Squares in a field with q^n elements

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SUMMARY

The discussion focuses on the conditions under which the equation \( y^2 = r \) has a solution in the field \( F \) with \( q^n \) elements, where \( q \) is an odd prime. It is established that \( y^2 = r \) has a solution if and only if \( r^m = 1 \), with \( m \) defined as \( m = \frac{q^n - 1}{2} \). The participants confirm that the non-zero elements of \( \mathbf{F}_{q^n} \) form a cyclic group of order \( 2m \), leading to the conclusion that if \( r^m = 1 \), then \( y^2 = r \) can be satisfied by \( r^{(q^n + 1)/2} \).

PREREQUISITES
  • Understanding of finite fields, specifically \( \mathbf{F}_{q^n} \)
  • Knowledge of cyclic groups and their properties
  • Familiarity with quadratic residues in fields
  • Basic algebraic manipulation involving exponents
NEXT STEPS
  • Study the properties of finite fields, focusing on \( \mathbf{F}_{q^n} \)
  • Learn about cyclic groups and their applications in number theory
  • Explore quadratic residues and non-residues in finite fields
  • Investigate the implications of the equation \( y^2 = r \) in different field structures
USEFUL FOR

Mathematicians, particularly those specializing in algebra and number theory, as well as students studying finite fields and their applications in cryptography and coding theory.

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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