$9^n \equiv 1 \mod 8$

Question

I would like someone to check this inductive proof (sketch)

The base case is clear. For the inductive step, it follows that $8 \mid 9^{n+1} - 9 = 9(9^n - 1)$ by the indutive hyp. So $9^{n+1} \equiv 9 \equiv 1 \mod 8$.

Feedback would be appreciated.

Answer 1

I'm assuming you mean what you say when you state your work as a proof "sketch".

The base case is clear. For the inductive step, it follows that $8 \mid 9^{n+1} - 9 = 9(9^n - 1)$ by the indutive hyp. So $9^{n+1} \equiv 9 \equiv 1 \mod 8$.

In your final write up, I'd suggest you "fill in" a bit of detail: e.g., to "walk through" the base case, at least stating that the congruence holds for $n=1$, or perhaps

"for $n = 1$, clearly, $9\equiv 1 \pmod 8$".

Then I suggest you make your inductive hypothesis explicit:

"Assume that it is true that $9^n \equiv 1 \pmod 8$,"

and then finish with, "for the inductive step....[what you wrote]"

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If your task was to prove the congruence holds using proof by induction on $n$, then you've done a fine job of sketching such a proof.

If you can use other strategies, then bonsoon's suggestion is worth considering:

"Or note that since $9 \equiv 1 \pmod 8$, we have $9^n\equiv 1^n = 1 \pmod 8.$"

Answer 2

Yes, that is correct. Alternatively, prove by induction the $\,\rm n$-ary congruence product rule

$$\rm\qquad\ \ a_k\equiv b_k\ \Rightarrow\ a_1\cdots\, a_n \equiv b_1\cdots\, b_n$$

by iterating the binary product rule $\rm\ a_k\equiv b_k\ \Rightarrow\ a_1 a_2 \equiv b_1 b_2,\:$ then specialize $\rm\:a_i \equiv 9,\ b_i\equiv 1$

Remark $\ $ Your proof can be viewed as special case of the obvious inductive proof that a sequence $\rm\:f_n\:$ is constant if successive values never change, i.e. if $\rm\:f_{n+1} \equiv f_n.$ Indeed, in your special case we have $\rm\:mod\ 8\!:\ f_{n+1} = 9^{n+1} = 9\, f_n \equiv f_n\:$ so the sequence is constant, hence $\rm\:f_n\equiv f_0\equiv 9^0\equiv 1.\:$