Irrational equation with 2 terms

Question

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?

Do you mean $\mathbb{R}\setminus\mathbb{Q}$? – almagest Feb 17 '18 at 19:17 — almagest, Feb 17 '18 at 19:17
@almagest yes, i’m sorry. I’m begginer with Latex – Radu Andrei Feb 17 '18 at 19:23 — Radu Andrei, Feb 17 '18 at 19:23

score 0 · Answer 1 · answered Feb 17 '18 at 18:29

0

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.

answered Feb 17 '18 at 18:29

Macavity

46,381

Do you know another solution? Without polynomials? I didn't studied them at school. – Radu Andrei Feb 17 '18 at 18:34
@RaduAndrei It may help to clarify what you are familiar with - will see if there is a simpler way, perhaps with induction which I assume you know. – Macavity Feb 17 '18 at 18:48
Yes, i know induction. I hope there is a way to solve this with induction – Radu Andrei Feb 17 '18 at 18:50
OK. Have put up an answer with induction, and it is simpler... – Macavity Feb 17 '18 at 19:34

score 0 · Accepted Answer · answered Feb 17 '18 at 19:35

0

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.

answered Feb 17 '18 at 19:35

Macavity

46,381

1

Thank you so much. I apreciate your effort – Radu Andrei Feb 17 '18 at 19:37

Irrational equation with 2 terms

2 Answers2