<s>[INST] Show that $({\sqrt{2}\!+\!1})^{1/n} \!+ ({\sqrt{2}\!-\!1})^{1/n}\!\not\in\mathbb Q$ 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.
 [/INST] 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$
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