Prove $x = \sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ is irrational

Question

Prove $x = \sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ is irrational.

I can prove that $x$ is irrational by showing that it's a root of a polynomial with integer coefficients and use rational root theorem to deduce it must be either irrational or an integer and then show it's not an integer, therefore, it must be irrational.

I was wondering what are the other methods to prove $x$ is irrational. I'd be very interested in seeing alternative proofs.

Answer 1

Let $y = \sqrt[100]{\sqrt{3} + \sqrt{2}}$. Then $x = y + {1 \over y}$. Suppose $x$ were some rational number $q$. Then $y^2 - qy + 1 = 0$. This means ${\mathbb Q}(y)$ is a field extension of ${\mathbb Q}$ of degree two, and every rational function of $y$ is in this field extension. This includes $y^{100} = \sqrt{3} + \sqrt{2}$, and $y^{-100} = \sqrt{3} - \sqrt{2}$. So their sum and difference is also in ${\mathbb Q}(y)$. Hence ${\mathbb Q}(\sqrt{2},\sqrt{3}) \subset {\mathbb Q}(y)$. But ${\mathbb Q}(\sqrt{2},\sqrt{3})$ is an extension of ${\mathbb Q}$ of degree 4, a contradiction.

You can make the above more elementary by showing successive powers of $y$ are always of the form $q_1y + q_2$ with $q_1$ and $q_2$ rational and eventually showing some $q_3\sqrt{2} + q_4\sqrt{3}$ must be rational, a contradiction.

Answer 2

If $y + y^{-1} = x$, then $y^n + y^{-n}$ is a polynomial $F_n(x)$ with integer coefficients in $x$. In particular, if $x$ is rational, then so is: $$F_{100}(x) = y^{100} + y^{-100} = (\sqrt{3}+\sqrt{2}) + (\sqrt{3}-\sqrt{2}) =2 \sqrt{3}.$$

The polynomials $F_{n}(x)$ are (up to scaling and renormalization) the Chebyshev polynomials. One has $F_0(x) = 2$, $F_1(x) = x$, $F_2(x) = x^2 - 2$, and $F_{n}(x) = x F_{n-1}(x) - F_{n-2}(x)$.

Answer 3

Note $\ \ \color{#c00}{\alpha+\bar\alpha,\ \alpha\:\!\bar\alpha}\in\Bbb Q\ \Rightarrow\ \color{#0a0}{\alpha^{n}+\bar\alpha^{\:\!n}}\in \Bbb Q\ $ for all $\,n\in\Bbb N\,$ by induction, since

$$ \alpha^{n+1}+\bar\alpha^{\:\!n+1} = (\color{#c00}{\alpha+\bar\alpha})\ (\alpha^n+\bar\alpha^{\:\!n})-\color{#c00}{\alpha\:\!\bar\alpha}\, (\alpha^{n-1}+\bar\alpha^{\:\!n-1})\quad$$

$\begin{align}\alpha = \sqrt[100]{\sqrt3+\!\sqrt2}\\ \bar\alpha = \sqrt[100]{\sqrt3-\!\sqrt2}\end{align}$ $\Rightarrow\,\color{#c00}{\alpha\:\!\bar\alpha} = 1\in\Bbb Q\ $ so $\, \color{#c00}{\alpha + \bar\alpha}\in\Bbb Q\,\Rightarrow\,2\sqrt{3} = \color{#0a0}{\alpha^{100} + \bar\alpha^{100}} \in\mathbb Q\, \Rightarrow\!\Leftarrow$

Generally $\,\alpha+\bar\alpha,\,\alpha\:\!\bar\alpha\in\Bbb Q\Rightarrow f(\alpha,\bar \alpha)\in \Bbb Q\,$ for any symmetric polynomial $f(x,y)\in\Bbb Q[x,y]\,$ since such a polynomial can be written as a polynomial in $\,\alpha+\bar\alpha,\ \alpha\:\!\bar\alpha\,$ with integer coefficients (Fundamental Theorem of Symmetric Polynomials). Above $\,f(x,y) = x^{100}+y^{100}$.

Look up Lucas-Lehmer sequences to learn more about such power sums of algebraic numbers.

Answer 4

We first claim that: $$x = a^{1/5} + a^{-1/5} \in \mathbb{Q} \implies a + a^{-1} \in \mathbb{Q}$$ To prove this we use some clever manipulation to find: $$(a^{1/5} + a^{-1/5})^5 = a + a^{-1} + 5(a^{1/5} + a^{-1/5})^3 - 5(a^{1/5} + a^{-1/5})$$ $$a + a^{-1} = x^5 - 5x^3 +5x \in \mathbb{Q}$$ Similiarly, we prove: $$x = a^{1/2} + a^{-1/2} \in \mathbb{Q} \implies a + a^{-1} \in \mathbb{Q}$$ Then if we let: $$a = \sqrt{3} + \sqrt{2}$$ and assume that: $$x = \sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}} \in \mathbb{Q}$$ we have: $$ x \in \mathbb{Q} \implies a^{1/100} + a^{-1/100} \in \mathbb{Q}$$ $$\implies a^{1/50} + a^{-1/50} \in \mathbb{Q} \implies a^{1/25} + a^{-1/25} \in \mathbb{Q}$$ $$\implies a^{1/5} + a^{-1/5} \in \mathbb{Q} \implies a + a^{-1} \in \mathbb{Q} \implies 2\sqrt{3} \in \mathbb{Q}$$ So, by contradiction: $$\sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}} \notin \mathbb{Q}$$