Prove that $\mathbb{F}_5[X]/(X^2+3)$ is a field with 25 elements Prove that the ring shown above is a field containing 25 elements.
Research effort:
$$\mathbb{F}_5[X]/(X^2+3)\cong (\mathbb{Z}[X]/5\mathbb{Z}[X])/ \overline{(X^2+3)} \cong \mathbb{Z}[X]/(5,X^2+3)$$
Modifying the term $3$, I tried to find $k$ such that $5k -3 = n^2$ for some $n \in \mathbb{N}$, in order to find a zero for the polynomial $X^2+3$. But there is no $k \in \{1, 2 , \cdots, 9 \}$, so that $k^2$ ends with a two or a seven,so we won't find a $k\geq 10$ that ends on $7$ or $2$ either, since $(10k +m)^2 = 10(10k^2+2km) +m^2$.
Maybe it's possible to modidy the other terms of the polynomial to find a root. In that case we could use a substitution homomorphism to prove that it's isomorphic to  $\mathbb{F}_5$. 
May intuition says that this is not possible however. Can you provide me a hint for this?
This means that we can not modify the polynomial to show that it has a root.

You probably saw a theorem that states that for any field $F$ and irreducible polynomial $P(X)\in F[X]$, the quotient ring $F[X]/(P(X))$ is a field. So, you need to establish that your polynomial $X^2+3$ is irreducible over the field $F=\mathbb F_5$. Since the polynomial has degree $2$, all you need to do is verify that the polynomial has not roots in the field. Since there are only $5$ elements in the field, it's quite easy to check them one by one.
