Does the element exist in the Galois Field? Let p be a prime positive integer and let a be an element of GF(p). Does there necessarily exist an element b of GF(p) satisfying b^2=a?
So, taking a element of GF(p), can we find a b element of GF(p) s.t. a=b^2
I know that C(complex) has this property: for every w element of C, there exists z element of C s.t. z^2=w.
I'm not absolutely sure how to use this, I'm assuming proof by contradiction????

Let $G$ be a finite group of order $n$.
If $G$ has a non-trivial element $a$ of order $2$, then when we square all the elements of $G$ we have $a^2=1^2=1$, we get at most $n-1$ different results (by counting), and there must be some element of $G$ which is not a square. This happens whenever $G$ has even order, so no group of even order has every element a square.
Your group has even order. 

Conversely if every element has odd order, then every element is a square. Suppose $x^p=1$ where $p\gt 1$ is odd, then $\left(x^{\frac {p+1}2}\right)^2=x$
