<s>[INST] 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.
 [/INST] 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 $k$.
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