<s>[INST] 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.
 [/INST] 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$\#2:$
$$f(n+1)-3f(n)=?$$
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