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In $\mathbb{R}$ for all $r$, $ x^2 = r $ or $ x^2 = -r $ has a solution. For which finite fields is this true?

I.e. for which finite fields has a solution at least one of the equations $ x^2 = r $ or $ x^2 = -r $ for arbitrary $ r\in \mathbb{F} $ ?

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Let $q=p^m,m\geq 1$, where $p$ is a prime number. If $p=2$, since $r=r^q=r^{2^m}(r^{2^{m-1}})^2$, $r$ is a square.

If $p$ is odd, one can assume that $ r\neq 0$. Then, it is well known that $\mathbb{F}_q^\times$ has exactly two different square classes.

If $q\equiv 3 [4]$, then $-1$ is not a square (classical fact), so $r$ and $-r$ two different square classes, and one of the two equations has a solution.

If $q\equiv 1 [4]$, $-1$ is a square and $r$ and $-r$ represents the same square class. Hence, the two equations both has solutions or none of them have solutions. Since there always exist an nonsquare $r$, the desired property does not hold in this case.

Conclusion. For all $r\in\mathbb{F}_q$, one of the equations $x^2=r$ or $x^2=-r$ has a solution if and only if $q$ is even or $q\equiv 3 [4]$.

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  • $\begingroup$ You didn't show that there is always an $r$ in $\Bbb F_q, q\equiv 1[4]$ with $x^2 = r$ having no solution, to justify the "only if" in the conclusion. Though perhaps you are interpreting the "or" as exclusive, while I am interpreting it inclusively. $\endgroup$ – Paul Sinclair Jan 28 at 19:06
  • $\begingroup$ True. I added few words. $\endgroup$ – GreginGre Jan 28 at 21:33
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Let $F$ be a finite field of $q$ elements. Then, for all $r$, $ x^2 = r $ or $ x^2 = -r $ has a solution in $F$ iff $(x^2-r)(x^2+r)=x^4-r^2$ always has a root in $F$ iff the set of 4th powers is the same as the set of squares.

In a cyclic group of order $n$, the set of $m$th powers is the same as the set of $d$th powers, where $d=\gcd(m,n)$. Therefore, the set of 4th powers is the same as the set of squares iff $(n,4)=(n,2)$.

The cyclic group $F^\times$ has $n=q-1$ elements. Therefore, the set of 4th powers is the same as the set of squares in $F$ iff $(q-1,4)=(q-1,2)$.

If $q$ is odd, then $(q-1,2)=2$. Thus $(q-1,4)=2$, which happens iff $q \equiv 3 \bmod 4$.

If $q$ is even, then $(q-1,4)=(q-1,2)=1$.

Bottom line: $q \not\equiv 1 \bmod 4$.

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For any field $F$ of with non-2 characteristic, either $x^2+1$ has no roots, or it has two. If it has two, called $\pm\sqrt{-1}$, then $x^2+r$ and $x^2-r$ for $r\neq0$ either both have roots, or none of them do (if $x_0$ is a root of one of them, then $\sqrt{-1}x_0$ is a root of the other).

If $x^2+1$ doesn't have roots, then $x^2+r$ and $x^2-r$ cannot both have roots (as the ratio between a root of one and a root of the other would be a root of $x^2+1$). I claim that if $F$ is finite of odd characteristic, exactly one of them has a root.

Proof of claim: Exactly half of all possible polynomials $x^2+r$ for non-zero $r$ have roots, as the squaring map is two-to-one. And if $x^2+r$ has a root, $x^2-r$ doesn't. Thus by a pigeonhole argument, if there is an $r\in F^\times$ where neither $x^2+r$ nor $x^2-r$ have roots in $F$, then there must be an $s\in F^\times$ where both $x^2+s$ and $x^2-s$ have roots, which is a contradiction.

So in conclusion, you're after finite fields where $x^2+1$ has no roots, which are exactly the fields $\Bbb F_{p^n}$ for $p\equiv 3\pmod 4$ prime and $n$ odd.

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    $\begingroup$ If $F$ has even characteristic, then $x^{2}+1$ has a unique root. I would say more about the implications of this, but this is really a problem that should be left to the OP. $\endgroup$ – Morgan Rodgers Jan 30 at 18:47
  • $\begingroup$ @MorganRodgers You're right. I specified. $\endgroup$ – Arthur Jan 30 at 18:54

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