$a_k = 8^{3k} + 8^{3k-1}+8^{3k-2} $

Question

Let $a_k = 8^{3k} + 8^{3k-1}+8^{3k-2} $, with $k$ being a natural number greater than $0$.

Determine all $k$ such that $6 \ | \ a_k-2$.

My attempt is to reduce $8$ as $8\equiv 2 \bmod 6$.

Then we can say that $2\equiv 2^{3k} + 2^{3k-1}+2^{3k-2} \bmod 6$.

Thus $2 \ | \ 2^{3k} + 2^{3k-1}+2^{3k-2}-2$.

Then we have to find $k$ such that $1\equiv 2^{3k} + 2^{3k-1}+2^{3k-2} \bmod 3$.

And we also have $3\ |\ 2^{3k}-1$ and $3 \ | \ 2^{3k-1}-1$ and $3 \ | \ 2^{3k-2}-2$.

Then $1\equiv 2^{3k} + 2^{3k-1}+2^{3k-2} \bmod 3$ for all $k$.

Thus the solution of the problems are all of the natural numbers greater than $0$.

Please help me with this question.

Note that $a_k = 73 \cdot 8^{3 k - 2} \equiv 2^{3 k - 2} \bmod 6$. — lhf, May 18 '23 at 12:22
You said we have $3\mid2^{3k}-1$, but that's only when $k$ is even, not when $k$ is odd — J. W. Tanner, May 18 '23 at 12:27
Just note that the powers of $2$ mod $6$ follow this pattern: $1,2,4,2,4,\dots$ — lhf, May 18 '23 at 12:31
By congruence arithmetic $\bmod 3!:\ a_k \equiv (-1)^{3k},$ so $,a_k \equiv 2\equiv -1 \iff k,$ is odd. $\ \ $ — Bill Dubuque, May 18 '23 at 21:24
OP wants to find $k$ such that $a_k\equiv\color{blue}1\pmod3$ — J. W. Tanner, May 18 '23 at 21:33

Itoz Darbien · Answer 1 · 2023-05-18T22:58:08.610

-1

It is known that $2 | a_{k}−2$.

So we only need to find k, such that $8^{3k}+8^{3k-1}+8^{3k-2}-2\equiv 0 \pmod 3$.

We have $$8^{3k}+8^{3k-1}+8^{3k-2}-2\equiv (-1)^{3k}+(-1)^{3k-1}+(-1)^{3k-2}+1=2(-1)^{3k}+(-1)^{3k-1}+1=(-1)^{3k}+1 \pmod 3$$

If $(-1)^{3k}+1 \equiv 0 \pmod 3$, we have $2 \nmid k$.

edited May 18 '23 at 22:58

answered May 18 '23 at 12:55

Itoz Darbien

1

Welcome to Mathematics Stack Exchange. Formatting note: use \pmod or \bmod – J. W. Tanner May 18 '23 at 14:14

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