Help with induction using divisibility. Need some help with this question for induction using divisibility: $8*19^n-2*3^{2n+2}$ is divisible by 10 for any positive integer.
What I have so far is proving the first value and $P_k$ and $P_{k+1}$
$P_1$ = $8*19^1-2*3^{2+2}$=-10; $P_1$ is true.
$P_k$ = $8*19^k-2*3^{2k+2}$= 10A
$P_{k+1}$ = $8*19^{k+1}-2*3^{2k+4}$ = 10B
I've just found myself stuck on the proof starting with:
$8*19^{k+1}-2*3^{2k+4}$=
Thanks for any help. If I could get an answer in the format of $P_k$ and $P_{k+1}$ that would greatly appreciated.
 A: In fact, it works for non-negative integers $n$. Let $u_n:=8\times 19^n-2\times 9^{n+1}$ so $u_0=-10$ and $u_{k+1}-9u_k=80\times 19^k$, so $10|u_0$ and $10|u_k\implies 10|u_{k+1}$.
A: Conceptually the inductive step follows very simply by multiplying the first two congruences below using  CPR = Congruence Product Rule, $ $ 
$$\begin{align}\bmod 10\!:\qquad\ \ \  \color{#c00}{19}\ &\equiv\ \color{#c00}{9}\\[.2em]
{\rm times}\ \ \ \ \ \ \ 8\cdot 19^{\large n}&\equiv 18\cdot 9^{\large n}\quad\ \ \ \ \ P(n)_{\phantom{|}}\\[.2em]
\hline
\Longrightarrow\ \ 8\cdot 19^{\large n+\color{#c00}1}&\equiv 18\cdot 9^{\large n+\color{#c00}1}\quad\ P(n\!+\!\color{#c00}1),
\ \ \rm by \ CPR^{\phantom{|^|}}\end{align}\qquad  $$
i.e. the proof is a special case of the  (inductive) proof of the Congruence Power Rule. Note how the use of congruences  highlights innate arithmetical structure allowing us to reduce the induction to an easy one $\,a\equiv b\,\Rightarrow\, a^n\equiv b^n,\,$ with obvious inductive step: multiply by $\,a\equiv b\,$ via the product rule.
Remark $ $  If you don't know congruences we can preserve this arithmetical essence by using an analogous product rule for divisibility (DPR), as explained here.
A: Let $P(n)$ be the statement that $10 \mid 8 \cdot 19^n - 2 \cdot 3^{2n + 2}$.
If we wish to establish that the statement holds for all nonnegative integers, we have to verify that $P(0)$ holds since that will be our base case.
Since $8 \cdot 19^0 - 2 \cdot 3^2 = 8 - 18 = -10 = 10 \cdot (-1)$, $P(0)$ holds.
If we wish to establish that the statement holds for all positive integers, we have to verify that $P(1)$ holds since that will be our base case.
Since $8 \cdot 19^1 - 2 \cdot 3^4 = 8 \cdot 19 - 2 \cdot 81 = 152 - 162 = -10 = 10 \cdot (-1)$, $P(1)$ holds.
Assume $P(k)$ holds for some positive integer $k$.  Then $10 \mid 8 \cdot 19^k - 2 \cdot 3^{2k + 2}$, so there exists $m \in \mathbb{Z}$ such that $8 \cdot 19^k - 2 \cdot 3^{2k + 2} = 10m$.
Let $n = k + 1$.  Then
\begin{align*}
8 \cdot 19^n - 2 \cdot 3^{2n + 2} & = 8 \cdot 19^{k + 1} - 2 \cdot 3^{2(k + 1) + 2}\\
& = 19 \cdot 8 \cdot 19^k - 2 \cdot 3^{2k + 4}\\
& = 19 \cdot 8 \cdot 19^k - 9 \cdot 2 \cdot 3^{2k + 2}\\
& = (10 + 9) \cdot 8 \cdot 19^k - 9 \cdot 2 \cdot 3^{2k + 2}\\
& = 10 \cdot 8 \cdot 19^k + 9(8 \cdot 19^k - 2 \cdot 3^{2k + 2})\\
& = 10 \cdot 8 \cdot 19^k + 9 \cdot 10m && \text{by the induction hypothesis}\\
& = 10(8 \cdot 19^k + 9m)
\end{align*}
Thus, if $10 \mid 8 \cdot 19^k - 2 \cdot 3^{2k + 2}$, $10 \mid 8 \cdot 19^{k + 1} - 2 \cdot 3^{2(k + 1) + 2}$.  Hence, $P(k) \implies P(k + 1)$.
Since $P(1)$ holds and $P(k) \implies P(k + 1)$ for each $k \geq 1$, $P(n)$ holds for each positive integer $n$.  The same steps that establish $P(k) \implies P(k + 1)$ also apply when $k$ is a nonnegative integer.
In fact, since $P(0)$ holds, $P(1)$ holds, and $P(k) \implies P(k + 1)$ for each $k \geq 1$, $P(n)$ holds for each nonnegative integer $n$.
Had we wished to prove that $P(n)$ holds for each nonnegative integer $n$, we could have skipped the step of establishing that $P(1)$ holds and instead assumed that $P(k)$ holds for some nonnegative integer $k$.
