Prove: $9\mid a^6+a^{6^2}+\ldots+ a^{6^{27}},\;a\in\mathbb N$

Question

Prove $$9\mid a^6+a^{6^2}+\ldots +a^{6^{27}},a\in\mathbb N$$

My attempt: \begin{aligned}&a \equiv 2\pmod{9}\implies a^{6} \equiv 2^{6} \equiv 1\pmod{9}\\&\begin{array}{l}{a \equiv 3\pmod{9}\implies a^{6} \equiv 3^{6} \equiv 0\pmod{9}} \\{a \equiv 4\pmod{9}\implies a^{6} \equiv\left(2^{6}\right)^{2} \equiv 1\pmod{9}}\\a\equiv 5\equiv -4\pmod{9}\implies a^6\equiv(-4)^6=4^6\equiv 1\pmod{9}\end{array}\end{aligned}

$$a^{6^i}=\left(a^6\right)^{6^{i-1}}\equiv [0,1]\pmod{9}\implies a^{6^i}=9k\;\lor\;a^{6^i}=9k+1,\;k\in\mathbb N_0$$ $$a^6+a^{6^2}+\ldots +a^{6^{27}}=\sum_{i=1}^{27}\left(a^6\right)^{6^{i-1}}=\sum_{i=1}^{27}(9k_i+[0,1])=\underbrace{\sum_{i=1}^{27}9k_i+[0,27]}_{\text{multiple of}\; 9}$$ //edited mistake [27,54]//

Is this correct and is there a more efficient method?

Answer 1

It is correct, but a bit long.

Another method will use Euler's theorem: $\;\varphi(9)=6$ so, if $a$ is not divisible by $3$, we have $a^6\equiv 1\mod 9$, and by an easy induction, $\;a^{6^k}=\bigl(a^6 \bigr)^{6^{k-1}}\equiv 1$. Therefore $$a^6+a^{6^2}+\ldots+ a^{6^{27}}\equiv\underbrace{1+1+\ldots+1}_{27 \:1\text{s}}\equiv 0\mod 9 $$ for this case.

On the other hand, if $a$ is divisible by $3$, $a^2\equiv 0\bmod 9$, hence $a^6\equiv 0$ too.

Answer 2

Observe that $$a^m(a^n-1)\equiv0\pmod9$$ for any integer $a$ if $m\ge2$ and $6\mid n$

So, for integer $k\ge1,$ $$a^{6^{k+1}}-a^6=a^6(a^{6^k}-1)\equiv0\pmod9$$

$$\implies\sum_{k=1}^na^{6^k}\equiv na^6\pmod9$$

Here $n=?$