Show that $({\sqrt{2}\!+\!1})^{1/n} \!+ ({\sqrt{2}\!-\!1})^{1/n}\!\not\in\mathbb Q$

Question

How could we prove that for every positive integer $n$, the number $$({\sqrt{2}+1})^{1/n} + ({\sqrt{2}-1})^{1/n}$$ is irrational?

I think it could be done inductively from a more general expression, but I don't know how.

I made an effort trying to solve it using many different methods.

Answer 1

Assume that for a certain positive integer $n$ $$ \big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\in \mathbb Q. $$ Clearly, $\,\big(\sqrt{2}-1\big)^{1/n}\cdot\big(\sqrt{2}+1\big)^{1/n}=1$, and thus $$ \left(\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\right)^2= \big(\sqrt{2}-1\big)^{2/n}+2+\big(\sqrt{2}+1\big)^{2/n} \in \mathbb Q, $$ and hence $$ \big(\sqrt{2}-1\big)^{2/n}+\big(\sqrt{2}+1\big)^{2/n} \in \mathbb Q. $$ Next $$ \left(\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\right)^3= \big(\sqrt{2}-1\big)^{3/n}+\big(\sqrt{2}+1\big)^{3/n}+3\left(\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\right), $$ and hence $$ \big(\sqrt{2}-1\big)^{3/n}+\big(\sqrt{2}+1\big)^{3/n}\in\mathbb Q. $$ Let $s_k=\big(\sqrt{2}-1\big)^{k/n}+\big(\sqrt{2}+1\big)^{k/n}$. Then $$ s_1^k=s_k+\binom{k}{1}s_{k-2}+\binom{k}{2}s_{k-4} +\cdots+\binom{k}{\lfloor k/2\rfloor}s_{k-2\lfloor k/2\rfloor}, $$ which means that we can inductively show that $$ s_1,s_2,\ldots,s_n\in\mathbb Q. $$ But $$ s_n=\big(\sqrt{2}-1\big)+\big(\sqrt{2}+1\big)=2\sqrt{2}\not\in\mathbb Q, $$ and hence $$ s_1=\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\not\in\mathbb Q. $$

Generalization. If $a,b\in\mathbb R$, such that $a,b, a+b\not\in\mathbb Q$ and $ab\in\mathbb Q$, then using exactly this method, we obtain that $a^{1/n}+b^{1/n}\not\in\mathbb Q$, for all $n\in\mathbb N$.

Answer 2

Use the fact that for every $n$ there exists a polynomial $P_n[x]$ monic of degree $n$ with integral coefficients so that we have the identity:

\begin{eqnarray*} t^n + \frac{1}{t^n} = P_n(t+ \frac{1}{t}) \end{eqnarray*}

The $P_n$'s are connected with the Chebyshev polynomial of first kind since for $t = e^{i \theta}$ we get $2 \cos n \theta = P_n (2 \cos \theta)$

\begin{eqnarray*} \cos n \theta &=& \frac{1}{2}P_n (2 \cos \theta) \\ \cos n \theta &=& T_n( \cos \theta) \end{eqnarray*} so \begin{eqnarray*} P_n(x) = 2 T_n(\frac{x}{2}) \end{eqnarray*}

For instance \begin{eqnarray*} T_3(x) = 4x^3 - 3 x \\ P_3(x) = 2 T_3(\frac{x}{2}) = x^3 - 3 x \end{eqnarray*}

One checks easily the identity:

\begin{eqnarray*} t^3 + \frac{1}{t^3} = P_3(t+\frac{1}{t}) \end{eqnarray*}

Back to our problem. We have $(\sqrt{2}-1)(\sqrt{2}+1) = 2 -1 = 1$ and so $(\sqrt{2}-1) = \frac{1}{\sqrt{2} + 1}$ and similarly for any power $$(\sqrt{2}-1)^{1/n} = \frac{1}{\sqrt{2} + 1)^{1/n}}$$.

Let $x =(\sqrt{2} + 1)^{1/n}$. Then $$ x + \frac{1}{x} = (\sqrt{2}+ 1)^{1/n} + (\sqrt{2} - 1)^{1/n}$$

Now if $x+\frac{1}{x}$ were rational then so would be $P_n(x+\frac{1}{x}) = x^n + \frac{1}{x^n} = \sqrt{2}+1 + \sqrt{2} - 1 = 2 \sqrt{2}$ which is not the case. We conclude that $x+\frac{1}{x} = (\sqrt{2}+ 1)^{1/n} + (\sqrt{2} - 1)^{1/n}$ is irrational.

In fact $(\sqrt{2}+ 1)^{1/n} + (\sqrt{2} - 1)^{1/n}$ is the unique positive root of the equation $(P_n[x])^2 - 8 =0$ or $ (T_n(\frac{x}{2}))^2 = 2$

Answer 3

Let $\alpha = (1+\sqrt{2})^{1/n}$, $\beta = (-1+\sqrt{2})^{1/n}$. Then $\alpha\beta = 1$, so $\alpha$ and $\beta$ are roots of the polynomial $X^2-(\alpha+\beta)X+1$.

If $\alpha+\beta\in\mathbb{Q}$, then $\alpha$ and $\beta$ lie in a quadratic extension of $\mathbb{Q}$. Since $\alpha^n = 1 + \sqrt{2}$, we have $\alpha\in\mathbb{Q}[\sqrt{2}]$, and since $\alpha$ is an algebraic integer, this implies that $1+\sqrt{2}$ is an $n$-th power in $\mathbb{Z}[\sqrt{2}]$.

We'll show that this can't happen. Suppose that $(a+b\sqrt{2})^n = 1+\sqrt{2}$. We may assume that $a>0$. Then $1=a^n + 2{n\choose 2} a^{n-2} b^2 + \cdots$ is a sum of at least two positive integers, contradiction.

Answer 4

Yet another answer: (credit goes 99% to lhf)

Let $\alpha = (\sqrt{2}+1)^{1/n}$, $\beta = (\sqrt{2}-1)^{1/n}$. Since $\alpha$ and $\beta$ are algebraic integers, $\alpha+\beta$ is an algebraic integer. Since a rational algebraic integer is an integer, it suffices to show that $\alpha+\beta$ is not an integer.

The problem is trivial for $n=1$, so we assume $n>1$.

We have $\alpha\beta=1$, so, by AM-GM, $\alpha+\beta > 2$ ($\alpha\neq\beta$, so we do not have equality). Furthermore, $\beta<1$ and $\alpha < (1+\sqrt{2})^{1/2} < 2$, so $\alpha+\beta<3$. Therefore $\alpha+\beta$ is not an integer, hence not rational.