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.

So we have to show that the equivalence class of any $aX + b$ modulo the ideal $(X^2 + 3)$, where one of $a, b \in \mathbb{F}_5$ is nonzero, has a multiplicative inverse. 
Think of $X$ as "$\sqrt{-3}$" in the quotient ring, and rationalize the denominator 
$$\frac1{a\sqrt{-3} + b} = \frac{b - a \sqrt{-3}}{b^2 - (-3)a^2}.$$ 
Note that $b^2 + 3a^2 \neq 0$ if one of $a, b$ is nonzero, essentially because $x^2 + 3 = 0$ has no solutions $x = b a^{-1}$ in $\mathbb{F}_5$. What this tells us is that 
$$\frac1{b^2 + 3a^2} (b - aX)$$ 
represents the inverse of $a X + b$ in the quotient ring, as you can check yourself. 
Oh yes: it's easy to see that this field has 25 elements, since any element $p(x)$ in $\mathbb{F}_5$ can be uniquely written in the form $b + aX + (3 + X^2)q(x)$, by using the Euclidean algorithm, and there are 25 possibilities for the pair $(a, b)$. 
