<s>[INST] Prove $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$ by induction Here is my attempted proof:
$\forall n \in \mathbb{N}$, let $S_n$ be the statement: $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
Base case: $S_1$: $\frac{2^{4(1)}-(-1)^1}{17} = \frac{16+1}{17} = 1 \in \mathbb{N}$
Inductive step: $\forall n \geq 1$, $S_n$ holds
$S_{n+1}$: $\frac{2^{4(n+1)}-(-1)^{n+1}}{17} = \frac{2^{4n} \cdot 2^4 +(-1)^{n}}{17} = \frac{2^{4n} \cdot (2^4+1) - 2^{4n} +(-1)^{n}}{17}$.
Now, we are left with $2^{4n} - \frac{2^{4n}-(-1)^n}{17}$. 
We know $\frac{2^{4n}-(-1)^n}{17}$, $2^{4n} \in \mathbb{N}$ and $2^{4n} > \frac{2^{4n}-(-1)^n}{17} \implies 2^{4n} - \frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
$$\tag*{$\blacksquare$}$$ 
Is my proof correct? I also feel there is a more elegant/smoother proof (assuming mine is correct).
 [/INST] Your proof is fine.
But note:
$(a - b)(\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k}) =$
$a(\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k}) - (\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k})=$
$(\sum\limits_{k=0}^{n-1}a^{k+1}b^{n-1 - k}) -(\sum\limits_{k=0}^{n-1}a^kb^{n-k}) = $
$(\sum\limits_{j=1}^n a^jb^{n-j})-(\sum\limits_{j=0}^{n-1}a^jb^{n-j})=$
$([\sum\limits_{j=1}^{n-1} a^jb^{n-j}]+a^nb^{n-n}) - (a^0b^{n-0}+[\sum\limits_{j=1}^{n-1} a^jb^{n-j}]) = $
$a^n - b^n$.
So for all integers $a,b$ and  $n\ge 1$ we will always have $a-b$ divides $a^n - b^n$
So $2^4 -(-1) = 17$ will always divide $(2^4)^n - (-1)^n$.
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