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Prove that $2^n3^{2n} -1$ is always divisible by $17$.
I am very new to proofs and i was considering using proof by induction but I am not sure how to. I know you have to start by verifying the statement is true for the integer 1 but I dont know where to go from there.
$$2^n\cdot3^{2n}-1 = 18^n-1 = (18-1)(\cdots) = 17(\cdots)$$
Based on the OP's statement that she's trying to do this inductively:
You want to prove that for all $n$, the statement "$2^n3^{2n}-1$ is divisible by $17$" is true.
The first thing to do is to notice that $2^n3^{2n}-1=18^n-1$.
Next, you need to prove your base case: that is, that plugging in $n=1$, the result is true.
Last, you need to show that if the result holds for a given $n$, it also holds for $n+1$; that is, assuming that $18^n-1$ is divisible by $17$, prove that $18^{n+1}-1$ is also divisible by $17$.
By way of a hint for this last part, consider writing $18^{n+1}=18\cdot 18^n=18^n+17\cdot 18^n$. Can you see any way to make use of this?
Hint: $2\cdot3^2=18$ and $18\equiv1\pmod{17}$
If you do not know that $$ a\equiv b\pmod{p}\implies a^n=b^n\pmod{p}\tag{1} $$ then $(1)$ can be proven by induction pretty easily.
