Understanding elements of $\mathbb Z[\sqrt p+ \sqrt q]$ 
Let $p,q$ be primes. Consider the ring $R=\mathbb Z[\sqrt p+ \sqrt q]$. It consists of all polynomials in $\sqrt p+\sqrt q$ with integer coefficients. I was told that we may assume that an element of $R$ can actually be considered as a  polynomial in $\sqrt p+ \sqrt q$ of degree $\le 3$. Why this is true? 

I was thinking about division with remainder but not sure what exactly I should divide by what, and in which ring (we don't know that $R$ is Euclidean).
 A: Imagine a polynomial $$a_n(\sqrt p + \sqrt q)^n + \dots + a_5(\sqrt p + \sqrt q)^5 + a_4(\sqrt p + \sqrt q)^4 + a_3(\sqrt p + \sqrt q)^3 + a_2(\sqrt p + \sqrt q)^2 + a_1(\sqrt p + \sqrt q) + a_0.$$ The idea here is that if we can express $(\sqrt p + \sqrt q)^4$ as a linear combination of terms with powers of $\sqrt p + \sqrt q$ less than or equal to 3, then we can continually substitute that into the expression to get rid of all the terms with degree 4 or greater. We start by substituting for the degree $n$ term, getting a new polynomial of degree $n-1$; then we substitute for the new degree $n-1$ term, etc., and we can continue substituting until the degree of the polynomial is less than 4.
So let's look for that linear combination:
$$ \begin{align}
(\sqrt p + \sqrt q)^2 &= p + 2\sqrt{pq} + q \\
2\sqrt{pq} &= (\sqrt p + \sqrt q)^2 - p - q \\
(\sqrt p + \sqrt q)^4 &= p^2 + 4p\sqrt{pq} + 6pq + 4q\sqrt{pq} + q^2 \\
&= p^2 + 2p((\sqrt p + \sqrt q)^2 - p - q) + 6pq + 2q((\sqrt p + \sqrt q)^2 - p - q) + q^2
\end{align}$$
A: Since $$(\sqrt{p}+\sqrt{q})^4-2(p+q)(\sqrt{p}+\sqrt{q})^2+(p-q)^2=0$$ if you have a polynomial in $\sqrt{p}+\sqrt{q}$ of a larger degree you can always lower the degree. 
For instance $(\sqrt{p}+\sqrt{q})^5+2p$ is $$(\sqrt{p}+\sqrt{q})\left (2(p+q)(\sqrt{p}+\sqrt{q})^2-(p-q)^2 \right )+2p$$ $$=2(p+q)(\sqrt{p}+\sqrt{q})^3-(p-q)^2(\sqrt{p}+\sqrt{q})+2p$$
which is a polynomial in $\sqrt{p}+\sqrt{q}$ of degree 3.
If you are familiar with quotient rings, notice that $$\mathbb{Z}[\sqrt{p}+\sqrt{q}]\cong\frac{\mathbb{Z}[x]}{(x^4-(p+q)x^2+(p-q)^2)}$$ so every element can be expressed as $$a_0+a_1x+a_2x^2+a_3x^3\mid a_0,a_1,a_2,a_3\in\mathbb{Z}$$ This is the same as saying that every element in $\mathbb{Z}[\sqrt{p}+\sqrt{q}]$ can be expressed as $$a_0+a_1(\sqrt{p}+\sqrt{q})+a_2(\sqrt{p}+\sqrt{q})^2+a_ 3(\sqrt{p}+\sqrt{q})^3\mid a_0,a_1,a_2,a_3\in\mathbb{Z}$$
