TITLE: Binary representation of powers of 3
QUESTION [10 upvotes]: I asked this question at Mathematics Stack Exchange but since I didn't got a satisfactory answer I decided to ask it here as well.
We write a power of 3 in bits in binary representation as follows.
For example $3=(11)$, $3^2=(1001)$ which means that we let the $k$-th bit from the right be $1$ if the binary representation of this power of 3 contains $2^{k-1}$, and $0$ otherwise.

Prove that the highest power of 3 that has a palindromic binary representation is $3^3 = (11011)$.
Prove that $3 = (11)$ is the only power of 3 with a periodic binary representation (in the sense that it consists of a finite sequence of $1$s and $0$s repeated two or more times, like "$11$" consists of two repetitions of the bitstring "$1$").

REPLY [5 votes]: Here is an alternative (and complete) proof of 2. Following Mark Sapir, we show that the only solution of
$$ \frac{2^{tm}-1}{2^m-1} = 3^r $$
in positive integer triplets $(t,m,r)$ is $(2,1,1)$ and $(2,3,2)$. I am sure this was written down before, by the way.
Our main tool is the observation that $\mathrm{ord}_{3^n}(4)=3^{n-1}$ when $n\geq 1$ (in other words $2$ is a primitive root modulo $3^n$), which implies that $v_3(4^k-1)=1+v_3(k)$. The observation follows from $4^{3^{n-2}}\equiv 1+3^{n-1}\pmod{3^n}$ when $n\geq 2$, which in turn can be proved by induction or by the binomial theorem. 
In the diophantine equation $r\geq 1$, hence $tm$ is even. If $m$ is even, then
$$ r=v_3\left (\frac{2^{tm}-1}{2^m-1}\right)=(1+v_3(tm))-(1+v_3(m))=v_3(t), $$
so that $t$ is divisible by $3^r$. This is a contradiction:
$$  3^r=\frac{2^{tm}-1}{2^m-1} > t \geq 3^r, $$
which proves that $m$ is odd and $t$ is even. In that case $4^m-1\mid 2^{tm}-1$, and
$$ \frac{2^{tm}-1}{4^m-1} = 3^s $$
for some $0\leq s\leq r$. As before
$$ s=v_3\left (\frac{2^{tm}-1}{4^m-1}\right)=(1+v_3(tm))-(1+v_3(m))=v_3(t), $$
so that $t$ is divisible by $3^s$. This is a contradiction when $t\geq 4$:
$$  3^s=\frac{2^{tm}-1}{4^m-1} > 4(t/2-1)\geq t \geq 3^s, $$
which proves that $t=2$. Now the original diophantine equation becomes
$$ 2^m+1=3^r,$$
where $m$ is odd. Then $1+v_3(m)=v_3(4^m-1)=r$, so that $m$ is divisible by $3^{r-1}$. It follows that $1+2^{3^{r-1}}\leq 3^r$. It is easy to see that this holds only for $r=1$ and $r=2$, which yields $m=1$ and $m=3$, respectively.