Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction) 
Prove $5^n + 2  \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$

Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2  \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2  \cdot 3^{n+1} - 3$
$$5^{n+1} + 2  \cdot 3^{n+1} - 3$$ $$5\cdot 5^n + 2\cdot 3\cdot 3^n - 3$$ $$ (5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n $$ $$ 5\cdot(5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n - 4\cdot(5^n + 2\cdot 3^n - 3)$$
$$ [5\cdot(5^n + 2\cdot 3^n - 3)]  - [4\cdot 3^n - 12]$$
The first term divides by 8 but I am not sure how to get the second term to divide by 8.

Hint: Try reducing your expression mod 8. For example, what is $5^n$ mod 8? Since $5 \equiv 5$ and $5^2 = 25 \equiv 1$, 
$5^n \equiv \begin{cases} 5, n \text{ odd} \\ 1, n \text{ even} \end{cases}$.
Now do the same for $3^n$, and add. 
(This solution doesn't use induction.)
