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?
I’m sure that she used the following factorization

$$\begin{align*}
2^{3n}-1&amp;=(2^3)^n-1\\
&amp;=8^n-1^n\\
&amp;=(8-1)(8^{n-1}\cdot 1^0+8^{n-2}\cdot 1^1+\dots+8^2\cdot 1^{n-3}+8^1\cdot 1^{n-2}+8^0\cdot 1^{n-1}\\
&amp;=7(8^{n-1}+8^{n-2}+\dots+8^2+8+1)\\
&amp;=7\sum_{k=0}^{n-1}8^k\;,
\end{align*}$$

which as Dylan Moreland pointed out is a special case of $$\begin{align*}a^n-b^n&amp;=(a-b)(a^{n-1}b^0+a^{n-1}b^1+\dots+a^1b^{n-2}+a^0b^{n-1})\\
&amp;=(a-b)\sum_{k=0}^{n-1}a^kb^{n-1-k}\end{align*}\;.$$

This is really just the formula for the sum of a finite geometric series in disguise:

$$\begin{align*}
\sum_{k=0}^{n-1}a^kb^{n-1-k}&amp;=b^{n-1}\sum_{k=0}^{n-1}\left(\frac{a}b\right)^k\\
&amp;=b^{n-1}\frac{\left(\frac{a}b\right)^n-1}{\frac{a}b-1}\\
&amp;=\frac{\frac{a^n}b-b^{n-1}}{\frac{a-b}b}\\
&amp;=\frac{a^n-b^n}{a-b}\;,
\end{align*}$$

and multiplying through by $a-b$ yields the factorization formula.
\begin{align*}
2^{3n}−1 &amp; =(2^3)^n−1\\
         &amp; =8^n−1^n\\
         &amp; =(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})\\
         &amp; =7\sum(8^{n−1}+8^{n−2}+⋯+8^2+8+1)\\
         &amp; =7\sum_{k=0}^{n−1} 8^k
\end{align*}

\begin{align*}
2^{3n}−1 &amp; =(2^3)^n−1\\
         &amp; = 8^n−1^n\\
         &amp; = (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})\\
          &amp; = 7(8^{n−1}+8^{n−2} + \cdots +8^2+8+1)\\
          &amp; = 7\sum_{k=0}^{n−1}8^k
\end{align*}