Hypothesis on quadratic residues and cyclic generators

Question

Playing around with some examples of quadratic residues, I've come up with the following:

Let $q = p^n$ be an odd prime power for some odd prime $p$ and some $n \in \mathbb{N}$. Suppose $\mathbb{F}_q^* = \langle \varepsilon \rangle$ for some generator $\varepsilon$. Consider $-1$ as an element of $\mathbb{F}_q^*$. Then the following hold:

1. If $q \equiv 1$ mod 4, then $-1 = a^2 \ $ for some $a \in \mathbb{F}_q^*$
2. If $q \equiv 3$ mod 4, then $-1 = b^2\varepsilon \ $ for some $b \in \mathbb{F}_q^*$

Anyone have any idea if it even holds true, or why? Any literature knocking about on something like this?

It seems the first case is well known if $n=1$, that is if $q$ is itself prime, but I don't know about generalizing - like I say it's just a pattern that I've noticed. Likewise the second case is just a pattern I've noticed (from some low prime power examples) and I'm wondering if it generalizes too? Thanks!

Answer 1

It is better to veiw this in a group theoretic sense.

Then it is easy to see that $-1=\varepsilon ^ {\dfrac{q-1}{2}}$.

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If $q-1 \overset{4}{\equiv} 0 \ , $ then $a=\varepsilon ^ {\dfrac{q-1}{4}}$ satisfies the relation $a^2=-1$.

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Now consider the case $q-1 \overset{4}{\equiv} 2 \ , $ suppose on contrary that there is $a$ such that $a^2=-1$, then $ord(a)=4$ and we can conclude that $4 \mid q-1$, which is a contradiction.

Also in this case we can easilly chack that $b=\varepsilon ^ {\dfrac{q-3}{4}}$ satisfies the relation $-1=b^2 \varepsilon$.

[Because we have $\dfrac{q-1}{2}=2*\dfrac{q-3}{4}+1$. ]

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It is better to veiw this in a group theoretic sense.

Let $G$ to be a finite cyclic multiplicative group with the element $-1$; i.e. an element of order $2$, so by lagrange theorem the order of the group is even. Let's denote the generator of $G$ by $\varepsilon$. Also let's call the order of $G$ to be $N$. It is easy to see that $-1=\varepsilon ^ {\dfrac{N}{2}}$.

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If $N \overset{4}{\equiv} 0 \ , $ then $a=\varepsilon ^ {\dfrac{N}{4}}$ satisfies the relation $a^2=-1$.

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Now consider the case $N \overset{4}{\equiv} 2 \ , $ suppose on contrary that there is $a$ such that $a^2=-1$, then $ord(a)=4$ and we can conclude that $4 \mid N$, which is a contradiction.

Also in this case we can easilly chack that $b=\varepsilon ^ {\dfrac{N-2}{4}}$ satisfies the relation $-1=b^2 \varepsilon$.

[Because we have $\dfrac{N}{2}=2*\dfrac{N-2}{4}+1$. ]