Squares in a field with q^n elements

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Discussion Overview

The discussion revolves around the conditions under which the equation \(y^2 = r\) has a solution in a field \(F\) with \(q^n\) elements, where \(q\) is an odd prime. The participants explore both directions of the proof regarding the relationship between the existence of solutions and the condition \(r^m = 1\), with a focus on mathematical reasoning and proofs.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the equation \(y^2 = r\) has a solution if and only if \(r^m = 1\), where \(m\) is defined as \((q^n - 1)/2\).
  • Another participant asks for clarification on the proof, specifically whether the direction that if \(y^2 = r\) has a solution then \(r^m = 1\) has been established.
  • A participant provides a partial proof for the direction that if \(y^2 = r\), then \(r^m = 1\), citing that the non-zero elements of the field form a cyclic group of order \(2m\).
  • Further, it is proposed that if \(r^m = 1\), then \(r^{(q^n + 1)/2}\) can be shown to be a solution to \(y^2 = r\), although this direction is still under discussion.

Areas of Agreement / Disagreement

Participants are engaged in a discussion with some agreement on the implications of the cyclic group structure of the field's non-zero elements, but there is no consensus on the complete proof or the validity of both directions of the argument.

Contextual Notes

The discussion involves assumptions about the properties of cyclic groups and the specific structure of the field \(F_{q^n}\), which may not be fully explored or agreed upon by all participants.

Prefer
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Let $F$ be a field with $q^n$ elements, where $q$ is an odd prime. Write $q^n=2m +1$ with $m \in \mathbb{N}.$

If $r \in F^{\times},$ show that the equation $y^2= r$ has a solution iff $r^m=1.$
 
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Hi Prefer,

Please show what your thoughts are on this problem or where you're stuck. Have you been able to prove one of the directions, e.g., if the equation $ y^2 = r $ has a solution, then $ r^m = 1$?
 
Euge said:
Hi Prefer,

Please show what your thoughts are on this problem or where you're stuck. Have you been able to prove one of the directions, e.g., if the equation $ y^2 = r $ has a solution, then $ r^m = 1$?

I have the direction you mention, so far:

The non-zero elements of $\mathbf F_{q^n}$ are a cyclic group of order $q^n-1=2m$. So, if $y^2=r$, we have $r^m=y^{2m}=1$.
 
Prefer said:
I have the direction you mention, so far:

The non-zero elements of $\mathbf F_{q^n}$ are a cyclic group of order $q^n-1=2m$. So, if $y^2=r$, we have $r^m=y^{2m}=1$.

Ok, great. For the reverse direction, assume $r^m = 1$. Since $q$ is odd, $q^n + 1$ is divisible by $2$. So we may consider $r^{(q^n + 1)/2}$. Show that this element is a solution to $y^2 = r$.
 

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