Show that $7\mid 2^{3n} -1 \,\,\,\forall n\in \mathbb N^+$

Question

Show that $7\mid 2^{3n} -1 \,\,\,\forall n\in \mathbb N^+$

Should I prove this by induction? If so, how should I go about it?

Answer 1

If you don't mind I'm gonna do some magic!

$$2^{3n}=(2^3)^n=8^n=(7+1)^n$$ Now use binomial expansion to get

$$\begin{align} (7+1)^n&=\binom{n}{0}7^n+\binom{n}{1}7^{n-1}+\cdots+\binom{n}{n-1}7+\binom{n}{n}7^0\\ &=\binom{n}{0}7^n+\binom{n}{1}7^{n-1}+\cdots+\binom{n}{n-1}7+1\\ (7+1)^n&=7k+1\\ \end{align}$$

So We have $$\begin{align}2^{3n}&=(7+1)^n\\ &=7k+1\\2^{3n}-1&=7k\end{align}$$

Hence

$$7\vert2^{3n}-1$$

Answer 2

Hint: $2^{3n} - 1 = (2^3)^n - 1^n = (2^3-1)((2^3)^{n-1} + ... + 1)$

Answer 3

Alternatively with some modulo arithmetic,

$$\begin{align}2^{3n} - 1 &= 8^n - 1\\ &\equiv 1^n - 1 \pmod 7\\ &= 0 \pmod 7 \end{align}$$