existence of $b$ such that $b^2=a$, in a finite field. I am working on some basic field question for a linear algebra class.Below is the question. I looked at the multiplication of the $\mathbb Z_7$ field, and I don't see any of the elements whose square is $3$ . Is this the correct way to look at it?
Let $p$ be a prime integer and $a\in\mathbb F_p$. Does there necessarily exist $b\in\mathbb F_p$ satisfying $b^2=a$. 
Thanks in advance for any insight.
 A: 3 has no square root in $\mathbb{F}_7$. You may use the Lagrange symbol to check this but it seems overkill to me. You can check it directly by evaluating all squares of $\mathbb{F}_7$.
Not every element of a finite field $\mathbb{F}_p$ has a square root, for any $p>2$. You can see it by a combinatorial argument: For each nonzero $a$, $a\neq -a$ and $a^2=(-a)^2$. Moreover, $a^2=b^2$ implies $a=\pm b$. Now we can conclude that the map $x\mapsto x^2$ is two-to-one map over $\mathbb{F}_p^\times$. Especially there are $(p-1)/2$ nonzero elements which has no square root.
A: 
Let $p$ be a prime integer and $a\in\mathbb F_p$. Does there necessarily exist $b\in\mathbb F_p$ satisfying $b^2=a$. 

The answer is no, and you've already found a counterexample.  Take $p=7$ and $a=3$.
\begin{align*}
  0^2 &= 0\\
  1^2 &= 1\\
  2^2 &= 4\\
  3^2 &= 9 \equiv 2 \pmod 7\\
  4^2 &= 16 \equiv 2 \pmod 7\\
  5^2 &= 25 \equiv 4 \pmod 7\\
  6^2 &= 36 \equiv 1 \pmod 7
\end{align*}
So there is no $b \in \Bbb F_7$ such that $b^2 = 3$.  There may be simpler counterexamples but one is all you need.
A: Let $\mathbb{F}$ be a finite field with characteristic $\neq2$ (so that $-1\neq1$), and consider the function $s:\mathbb{F}\to\mathbb{F}$ given by $s(b):=b^2$. As $s(-1)=s(1)=1$ it follows that $s$ cannot be injective, which (as the domain and codomain of $s$ have the same finite size) implies that $s$ cannot be surjective; i.e. there is some $a\in\mathbb{F}$ such that $s(b)=b^2\neq a$ for every $b\in\mathbb{F}$.
