Let $ n \in \mathbb{N}, n \ge 2$.
Prove that
$\sqrt[n] {\sqrt{2018}+\sqrt{2017}}$ $+$ $\sqrt[n]{\sqrt{2018}-\sqrt{2017}} $ $\in \mathbb{R}/\mathbb{Q}$
Any ideas? I noticed that first member multiply with the second is equal with $1$. I tried to use Viette relations, but i can't finish the solution. Can anyone help me?
On
Let $a=\sqrt{2018}+\sqrt{2017}$ and $b =\sqrt{2018}-\sqrt{2017}$. As you noticed, $ab=1$ and $a+b=2\sqrt{2018} \not \in \mathbb Q$.
Suppose $u=a^{1/n} + b^{1/n} \in \mathbb Q$. Then we have $v = a^{1/n}\cdot b^{1/n}=1\in \mathbb Q$ also. As symmetric polynomials can be always expressed in terms of elementary symmetric polynomials with rational coefficients, $a+b$ must be expressible in terms of $u, v$, so $a+b$ must also be rational. Hence, there is a contradiction.
It is enough to prove the following statement: If $a+b \in \mathbb Q$ with $ab=1$, then $f(n)=a^n+b^n \in \mathbb Q$, for $n \in \mathbb N$.
Let us use induction. The statement is trivial for $n=1$.
Suppose it holds true for all $n\leqslant m$. Then, from $$(a+b)^{m+1}=f(m+1)+\sum_{k=1}^{\lfloor m/2\rfloor}\binom{m+1}k f(m+1-2k)$$ it follows that $f(m+1)$ must also be rational, hence we have our proof by strong induction.