Suppose 1/Pi * Arccos(1/3)=p/q where p and q are integers. Then
e^{(p/q)Pi*i}=1/3+(8^{1/2}/3)**i**
where the coefficient of **i** is determined by the Pythagorean theorem.
Raising both sides to the power of 2q, we see
1=e^{2Pi*p*i}=(1/3+(8^{1/2}/3)**i**)^{2q}
This establishes the hint. I now claim that
(1/3+(8^{1/2}/3)**i**)^{n}=a_{n}/3^{n+1}+
**i**2^{1/2}b_{n}/3^{n+1}
where a_{n} and b_{n} are integers.
Clearly this holds for n=1. Suppose it holds for n-1. Then
(1/3+(8^{1/2}/3)**i**)^{n}=(a_{n-1}/3^{n}+
**i**2^{1/2}b_{n-1}/3^{n})(1/3+(8^{1/2}/3)**i**)
Multiplying out the right hand side, one arrives at the formulas
a_{n}=a_{n-1}-4b_{n-1}

b_{n}=b_{n-1}+2a_{n-1}
Hence, a_{n} and b_{n} are both integers.
Finally, I would like to show that b_{n} can never be zero, which
will be sufficient. To see this, we will show that it is not zero modulo 3.
Note that, a_{1}=1 and b_{1}=2=-1 modulo 3. Plugging into
the above recursions, we see that a_{2}=-1 mod 3 and b_{2}=1
mod 3. Plugging in once more we get
a_{3}=1 mod 3 and b_{3}= -1 mod 3, whereupon the cycle
repeats. In particular, b_{n} is never congruent to zero modulo three.