New user? Sign up

Existing user? Log in

Let \(x\) be a real number such that \(x + x^{-1}\) is an integer. Prove that \(x ^n + x^{-n}\) is an integer, for all positive integer \(n\).

Note by Dev Sharma 2 years, 10 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

It's probably easiest to use strong induction. The statement is given for \(n = 1.\) Now suppose \(x^{k} + x^{-k}\) is an integer for all \(k \le n.\) We have that

\(\left(x^{n} + \dfrac{1}{x^{n}}\right) \left(x + \dfrac{1}{x}\right) = x^{n+1} + \dfrac{1}{x^{n-1}} + x^{n-1} + \dfrac{1}{x^{n+1}}\)

\(\Longrightarrow x^{n+1} + \dfrac{1}{x^{n+1}} = \left(x^{n} + \dfrac{1}{x^{n}}\right)\left(x + \dfrac{1}{x}\right) - \left(x^{n-1} + \dfrac{1}{x^{n-1}}\right),\)

which is an integer by the induction assumption. This completes the proof by strong induction.

Log in to reply

Nice one sir :)

Thanks for the proof :)

nice

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIt's probably easiest to use strong induction. The statement is given for \(n = 1.\) Now suppose \(x^{k} + x^{-k}\) is an integer for all \(k \le n.\) We have that

\(\left(x^{n} + \dfrac{1}{x^{n}}\right) \left(x + \dfrac{1}{x}\right) = x^{n+1} + \dfrac{1}{x^{n-1}} + x^{n-1} + \dfrac{1}{x^{n+1}}\)

\(\Longrightarrow x^{n+1} + \dfrac{1}{x^{n+1}} = \left(x^{n} + \dfrac{1}{x^{n}}\right)\left(x + \dfrac{1}{x}\right) - \left(x^{n-1} + \dfrac{1}{x^{n-1}}\right),\)

which is an integer by the induction assumption. This completes the proof by strong induction.

Log in to reply

Nice one sir :)

Thanks for the proof :)

Log in to reply

nice

Log in to reply