1 Answer

Llermaly · Answer 1 · 2023-06-07T05:05:54+0000

I suppose the claim is
$5 \mid 6^n - 1$ for
$n\in\Bbb N$ .

When
n=1 , we have
6^1 - 1 = 6 - 1 = 5 , and of course 5 divides 5.

Assume the claim holds for
n=k , that
$5 \mid 6^k - 1$ . We want to use this to show it holds for
n=k+1 , that
$5 \mid 6^(k+1) - 1$ .

We have

$6^(k+1) - 1 = \left(6^(k+1) - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5$

Since
$5 \mid 6^k - 1$ , we can write
$6^k - 1 = 5\ell$ for some integer
$\ell$ . Then

$6^(k+1) - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)$

which is clearly divisible by 5. QED

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