In a finite field product of non-square elements is a square I came across one problem in a finite field as follows:

Let $F$ be a finite field. Show that if $a, b\in F$ both are non-squares, then $ab$ is a square. 

I wanted to prove it by using the idea of Biquadratic field extension. But there is no biquadratic extension over finite fields. Please, any hints for proving above fact? Thanks. 
 A: Hint: I assume, $a,b,\neq 0$ otherwise it is obvious. $F^*$ is cyclic. Suppose it is generated by $x$, you can write $a=x^n, b=x^m$, $n,m$ odd. Then $n+m$ is even.
A: Zero is a square, hence we may assume that both $a$ and $b$ belong to $\mathbb{F}^*$.
$\mathbb{F}^*$ is a cyclic group with order $q-1$: if the characteristic is $2$, every element of $\mathbb{F}$ is a square and there is nothing to prove. Otherwise, we may assume that $\mathbb{F}^*$ is generated by some element $g$, and in such a case the quadratic residues in $\mathbb{F}^*$ are the elements of the form $g^{\text{even}}$ and the non-quadratic residues are the elements of the form $g^{\text{odd}}$. Since $\text{odd}+\text{odd}=\text{even}$, the product of two non-quadratic residues is a quadratic residue, i.e. the Legendre symbol is multiplicative.
A: Egreg's argument is probably the simplest, followed by using cyclicity of the multiplicative group as outlined by Tsemo and Jack. The following approach might be closer to your first idea - utilizing the fact that a finite field has no biquadratic extensions. 
So my answer needs the fact that a finite field $F$ has a unique quadratic extension (up to an $F$-isomorphism).

We are given two elements $a,b\in F$ that don't have a square root in $F$. Therefore both extension fields $F(\sqrt a)$ and $F(\sqrt b)$ are quadratic, and hence isomorphic over $F$. This means that we can find a square root of $b$
inside $F(\sqrt a)$. More precisely, it implies that there exist elements $c_1,c_2\in F$ such that
$$
(c_1+c_2\sqrt a)^2=b.\qquad(*)
$$
Expanding the left hand side gives
$$
(c_1^2+ac_2^2)+2c_1c_2\sqrt a=b.
$$
Because $1,\sqrt a$ is a basis for the extension, and the right hand side is an element of $F$ we can conclude that we must have $2c_1c_2=0$. There are three 
ways how this could happen:


*

*If $2=0$, then we are in characteristic two, squaring is injective (see Jack's answer, or infer from Egreg's answer that squaring has trivial kernel), and therefore all the elements of $F$ are squares.

*If $c_2=0$, then $(*)$ is saying that $c_1^2=b$ contradicting the assumption that $b$ has no square root in $F$.

*That leaves the possibility $c_1=0$. So there exists an element $c_2\in F$ such that $ac_2^2=b$. This implies that
$$(ac_2)^2=a(ac_2^2)=ab$$
and therefore in this case $ab$ is the square of an element of $F$.

A: If the characteristic of $F$ is $2$, then $x\mapsto x^2$ is an automorphism of $F$ and therefore every element is a square. So we can assume the characteristic is an odd prime.
Consider the multiplicative group $F^*$ of nonzero elements; the map $x\mapsto x^2$ is a group endomorphism of $F^*$ with kernel $\{1,-1\}$. Therefore the image $H$ of this map is a subgroup satisfying $|H|=|F^*|/2$; this amounts to saying that $H$, which is the set of all squares in $F^*$, has index $2$. Therefore $F^*/H$ is a two-element group and the statement follows.
A: For my abstract algebra class (which was rings-first to an extent that I didn't ever get to defining a group...), I have written up a proof that uses neither groups nor field extensions. It is spread across two assignments:


*

*Exercise 5 (d) on UMN Spring 2019 Math 4281 homework set #6 shows that if $\mathbb{F}$ is a finite field with $2 \cdot 1_{\mathbb{F}} \neq 0_{\mathbb{F}}$, then $\mathbb{F}$ contains exactly $\dfrac{1}{2}\left(\left|\mathbb{F}\right|+1\right)$ many squares. This is proven by a double counting argument: Every nonzero square is a square of exactly two different elements of $\mathbb{F}$ (the word "different" here relies on $2 \cdot 1_{\mathbb{F}} \neq 0_{\mathbb{F}}$), and the square $0$ is a square of only one element of $\mathbb{F}$.

*Exercise 4 (c) on UMN Spring 2019 Math 4281 midterm #3 shows that if $\mathbb{F}$ is a finite field with $2 \cdot 1_{\mathbb{F}} \neq 0_{\mathbb{F}}$, then any product of two nonsquares in $\mathbb{F}$ is a nonsquare. This is proven in a somewhat tricky way, by first arguing that a product of squares is a square and that a product of a nonzero square with a nonsquare is a nonsquare, and finally making conclusions by counting (using the previously proven fact that $\mathbb{F}$ contains exactly $\dfrac{1}{2}\left(\left|\mathbb{F}\right|+1\right)$ many squares).

*Finally, Theorem 4.1 (c) in Section 4.4 of UMN Spring 2019 Math 4281 midterm #3 proves that in any finite field $\mathbb{F}$, the product of two nonsquares is a square. Indeed, if $2 \cdot 1_{\mathbb{F}} \neq 0_{\mathbb{F}}$, then this was already proven; but if $2 \cdot 1_{\mathbb{F}} = 0_{\mathbb{F}}$, then there are no nonsquares in $\mathbb{F}$ to begin with, and the claim is moot.
All of the proofs here use just basic properties of fields and the pigeonhole principle.
