<s>[INST] Use mathematical induction to prove that for all integers $n \ge 2$, $2^{3n}-1$ is not prime I had a homework due yesterday with this problem.
The TA did the problem last week in discussion but I didn't understand it.
She pulled out a $7k$ almost immediately, and I have no idea from where.
It was like, it wasn't prime if $2^{3n}-1$ was 7 times some constant. I understand that that would make $2^{3n}-1$ not prime, but I don't understand how she just used "7".
Where did she get that from? I was thinking, maybe it was like $2^3 \cdot 2^n-1$... which is $8 \cdot 2^n-1$... but you can't just do $8-1$. How?
 [/INST] \begin{align*}
2^{3n}−1 & =(2^3)^n−1\\
         & =8^n−1^n\\
         & =(8−1)(8^{n−1} \cdot 1^0+8^{n−2} \cdot 1^1+⋯+8^2 \cdot 1^{n−3}+8^1 \cdot 1^{n−2}+8^0 \cdot 1^{n−1})\\
         & =7\sum(8^{n−1}+8^{n−2}+⋯+8^2+8+1)\\
         & =7\sum_{k=0}^{n−1} 8^k
\end{align*}
\begin{align*}
2^{3n}−1 & =(2^3)^n−1\\
         & = 8^n−1^n\\
         & = (8−1)(8^{n−1} \cdot 1^0+8^{n−2}\cdot 1^1+ \cdots +8^2⋅1^{n−3}+ 8^1 \cdot 1^{n−2}+ 8^0⋅1^{n−1})\\
          & = 7(8^{n−1}+8^{n−2} + \cdots +8^2+8+1)\\
          & = 7\sum_{k=0}^{n−1}8^k
\end{align*}
</s>