MHB Squares in a field with q^n elements

[Prefer]

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 $r^m=1.$

 

[Euge]

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$?

 

[Prefer]

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 $r^m=y^{2m}=1$.

 

[Euge]

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 $y^2 = r$.