Sum of nth roots of unity equal to Sqrt[n]

Asked Nov 13 '18 at 09:59

Active Nov 13 '18 at 09:59

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Let $n >1$ be an odd integer.

I would like to know how to write (if it is possible) $\sqrt{n}$ as a integral linear combination of $n$th roots of unity.

I have figured it out for $n$ congruent to 1 mod 4, but am not sure about the case where $n$ is congruent to 3 mod 4.

Any references/guidance appreciated!

asked Nov 13 '18 at 09:59

ChockaBlock

1

How did you figure it out for $n \equiv1 \mod 4$? A brief explanation suffices. – Sarvesh Ravichandran Iyer Nov 13 '18 at 10:00
See https://math.stackexchange.com/a/282779/589 for $n$ prime – lhf Nov 13 '18 at 10:03
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A corollary of a theorem by Gauss shows that for $n \equiv 1$ mod 4, $\sqrt{n} = 1+ (z+ z^4+z^9 + \dots z^{(n-1)^2})$. For $n \equiv 3$ mod 4, this gives a way to construct $i\sqrt{n}$, but not $\sqrt{n}$. – ChockaBlock Nov 13 '18 at 10:09

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