Determine the number of solutions for $x^2 = c$ such that $c \in \mathbb{F}$ I am working on the problem from number system class taught by the professor who shares his knowledge about the relation of abstract algebra and number systems.  Here is the problem:
Assume $c$ and $\mathbb{F}$ are arbitrary.  For $c \in \mathbb{F}$, how many distinct solutions can the equation $x^2 = c$ have?  Be specific about what properties of $c$ lead to what number of solution and justify your reasoning using properties a field must have.
Thoughts for the Problem
I believe that there is a solution when $c = 0$ for any field.  If $c = 1$, and we have $\mathbb{F}_2$, then we obtain exactly one solution.  BUT if $\mathbb{F}$ is any arbitrary field and $c = 1$, then we have two solutions.  This situation could occur the same way for any arbitrary field and $c \geq 2$, being an integer.  The thing is: what if $c$ is either trascendental or irrational (like roots number)?  Is it possible for the equation to have solutions like this?  I am having a hard time, approaching this problem.

Equivalently, we are talking about the solutions over $x \in \mathbb{F}$ to $x^2 - c= 0$, where $c\in \mathbb{F}$.
In any case, one of two things happens: either there is no solution, or there is a solution $d$ such that $d^2 = c$.  In such a case, we may write
$$
x^2 - c = (x+d)(x-d)
$$ 
So that $x^2 = 0$ if and only if $x = d$ or $x = -d$.  If (and only if) $d = -d$, our solution is unique.  Otherwise, there are two solutions.
Note that 
$d = -d \iff 2d = 0$.  So, if $c$ has a unique square root, then either $c = 0$ or there is an element $d \in \mathbb{F}$ for which $2d = 0$.
I'm not sure what you mean with the statement "what if $c$ is transcendental or irrational". What about it?
