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## nt.quantity idea – linear independence of $error(ok pi / m)$

Note: Fedor and Vladimir have already answered the query, however this can be a partial respond within the different course, underneath a stronger speculation. (This respond, which I had earlier deleted, has been edited in response to some valid feedback.)

If $m$ is queer and square-free, then the pretense of the OP holds.

Let $S$ breathe the clique ${ok: 1 leq ok leq m/2, (ok, m) = 1}$. If $sum_{ok in S} a_k error(kpi/m) = 0$ for some rationals $a_k$, then

$$sum_{ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) = 0.$$

Here $N = 2m$ is square-free, and in that illustration the raw $N$-th roots of union are linearly unbiased over $mathbb{Q}$ (behold this mathstackexchange dialogue: https://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension).

For $N = 2m$, any queer $ok in S$ is prime to $N$ and therefore $e^{ok pi i/m}, e^{-kpi i/m}$ are raw $N$-th roots of union. If $ok in S$ is plane, then $m + ok$ is queer and prime to $m$ and thus to $N$, so $e^{(m + ok)pi i/m} = -e^{kpi i/m}$ is too raw $N$-th root of union, as is its conjugate $e^{(m – ok)pi i/m}$; recognize the $m-k$ equivocate in ${j: m/2 leq j leq m, gcd(j,m) = 1}$ which is disjoint from $S$. Then

$$commence{array}{lll}

sum_{ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) & = & sum_{textual content{queer}; ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) + sum_{textual content{plane}; ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m})

& = & sum_{textual content{queer}; ok in S} a_k(e^{kpi i/m} – e^{-kpi i/m}) + sum_{textual content{plane}; ok in S} a_k(e^{(m-k)pi i/m} – e^{-(m-k)pi i/m})

aim{array}$$

the place all of the raw roots of union showing within the final expression are manifestly discrete. By linear independence of the raw roots, if that linear mixture is zero, then $a_k = 0$ for all $ok$, as required.

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