Suppose 1/Pi * Arccos(1/3)=p/q where p and q are integers. Then
where the coefficient of i is determined by the Pythagorean theorem. Raising both sides to the power of 2q, we see
This establishes the hint. I now claim that
(1/3+(81/2/3)i)n=an/3n+1+ i21/2bn/3n+1
where an and bn are integers. Clearly this holds for n=1. Suppose it holds for n-1. Then
(1/3+(81/2/3)i)n=(an-1/3n+ i21/2bn-1/3n)(1/3+(81/2/3)i)
Multiplying out the right hand side, one arrives at the formulas
Hence, an and bn are both integers. Finally, I would like to show that bn can never be zero, which will be sufficient. To see this, we will show that it is not zero modulo 3. Note that, a1=1 and b1=2=-1 modulo 3. Plugging into the above recursions, we see that a2=-1 mod 3 and b2=1 mod 3. Plugging in once more we get a3=1 mod 3 and b3= -1 mod 3, whereupon the cycle repeats. In particular, bn is never congruent to zero modulo three.