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.

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.

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.

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.

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 $F$.