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.
 A: As $3^n$ is an odd number, $4\cdot 3^n\equiv 4~(mod~8)$, also $12\equiv 4~(mod~8)$.
A: $$4 \cdot 5^n+4 \cdot 3^n$$ is clearly divisible by 8 cause once you take the 4 out it is the sum of two odd numbers.
A: It is $4\cdot 3^n-12=4(3^n-3)$, since $3^n-3$ is even $2\mid 3^n-3$. Hence $8\mid 4\cdot 3^n-12$.
To make it clear: $3^n-3$ is even, since for $n\geq 1$ it is $3^n$ odd, and 
"odd-odd=even" Since $(2k+1)-(2l+1)=2k-2l=2(k-l)$ which is even.
A: If $f(m)=5^m+2\cdot3^m-3,$
$$f(n+2)-f(n)=5^n(5^2-1)+2(3^2-1)3^n$$ which is clearly divisible by $8$
$\implies8|f(n+2)\iff8|f(n)$
Now establish the base cases $f(0),f(1)$
A: 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.)
A: If $f(m)=5^m+2\cdot3^m-3,$
Let use eliminate either of $5^n$ or $3^n$
Method$\#1:$
$$f(n+1)-5f(n)=3^n(6-10)-(3-3\cdot5)=4(3-3^n)$$
Now as $3^n$ is odd, for integer $n\ge0,3-3^n$ is even
$\implies f(n+1)-5f(n)$ is divisible by $8$
$\implies8|f(n+1)\iff8|f(n)$ as $8\nmid5$
Now establish the base case $f(0)$
Method$\#2:$
$$f(n+1)-3f(n)=?$$
