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Here is an extended hint for proving 2 (an almost complete proof is in the Update below). If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be a power of 3 (as a divisor of $3^s$). But it looks like $q_n(2)$ is never 0 modulo 9 (this should be possible to prove rigorously but I do not have time). Hence $q_{tm}(2)$ must be equal to 3 or 1 which gives a bound on $s$.

Update 1 Since $2^3=-1 \mod 9$, we only need remainders of cyclotomic polynomials modulo $x^3+1$ with coefficients modulo 9. Here are all 50 of them $$ \begin{array}{l} 1,3,{x}^{2},8x,8{x}^{2},1+8x,5+5{x}^{2},x+1,x+8,x+8 {x}^{2},\\\ 4x+5{x}^{2},{x}^{2}+1,1+8{x}^{2}+8x,2+3x+2{x}^{2}, 2+5x+4{x}^{2},2+8x+{x}^{2},2+{x}^{2}+7x,\\\ 2+2{x}^{2}+8x,2+6 {x}^{2}+7x,3+x+8{x}^{2},3+5x+3{x}^{2},\\\ 3+6x+2{x}^{2},3+2 {x}^{2}+7x,4+6x+3{x}^{2},4+7x+4{x}^{2},4+2{x}^{2}+5x,4 +3{x}^{2}+5x,\\\ 4+4{x}^{2}+6x,5+2x+7{x}^{2},5+4x+4{x}^{2} ,5+4{x}^{2}+5x, \\\ 6+2x+6{x}^{2}, \\\ 6+4x+5{x}^{2},6+7x+2{x}^{ 2},6+5{x}^{2}+3x,6+6{x}^{2}+4x,\\\ 6+7{x}^{2}+3x,7+2x+6{x} ^{2},7+4x+7{x}^{2},\\\ 7+6{x}^{2}+3x,8+2x+7{x}^{2},8+6x+8{ x}^{2},8+8x+{x}^{2}, \\\ 8+7{x}^{2}+x,8+8{x}^{2}+2x,{x}^{2}+x+1,{x} ^{2}+8x+1 \end{array}$$ None of these polynomials become 0 when evaluated at $2 \mod 9$.

As for the last claim about the size of $q_{mt}(2)$: every root of the cyclotomic polynomial is on the unit circle, so the values at 2 grow with the degree.

This almost completes the proof of 2 (one needs to show that the set of 50 polynomials is complete, but that can be done by induction).

Update 2. A simplification. If $q_m(2)$ is $0 \mod 9$, then $2^m-1=0 \mod 9$. Hence one only need to consider $m=6k$. It looks like $q_{6k}(2)=1 \mod 3$ as long as $k>1$. Thus we can consider $q_{6k}(2) \mod 3$. Hence it is enough to look at $q_{6k}(x)$ modulo $x^2-1$ since $2^2=1 \mod 3$. In that case for $k> 1$, the only remainder modulo $x^2-1$ ($\mod 3$) is 1. So $q_{6k}(2)=1 \mod 3$ if $k> 1$.

Here is an extended hint for proving 2 (an almost complete proof is in the Update below). If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be a power of 3 (as a divisor of $3^s$). But it looks like $q_n(2)$ is never 0 modulo 9 (this should be possible to prove rigorously but I do not have time). Hence $q_{tm}(2)$ must be equal to 3 or 1 which gives a bound on $s$.

Update Since $2^m-1=0 \mod 9$ only when $m=0 \mod 6$. it is enough to consider $q_{6k}(2)$. Since $2^3=-1 \mod 9$, we only need remainders of cyclotomic polynomials modulo $x^3+1$ with coefficients modulo 9. Here are all 50 of them $$ \begin{array}{l} 1,3,{x}^{2},8x,8{x}^{2},1+8x,5+5{x}^{2},x+1,x+8,x+8 {x}^{2},\\\ 4x+5{x}^{2},{x}^{2}+1,1+8{x}^{2}+8x,2+3x+2{x}^{2}, 2+5x+4{x}^{2},2+8x+{x}^{2},2+{x}^{2}+7x,\\\ 2+2{x}^{2}+8x,2+6 {x}^{2}+7x,3+x+8{x}^{2},3+5x+3{x}^{2},\\\ 3+6x+2{x}^{2},3+2 {x}^{2}+7x,4+6x+3{x}^{2},4+7x+4{x}^{2},4+2{x}^{2}+5x,4 +3{x}^{2}+5x,\\\ 4+4{x}^{2}+6x,5+2x+7{x}^{2},5+4x+4{x}^{2} ,5+4{x}^{2}+5x, \\\ 6+2x+6{x}^{2}, \\\ 6+4x+5{x}^{2},6+7x+2{x}^{ 2},6+5{x}^{2}+3x,6+6{x}^{2}+4x,\\\ 6+7{x}^{2}+3x,7+2x+6{x} ^{2},7+4x+7{x}^{2},\\\ 7+6{x}^{2}+3x,8+2x+7{x}^{2},8+6x+8{ x}^{2},8+8x+{x}^{2}, \\\ 8+7{x}^{2}+x,8+8{x}^{2}+2x,{x}^{2}+x+1,{x} ^{2}+8x+1 \end{array}$$ None of these polynomials become 0 when evaluated at $2 \mod 9$.

As for the last claim about the size of $q_{mt}(2)$: every root of the cyclotomic polynomial is on the unit circle, so the values at 2 grow with the degree.

This almost completes the proof of 2 (one needs to show that the set of 50 polynomials is complete, but that can be done by induction).

Here is an extended hint for proving 2 (an almost complete proof is in the Update below). If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be a power of 3 (as a divisor of $3^s$). But it looks like $q_n(2)$ is never 0 modulo 9 (this should be possible to prove rigorously but I do not have time). Hence $q_{tm}(2)$ must be equal to 3 or 1 which gives a bound on $s$.

Update Since $2^3=-1 \mod 9$, we only need remainders of cyclotomic polynomials modulo $x^3+1$ with coefficients modulo 9. Here are all 47 of them $$ \begin{array}{l} 1,3,{x}^{2},8x,8{x}^{2},1+8x,5+5{x}^{2},x+1,x+8,x+8 {x}^{2},\\\ 4x+5{x}^{2},{x}^{2}+1,1+8{x}^{2}+8x,2+3x+2{x}^{2}, 2+5x+4{x}^{2},2+8x+{x}^{2},2+{x}^{2}+7x,\\\ 2+2{x}^{2}+8x,2+6 {x}^{2}+7x,3+x+8{x}^{2},3+5x+3{x}^{2},\\\ 3+6x+2{x}^{2},3+2 {x}^{2}+7x,4+6x+3{x}^{2},4+7x+4{x}^{2},4+2{x}^{2}+5x,4 +3{x}^{2}+5x,\\\ 4+4{x}^{2}+6x,5+2x+7{x}^{2},5+4x+4{x}^{2} ,5+4{x}^{2}+5x, \\\ 6+2x+6{x}^{2}, \\\ 6+4x+5{x}^{2},6+7x+2{x}^{ 2},6+5{x}^{2}+3x,6+6{x}^{2}+4x,\\\ 6+7{x}^{2}+3x,7+2x+6{x} ^{2},7+4x+7{x}^{2},\\\ 7+6{x}^{2}+3x,8+2x+7{x}^{2},8+6x+8{ x}^{2},8+8x+{x}^{2}, \\\ 8+7{x}^{2}+x,8+8{x}^{2}+2x,{x}^{2}+x+1,{x} ^{2}+8x+1 \end{array}$$ None of these polynomials become 0 when evaluated at $2 \mod 9$.

As for the last claim about the size of $q_{mt}(2)$: every root of the cyclotomic polynomial is on the unit circle, so the values at 2 grow with the degree.

This almost completes the proof of 2 (one needs to show that the set of 47 polynomials is complete, but that can be done by induction).

Here is an extended hint for proving 2 (an almost complete proof is in the Update below). If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be a power of 3 (as a divisor of $3^s$). But it looks like $q_n(2)$ is never 0 modulo 9 (this should be possible to prove rigorously but I do not have time). Hence $q_{tm}(2)$ must be equal to 3 or 1 which gives a bound on $s$.

Update 1 Since $2^3=-1 \mod 9$, we only need remainders of cyclotomic polynomials modulo $x^3+1$ with coefficients modulo 9. Here are all 50 of them $$ \begin{array}{l} 1,3,{x}^{2},8x,8{x}^{2},1+8x,5+5{x}^{2},x+1,x+8,x+8 {x}^{2},\\\ 4x+5{x}^{2},{x}^{2}+1,1+8{x}^{2}+8x,2+3x+2{x}^{2}, 2+5x+4{x}^{2},2+8x+{x}^{2},2+{x}^{2}+7x,\\\ 2+2{x}^{2}+8x,2+6 {x}^{2}+7x,3+x+8{x}^{2},3+5x+3{x}^{2},\\\ 3+6x+2{x}^{2},3+2 {x}^{2}+7x,4+6x+3{x}^{2},4+7x+4{x}^{2},4+2{x}^{2}+5x,4 +3{x}^{2}+5x,\\\ 4+4{x}^{2}+6x,5+2x+7{x}^{2},5+4x+4{x}^{2} ,5+4{x}^{2}+5x, \\\ 6+2x+6{x}^{2}, \\\ 6+4x+5{x}^{2},6+7x+2{x}^{ 2},6+5{x}^{2}+3x,6+6{x}^{2}+4x,\\\ 6+7{x}^{2}+3x,7+2x+6{x} ^{2},7+4x+7{x}^{2},\\\ 7+6{x}^{2}+3x,8+2x+7{x}^{2},8+6x+8{ x}^{2},8+8x+{x}^{2}, \\\ 8+7{x}^{2}+x,8+8{x}^{2}+2x,{x}^{2}+x+1,{x} ^{2}+8x+1 \end{array}$$ None of these polynomials become 0 when evaluated at $2 \mod 9$.

As for the last claim about the size of $q_{mt}(2)$: every root of the cyclotomic polynomial is on the unit circle, so the values at 2 grow with the degree.

This almost completes the proof of 2 (one needs to show that the set of 50 polynomials is complete, but that can be done by induction).

Update 2. A simplification. If $q_m(2)$ is $0 \mod 9$, then $2^m-1=0 \mod 9$. Hence one only need to consider $m=6k$. It looks like $q_{6k}(2)=1 \mod 3$ as long as $k>1$. Thus we can consider $q_{6k}(2) \mod 3$. Hence it is enough to look at $q_{6k}(x)$ modulo $x^2-1$ since $2^2=1 \mod 3$. In that case for $k> 1$, the only remainder modulo $x^2-1$ ($\mod 3$) is 1. So $q_{6k}(2)=1 \mod 3$ if $k> 1$.

Here is an extended hint for proving 2 (an almost complete proof is in the Update below). If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be a power of 3 (as a divisor of $3^s$). But it looks like $q_n(2)$ is never 0 modulo 9 (this should be possible to prove rigorously but I do not have time). Hence $q_{tm}(2)$ must be equal to 3 or 1 which gives a bound on $s$.

Update Since $2^6=1 \mod 9$, we only need remainders of cyclotomic polynomials modulo $x^3+1$ with coefficients modulo 9. Here are all 28 of them $$ \begin{array}{l}1,3,{x}^{2},8\,x,1+8\,x,x+1,x+8,4x+5{x}^{2},{x}^{2}+1,\\\1+8{x}^{2}+8x,2+3x+2{x}^{2},2+8x+{x}^{2},2+{x}^{2}+7x,\\\2+2{x }^{2}+8x,3+2{x}^{2}+7x,\\\4+6x+3{x}^{2},4+7x+4{x}^{2},4+3{x}^{2}+5x,\\\5+2x+7{x}^{2},5+4x+4{x}^{2},5+4{x}^{2}+5x,\\\6+4x+5{x}^{2},6+5{x}^{2}+3x,7+2x+6{x}^{2},\\\ 7+6{x}^{2}+3x,8 +7{x}^{2}+x,{x}^{2}+x+1,{x}^{2}+8x+1\end{array}$$ None of these polynomials become 0 when evaluated at $2 \mod 9$.

As for the last claim about the size of $q_{mt}(2)$: every root of the cyclotomic polynomial is on the unit circle, so the values at 2 grow with the degree.

This almost completes the proof of 2 (one needs to show that the set of 28 polynomials is complete, but that can be done by an easy induction).

Here is an extended hint for proving 2 (an almost complete proof is in the Update below). If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be a power of 3 (as a divisor of $3^s$). But it looks like $q_n(2)$ is never 0 modulo 9 (this should be possible to prove rigorously but I do not have time). Hence $q_{tm}(2)$ must be equal to 3 or 1 which gives a bound on $s$.

Update Since $2^3=-1 \mod 9$, we only need remainders of cyclotomic polynomials modulo $x^3+1$ with coefficients modulo 9. Here are all 47 of them $$ \begin{array}{l} 1,3,{x}^{2},8x,8{x}^{2},1+8x,5+5{x}^{2},x+1,x+8,x+8 {x}^{2},\\\ 4x+5{x}^{2},{x}^{2}+1,1+8{x}^{2}+8x,2+3x+2{x}^{2}, 2+5x+4{x}^{2},2+8x+{x}^{2},2+{x}^{2}+7x,\\\ 2+2{x}^{2}+8x,2+6 {x}^{2}+7x,3+x+8{x}^{2},3+5x+3{x}^{2},\\\ 3+6x+2{x}^{2},3+2 {x}^{2}+7x,4+6x+3{x}^{2},4+7x+4{x}^{2},4+2{x}^{2}+5x,4 +3{x}^{2}+5x,\\\ 4+4{x}^{2}+6x,5+2x+7{x}^{2},5+4x+4{x}^{2} ,5+4{x}^{2}+5x, \\\ 6+2x+6{x}^{2}, \\\ 6+4x+5{x}^{2},6+7x+2{x}^{ 2},6+5{x}^{2}+3x,6+6{x}^{2}+4x,\\\ 6+7{x}^{2}+3x,7+2x+6{x} ^{2},7+4x+7{x}^{2},\\\ 7+6{x}^{2}+3x,8+2x+7{x}^{2},8+6x+8{ x}^{2},8+8x+{x}^{2}, \\\ 8+7{x}^{2}+x,8+8{x}^{2}+2x,{x}^{2}+x+1,{x} ^{2}+8x+1 \end{array}$$ None of these polynomials become 0 when evaluated at $2 \mod 9$.

As for the last claim about the size of $q_{mt}(2)$: every root of the cyclotomic polynomial is on the unit circle, so the values at 2 grow with the degree.

This almost completes the proof of 2 (one needs to show that the set of 47 polynomials is complete, but that can be done by induction).