Prove That $6^n -1$ is composite

Question

Prove That $6^n -1$ is composite $\forall n>1 \in \mathbb{N}$

http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b — lab bhattacharjee, Nov 24 '16 at 11:17
What is the "unit digit" of any such $6^{n}$? What is it once you subtract the 1? What can you say about all numbers with such last digit? — AlphaNumeric, Nov 24 '16 at 11:19

score 0 · Accepted Answer · answered Nov 24 '16 at 11:22

0

Hint: Use $$(a^n-b^n)=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1}).$$

Another method would be to see that $6 \equiv 1 (\mod 5)$ and $6^n \equiv1^n(\mod 5) \equiv 1(\mod 5)$. Now subtract 1 and see what happens.

answered Nov 24 '16 at 11:22

MrYouMath

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