To find a field with $p^{2}$ elements, where $p$ is prime 
Show that there exists a finite field with $p^{2}$ elements for every prime $p\in\mathbb{N}$.

What I thought is that if I find some irreducible polynomial of degree two over $\mathbb{Z}_p[x]$, then I will be done. But I had find it for some particular prime $2,3,5.$ But how to find irreducible polynomial of degree two in general in $\mathbb{Z_p[x]}$?
 A: Let $p=2$. Note that $x^2+x+1$ is irreducible over the two-element field, for it has no roots in that field.
Now let $p$ be odd. Let $\mathbb F_p$ be the $p$-element field. We have $(-a)^2=a^2$, and if $a\ne 0$ then $-a\ne a$. Thus  there are at most $\frac{p-1}{2}$ non-zero elements of $\mathbb F_p$ that are squares of elements of $\mathbb F_p$. (Actually, there are exactly $\frac{p-1}{2}$, but we don't need that.)
So there are at least $\frac{p-1}{2}$ elements of $\mathbb F_p$ that are not the square of any element of $\mathbb F_p$. Let $b$ be a non-square in $\mathbb F_p$. Then $x^2-b$ is irreducible over $\mathbb F_p$.
A: How many polynomials are there of degree 2?
How many reducible polynomials are there of degree 2?
Let the leading coefficient in both cases be 1.
A: Though it is possible to find fields of order $p^2$ for a fixed $p$ by what you have done  but to prove it for an arbitrary field you need to use a concept popularly known as Splitting field.
Once you get that(I don't know whether you are accustomed to it or not) you need to consider the splitting field of the polynomial
$x^{p^2}-x $  over $\Bbb Z_p$ and you are done.
