Prove that if n is a positive integer, then $5 | (6^n-1)$

Question

Is this always true? Is it always true for any consecutive integers (change 5 and 6 for any consecutive numbers).

Answer 1

We have the polynomial identity:

$x^n - 1 = (x - 1) \displaystyle \sum_1^n x^{i - 1} = (x - 1)(x^{n - 1} + x^{n - 2} + \ldots + x + 1), \tag 1$

which holds for all $1 \le n \in \Bbb N$; setting

$x = 6 \tag 2$

yields

$6^n - 1 = (6 - 1) \displaystyle \sum_1^n 6^{i - 1} = 5\sum_1^n 6^{i - 1}, \tag 3$

whence

$5 \mid 6^n - 1, \; \forall n \ge 1. \tag 4$

Note Added in Edit, Sunday 7 April 2019 12:03 AM PST: Note that in fact our identity (1) allows us to conclude that for any integer $z$,

$(z - 1) \mid z^n - 1; \tag 5$

just take $x = z$; so, for example

$12 \mid (13)^n - 1, \; \forall 1 \le n \in \Bbb N, \tag 6$

und so weiter . . . End of Note.

Answer 2

$6 \equiv 1\ (\text {mod}\ 5) \implies 6^n \equiv 1\ (\text {mod}\ 5) \implies 6^n-1 \equiv 0\ (\text {mod}\ 5).$

Answer 3

$6^n=(1+5)^n=1+5n+\cdot\cdot\cdot+5^n=1+5k$

Answer 4

For any integer $a$ and its consecutive integer $a+1$, it is the case that $a+1\equiv 1 \mod a$

$(a+1)^n-1 \equiv 1^n-1 \equiv 0 \mod a$. This means the same as $a\mid (a+1)^n-1$

So what is true for $5$ and $6$ is true for all consecutive integers.