Summing Up Certain Roots of Unity

We can have $A = 0, 2, 4, 5, 6, 8, 10$ , for a total of 7 values.

The trickier part is explaining why these are the only solutions, and how we can generalize the approach. Read on for how to do this.

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We will answer the general question for $n = 2p$ , where $p$ is an odd prime. In this case, we have $p = 5$ .

Claim: The number of values of $A$ is $p + 2$ . These correspond to the $p+1$ even numbers $0, 2, \ldots, 2p$ and the odd number $p$ .

Proof: First, we show why those numbers work.
In the even case, for $A = 2k$ , observe that since $\omega^p = - 1$ , thus $\omega^i = - \omega^{p+i}$ . We can set $a_1 = a_2 = \ldots = a_k = 1, a_{p+1} = a_{p+2} = \ldots = a_{p+k} = 1$ , and $a_i = 0$ otherwise. Then, since $\omega^i + \omega^{p+i} = 0$ , thus the terms can be paired up and sum to zero.
In the odd case, for $A = p$ , we set $a_2 = a_4 = a_6 = \ldots = a_{2p} = 1$ , and $a_i = 0$ otherwise. Notice that $\omega^{2k}$ correspond to the $k$ roots of unity.

Now, we will show why these are the only numbers which work.
Let $x = \omega ^2$ . Then, $x$ is a primitive $p$ root of unity.

Claim: The only integer solutions to $\sum_{i=0}^{p-1} b_i x^i = 0$ is when $b_i$ are all equal. We will not prove this claim here, but it arises because the minimal polynomial of $x$ is $1 + x + x^2 + \ldots + x^{p-1} = 0$ , hence all of $b_i$ must be equal.

Now, rewrite the original equation as:

$(a_0 -a_p) 1 + (a_2 - a_{p+2}) x + (a_4 - a_{p+4}) x^2 + \ldots + (a_{2p-2} - a_{p + (2p-2) } ) x^{p-1} = 0,$

where the indices are evaluated modulo $2p$ , meaning that $a_{3p-2} = a_{p-2}$ .

Then, by the previous claim, the only solutions occur when these values of $a_{2k} - a_{2k+p}$ are all equal.
Since $a_i \in \{0, 1 \}$ , hence $a_{2k} - a_{2k+p} \in \{ 0, 1, -1 \}$ .
In the event that $a_{2k} - a_{2k+p} = 1$ , we must have $a_{2k} = 1, a_{2k+p} = 0$ , which gives $A = p$ .
In the event that $a_{2k} - a_{2k+p} = -1$ , we must have $a_{2k} = 0, a_{2k+p} = 1$ , which gives $A = p$ .
In the event that $a_{2k} - a_{2k+p} = 0$ , we must have $a_{2k} = a_{2k+p}$ , which gives that $A$ is even.

Hence, these are the only numbers which would work.

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This solution can be generalized using group theory , specifically Galois theory. It does exactly what we need with the coefficients, and yields that the only values of $A$ which work are when $\gcd(n, A ) \neq 1$ . In particular, for $n = 15$ , the only possible values of $A$ are indeed $0, 3, 6, 9, 12, 15, 5, 10$ .

I do not yet know how of an elementary solution to deal with $n = pq$ . The minimal polynomial trick no longer works, because with $x = \omega ^p$ , we have coefficients of the form $a_i + a_{i+q} \omega^q + a_{i+2q} \omega^{2q} + \ldots$ , which need not be an integer (and in fact are often complex valued).

This is a simplified solution based on Calvin's fuller explanation, to help better understand it

Since ${a}_{i}$ is either $0$ or $1$ , then we can imagine adding unit vectors from the center to the vertices of a decahedron. We want to know which combinations of vectors add up to $0$ . It immediately becomes obvious that $0$ and $10$ vectors will add up to nothing, and that any opposite pairs of vectors will also add up to nothing, and that $5$ vectors equally spaced, i.e., a pentagon, will add up to nothing. Meanwhile, obviously a single vector remains a single vector, which means its complement of $9$ vectors will also add up to a single vector. We're then left with $3$ vectors to look at (and its complement, $7$ vectors)---and it's easy to see that for $3$ vectors to add up to nothing, they have to be at angles of $120$ degrees apart, which is impossible here (which means that its complement of $7$ vectors is also impossible).

Hence we've got $0, 2, 4, 5, 6, 8, 10$ vectors that can add to nothing, while $1, 3, 7, 9$ vectors cannot add to nothing.

Here's a graphic showing $3$ red (or $7$ blue) vectors, neither combination adding up to a null vector

1) We can easily see that the equation is valid for, $a_{i}=0$ $\Rightarrow (A=0)$ is valid

2) We know that,

$\large\omega^{10}=1$ $\Rightarrow \large\omega^{10}-1=0$

thus $\dfrac{\large\omega^{10}-1}{\omega-1}=\sum_{i=0}^{9} \large\omega^{i}=0$ $\Rightarrow (A=10)$ is valid

3) $\large\omega^{5}=-1$ $\Rightarrow \large \omega^{5}+1=0$ $\Rightarrow (A=2)$ is valid

multiplying throughout by $\omega$ we get $4$ more pairs

$\large\omega^{6}+\large\omega=0$

$\large\omega^{7}+\large\omega^{2}=0$

$\large\omega^{6}+\large\omega^{3}=0$

$\large\omega^{6}+\large\omega^{4}=0$

4) Adding distinct pairs from the set of 5 pairs above would give another valid case, thus we can see that it is valid for $(A=4,6,8)$

5) $(\large\omega^{5})^2=1$ $\Rightarrow \dfrac{(\large\omega^{2})^5-1}{\large\omega^{2}-1} =1+\large\omega^{2}+\large\omega^{4}+\large\omega^{6}+\large\omega^{8}=0$ $\Rightarrow (A=5)$ is valid

6) now we need to show it is impossible for $A=1,3,7,9$

i) Case $1:(A=1)$

we know that $\large\omega^{10}=1$ so, $\large\omega^{n}\neq 0$

thus we have no solution for $A=1$

ii) Case $2:(A=3)$

assume it is valid for $A=3$

then $\large\omega^{a}+\large\omega^{b}+\large\omega^{c}=0$ where a,b,c are distinct and $0\leq a,b,c \leq 9$ assume a to be the smallest then we have

$\large\omega^{a}(1+\large\omega^{b-a}+\large\omega^{c-a})=0 \Rightarrow \large\omega^{b-a}+\large\omega^{c-a}=-1$ , $(b-a)$ and $(c-a)$ are distinnct.

since the RHS is purely real and negative we need to check only 2 cases,namely,

$\large\omega^6+\large\omega^4$ and $\large\omega^7+\large\omega^3$

since both of these doesn't satisfy the equation there is no solution for $A=3$

iii) Case $3:(A=7,9)$

we can see that

$\large\omega^{5+x}+\large\omega^{x}=0$

when we select 7 distinct powers of $\large\omega$ we will always have 2 pairs that add upto 0 ,

thus it reuces to the case where $A=3$

similary when we select 9 distinct powers of $\large\omega$ we will always have 4 pairs that add upto 0 ,

thus it reuces to the case where $A=1$

since we already proved that $A\neq1,3$ ,we can see that $A\neq7,9$

thus the possible distinct values of $A$ are $A=0,2,4,5,6,8,10$

the number of distinct values of A are $\boxed{7}$