Sum of nth roots of unity equal to Sqrt[n]
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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!
number-theory complex-numbers
asked Nov 13 at 9:59
Melissium
856
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up vote
0
down vote
favorite
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!
number-theory complex-numbers
asked Nov 13 at 9:59
Melissium
856
1
How did you figure it out for $n equiv1 mod 4$? A brief explanation suffices.
– астон вілла олоф мэллбэрг
Nov 13 at 10:00
See math.stackexchange.com/a/282779/589 for $n$ prime
– lhf
Nov 13 at 10:03
1
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 $isqrt{n}$, but not $sqrt{n}$.
– Melissium
Nov 13 at 10:09
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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!
number-theory complex-numbers
asked Nov 13 at 9:59
Melissium
856
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!
number-theory complex-numbers
number-theory complex-numbers
asked Nov 13 at 9:59
Melissium
856
asked Nov 13 at 9:59
Melissium
856
asked Nov 13 at 9:59
Melissium
856
asked Nov 13 at 9:59
Melissium
856
asked Nov 13 at 9:59
Melissium
856
856
1
How did you figure it out for $n equiv1 mod 4$? A brief explanation suffices.
– астон вілла олоф мэллбэрг
Nov 13 at 10:00
See math.stackexchange.com/a/282779/589 for $n$ prime
– lhf
Nov 13 at 10:03
1
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 $isqrt{n}$, but not $sqrt{n}$.
– Melissium
Nov 13 at 10:09
add a comment |
1
How did you figure it out for $n equiv1 mod 4$? A brief explanation suffices.
– астон вілла олоф мэллбэрг
Nov 13 at 10:00
See math.stackexchange.com/a/282779/589 for $n$ prime
– lhf
Nov 13 at 10:03
1
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 $isqrt{n}$, but not $sqrt{n}$.
– Melissium
Nov 13 at 10:09
1
1
How did you figure it out for $n equiv1 mod 4$? A brief explanation suffices.
– астон вілла олоф мэллбэрг
Nov 13 at 10:00
How did you figure it out for $n equiv1 mod 4$? A brief explanation suffices.
– астон вілла олоф мэллбэрг
Nov 13 at 10:00
See math.stackexchange.com/a/282779/589 for $n$ prime
– lhf
Nov 13 at 10:03
See math.stackexchange.com/a/282779/589 for $n$ prime
– lhf
Nov 13 at 10:03
1
1
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 $isqrt{n}$, but not $sqrt{n}$.
– Melissium
Nov 13 at 10:09
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 $isqrt{n}$, but not $sqrt{n}$.
– Melissium
Nov 13 at 10:09
add a comment |
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How did you figure it out for $n equiv1 mod 4$? A brief explanation suffices.
– астон вілла олоф мэллбэрг
Nov 13 at 10:00
See math.stackexchange.com/a/282779/589 for $n$ prime
– lhf
Nov 13 at 10:03
1
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 $isqrt{n}$, but not $sqrt{n}$.
– Melissium
Nov 13 at 10:09