How to show that $\sqrt{2}+\sqrt{3}$ is algebraic?

Let $x=\sqrt{2}+\sqrt{3}$. Note that $x^2=2+2\sqrt{6}+3$ and therefore $x^2-5=2\sqrt{6}$. We can square both sides of this equation and obtain $(x^2-5)^2=24$. You should now be able to show that $x$ is an algebraic number (over $\mathbb{Q}$). (In fact, it is instructive to expand this equation and obtain a monic polynomial $p$ of degree $4$ such that $p(x)=0$.)

The following exercises are relevant:

Example-Based Exercises:

Exercise 1: Prove that the following numbers are algebraic (over $\mathbb{Q}$):

(a) $\sqrt{2}+\sqrt{5}$

(b) $\sqrt{2}+\sqrt{4}$

(c) $\sqrt{2}+\sqrt[3]{3}$

The degree of an algebraic number $x$ is defined to be the least positive integer $n$ such that there is a monic polynomial $p$ of degree $n$ with $p(x)=0$. Compute the degrees of each of the algebraic numbers above (in (a), (b) and (c)). Prove that your answer is correct in each case.

Exercise 2: Let $p_1,\dots,p_n$ be distinct (positive) prime numbers. Prove that the number $x=\sqrt{p_1}+\cdots+\sqrt{p_n}$ is algebraic. What is the degree of $x$?

Exercise 3: Let $x$ be an algebraic number and let $q$ be a rational number. Prove that the degree of $x$ equals the degree of $x+q$.

Exercise 4: Let $x=\cos(\frac{2\pi}{n})$ for some positive integer $n$. Is $x$ algebraic (over $\mathbb{Q}$)? If not, prove it, and if so find the degree of $x$. (Your answer will depend on $n$.)

Theoretical Exercises:

Exercise 5: Let $x\in\mathbb{C}$ and define $\mathbb{Q}[x]=\{f(x):f\text{ is a polynomial with rational coefficients}\}$.

(a) Prove that $\mathbb{Q}[x]$ is a finite dimensional vector space over $\mathbb{Q}$ if and only if $x$ is algebraic over $\mathbb{Q}$. (Hint (for one direction): if $x$ is algebraic over $\mathbb{Q}$ and if $a_0+a_1x+\cdots+a_{n-1}x^{n-1}+a_nx^n=0$ where $a_i\in\mathbb{Q}$ for all $0\leq i\leq n$, show that $\{1,x,x^2,\dots,x^{n-1}\}$ spans $\mathbb{Q}[x]$ as a $\mathbb{Q}$-vector space.)

(b) Prove that $\mathbb{Q}[x]$ is a field if $x$ is algebraic over $\mathbb{Q}$. (Hint: it suffices to prove that $x$ has a multiplicative inverse. Let $p$ be a monic polynomial of minimal degree (with coefficients in $\mathbb{Q}$) such that $p(x)=0$. Prove that $p(0)\neq 0$ if $x\neq 0$.)

(c) Prove that if $\mathbb{Q}\subseteq K\subseteq L$ is a tower of fields and if $K$ is a finite dimensional $\mathbb{Q}$-vector space, $L$ is a finite dimensional $K$-vector space, then $L$ is a finite dimensional $\mathbb{Q}$-vector space.

(d) Prove that the sum and product of two algebraic numbers (over $\mathbb{Q}$) is algebraic. (Hint: if $x$ and $y$ are algebraic, use (c) to show that $\mathbb{Q}[x,y]=\{f(x,y):f\text{ is a polynomial in two variables with rational coefficients}\}$ is a finite dimensional $\mathbb{Q}$-vector space. (a) is relevant.)

Challenge: If $x$ and $y$ are algebraic numbers (over $\mathbb{Q}$), can you find an explicit polynomial $p$ with rational coefficients such that $p(x+y)=0$? Can you find an explicit polynomial $q$ with rational coefficients such that $q(xy)=0$?

I hope this helps!

We show how to exploit symmetries to produce an answer. Although we only deal with the specific question, we deal with it in a structured way that at least hints at generalization.

Note that $\sqrt{2}+\sqrt{3}$ is the solution of $x-(\sqrt{2}+\sqrt{3})=0$. But $\sqrt{2}$ and $-\sqrt{2}$ are friends and like to travel in pairs, as do $\sqrt{3}$ and $-\sqrt{3}$.

In acknowledgement of this, let $$P(x)=(x-(\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}-\sqrt{3}))(x-(\sqrt{2}-\sqrt{3})).$$

Intuitively speaking at least, $P(x)$ should have integer coefficients. Why? First observe that the coefficients of a polynomial are symmetric functions of the roots.

Now imagine "simplifying" $P(x)$. Each coefficient of $P(x)$ must be of the shape $$A+B\sqrt{2}+ C\sqrt{3} + D\sqrt{2}\sqrt{3},$$ where $A$, $B$, $C$, and $D$ are integers. But this expression must remain unchanged if we replace $\sqrt{2}$ by $-\sqrt{2}$ and/or $\sqrt{3}$ by $-\sqrt{3}$, so we should have $B=C=D=0$. Thus every coefficient of $P(x)$ should be an integer.

Until we back up this kind of intuition with general theorems, our discussion has a speculative character. However, in this case, we can easily multiply out $P(x)$ and check directly that it has integer coefficients. The product of the first two terms is $x^2-2\sqrt{3}\,x+1$, and the product of the next two terms is $x^2+2\sqrt{3}\,x +1$. Thus $$P(x)=x^4-10x^2+1,$$ and our intuition has been fully verified.

Comment: All the speculative elements above can be put on a firm foundation. With the proper theoretical background in place, we would not have needed to compute at all. "Symmetries" play a central role in the study of algebraic numbers. This was a key insight of Lagrange, which led ultimately to the breakthroughs by Abel and Galois.

Call $x= \sqrt{2} + \sqrt{3}$, then by your equation we have $$x^{2} = 5 +\sqrt{24} \Longrightarrow (x^{2}-5)^{2}=24$$

For a proof of sum of $2$ algebraic numbers being algebraic, please see: