Question

Prove that 7 is a factor of $2^{3n}$ - 1 for all natural numbers n.

Answer 1

Let P(n): 7 is a factor of $2^{3n}$ - 1 be the given statement
Step 1: When n = 1
$2^{3\left(1\right)}$ - 1 = 7 and 7 is a factor of itself
$\therefore$ P(n) be true for n = k.
Step 2: Let P(n) be true for n = k.
$\Rightarrow$ 7 is a factor of $2^{3k}$ - 1.
$\Rightarrow$ $2^{3k}$ - 1 = 7M, where M $ϵ$ N.
$\Rightarrow$ $2^{3k}$ = 7M + 1 $\to$ (1)
Now consider $2^{3(k+1)}$ - 1 = $2^{3k+3}$ - 1 = $2^{3k}$. $2^3$ - 1
= 8(7M + 1) 1 (using (1)) = 56M + 7 (As $2^{3k}$
= (7M + 1)
$\therefore$ $2^{3(k+1)}$ - 1 = 7 (8m + 1)
$\Rightarrow$ 7 is factor of $2^{3(k+1)}$ - 1
$\Rightarrow$ P(n) istrue for n = k + 1
$\therefore$ By the principle of mathematical induction,
P(n) is true for all natural numbers n.
Hence, 7 is a factor $2^{3n}$ - 1 for all n $ϵ$ N.