If $\mathbb{F}$ is a finite field such that for every $a \in \mathbb{F}$ the equation $x^2=a$ has a solution in $\mathbb{F}$ 
Let $\mathbb{F}$ be a finite field such that for every $a \in \mathbb{F}$ the equation $x^2=a$ has a solution in $\mathbb{F}$. Then $\mathbb{F} \simeq \mathbb{F}_{2^n}$.

I have tried this for the fields $\mathbb{F}_{2}$, $\mathbb{F}_{4}$ and $\mathbb{F}_{8}$ but I cant see a pattern. How do I prove this and also is there some general theory for the case when $x^r=a$ $(r \in \mathbb{N})$ has solutions for all $a$ in the field $\mathbb{F}$??
I am working on a problem set (Number theory) and apparently there are many problems relating to solutions of polynomials of the form  $x^k=a$ over finite fields. 
 A: Since the order of a finite field is always the power of a prime $p,$ we only need to show that the characteristic of $\mathbb F$ is $2.$
Consider the group homomorphism $f:\Bbb F^*\rightarrow\Bbb F^*$ sending $x$ to $x^2.$ If the characteristic of $\Bbb F$ is $p\ne2,$ then the kernel of $f$ contains $1$ and $-1,$ which are different elements in $\Bbb F^*,$ thus $f$ cannot be surjective, contradicting our hypothesis, therefore $p=2$ and $\Bbb F\cong\Bbb F_{2^n}.$
For general $r,$ per chance you can try considering the homomorphism sending $x$ to $x^r,$ and then consider its kernels, images, etc?  
Hope this helps.
A: An alternative approach. $\mathbb{F}^*$ is a cyclic group generated by some $g\in\mathbb{F}^*$: let $M=\left|\mathbb{F}^*\right|$.
All the elements of $\mathbb{F}^*$ are listed here:
$$\{g,g^2,g^3,\ldots, g^M\} $$
and the map $\psi:x\to x^2$ sends the element $g^a$ into $g^{2a\pmod{M}}$: it follows that if $M$ is even $\psi$ is not surjective, contradicting the statement. So we have that $M$ is odd, i.e. $\left|\mathbb{F}\right|$ is even, i.e.
$$ \mathbb{F}\simeq \mathbb{F}_{2^n}. $$
