Is the absolute value of the sum of six of the 16th roots of unity ever a nonzero integer? Let $\zeta_1, \ldots \zeta_{16}$ be the $16$th roots of unity. For the proper subset $J \subset \{1,2,\dots,16\}$ and $|J|=6$, can the following sum ever be satisfied for an integer not equal to zero?
$$\left\lvert\sum_{j\in J} \zeta_j\right\rvert=C \in \mathbb{N}^+.$$
I think the answer is negative but have been unable to finish the following argument. Squaring both sides yields the real sum
$$\sum_{j,k\in J} \zeta_j \zeta_k^* = \sum_{j,k} \cos\left(\frac{2\pi(j-k)}{16}\right) = C^2  $$
which after removing the conjugate product terms yields
$$ \sum_{j\neq k} \zeta_j \zeta_k^* = C^2 - |J| . $$
I was thinking of then proceeding using the cardinality of $J$ and that $ \cos\left(\frac{2\pi}{16}\right)$ is irrational to show that no such sum is possible but keep going in circles. I suspect this question is very simple using algebraic number theory but I am still a novice there. Any suggestions on how to move forward would be appreciated, particularly using number theory.
 A: Not a solution yet. For now I prove that $C=2$ is the only possibility, and rewrite this as a "combinatorial" problem. I hope to come back to it sooner rather than later.

Everything takes place inside the number field $K=\Bbb{Q}(2\cos(\pi/8))$, the real subfield of the sixteenth cyclotomic field. Let us denote $\zeta=e^{\pi i/8}$ and
$s_k=\zeta^k+\zeta^{-k}$, $k=0,1,\ldots,15$. It is straightforward to show that
$\mathcal{B}=\{1,s_1,s_2,s_3\}$ is a $\Bbb{Q}$-basis of $K$. We have the obvious relations
$$
\begin{aligned}
s_4&=0,\\
s_j&=-s_{8-j}, j=1,2,3,\\
s_j&=-s_{j+8}, j=0,1,\ldots,7
\end{aligned}
$$
that allow us to write all the numbers $s_j$, $0\le j<16$, in terms of the basis elements.
Let $J=\{i_1,i_2,\ldots,i_6\}$ be the chosen six exponents of $\zeta$. Without loss of generality we can assume that they are sorted, so $0\le i_1<i_2<i_3<\cdots<i_6\le15$. The first observation we can make is that if
$J$ is a solution, then so is $J-i_1=\{i_k-i_1\mid k=1,\ldots,6\}$ because we only rotate the sum in the complex plane by $\pi i_1/8$. So without loss of generality we can assume that $i_1=0$.
As already calculated in the question body, we have
$$
C^2=6-\sum_{j,k\in J, k<j}s_{j-k}.
$$
For all $\ell=1,2,\ldots,15$ let's denote by
$$
N_\ell=\#\{(j,k)\in J\times J, k<j, j-k=\ell\}
$$
the number of pairs $(j,k)\in J\times J$ such that $j-k=\ell$.
In light of all of the above, we arrive at
$$
C^2=(6-2N_8)+(N_1+N_{15}-N_7-N_9)s_1+(N_2+N_{14}-N_6-N_{10})s_2+(N_3+N_{13}-N_5-N_{11}) s_3.
$$
Because $\mathcal{B}$ is a basis it follows that $6-2N_8$ must be a perfect square, and that the coefficients of $s_j, j=1,2,3$, must all vanish. Because $C=0$ is not interesting, and $C^2\le6$, this forces $C=2$ and $N_8=1$.

The problem is then to prove that it is impossible to find a set $J$ such that the constraints $N_8=1$, $N_1+N_{15}=N_7+N_9$, $N_3+N_{13}=N_5+N_{11}$, and
$N_2+N_{14}=N_6+N_{10}$ all hold.


*

*The second observation I make is that by rotational invariance we can now actually assume that both $0$ and $8$ are in $J$.

*The next observation is that it is impossible for the
set $J$ to consist of only even numbers. For otherwise the numbers in
$J$ fall into at most four distinct residue classes modulo eight,
forcing $N_8\ge2$.

*A corollary of the previous bullet is that without loss of generality we can assume that $1\in J$. This is because the condition on the sum is clearly invariant under the Galois group $G=Gal(\Bbb{Q}(\zeta)/\Bbb{Q})$, and $G$ acts transitively in the set $\{\zeta^j\mid \text{$j$ odd}\}$. Furthermore, all the elements of $G$ have $\zeta^0$ and $\zeta^8$  as their fixed points.


If we end up doing a brute force search, knowing that it suffices to check the sets containing all of $0,1,8$ but not containing $9$ will reduce the complexity of the search.

