $5$. Hence, there exists $m \in \mathbb{Z}$ such that $21^k - 21 = 5m$. (Hence, $21^k = 21 + 5m$).
Induction Step: We have: \begin{align*} 21^{k+1} - 21 & = 21^k \cdot 21 - 21 \\ & = (21 + 5m) 21 - 21 \\ & = 441 + 5(21m) - 21 \\ & = 420 + 5(21m) \\ & = 5(84) + 5(21m) \\ & = 5(84 + 21m). \end{align*}