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/3ⁿ⁺¹+ i2^1/2b_n/3ⁿ⁺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-1/3ⁿ+ i2^1/2b_n-1/3ⁿ)(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, 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, b_n is never congruent to zero modulo three.