# Finite field with real-like square root [closed]

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 $$r\in \mathbb{F}$$ ?

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 $$q\equiv 3 [4]$$.

• 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. – Paul Sinclair Jan 28 at 19:06
• True. I added few words. – GreginGre Jan 28 at 21:33

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

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.

• 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. – Morgan Rodgers Jan 30 at 18:47
• @MorganRodgers You're right. I specified. – Arthur Jan 30 at 18:54