New user? Sign up

Existing user? Sign in

Determine all pairs \((r,s)\) of integers which satisfy: \( r^3+s^3+3rs=53 \).

Note by Tomás Carvalho 1 year, 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:

\(\large r^3+s^3+\color{red}1 = 53+\color{red}1-3rs\)

\(\large \implies r^3+s^3+1\ge 3rs \implies rs\le\frac{54}{6} = 9\)

By WLOG we assume \(r\ge s\) & we get \(s^2 \le rs \le 9 \implies |s|\le3 \implies -3\le s\le3\)

Since having \(r\ge s\ge-3\) we bound \((r,s) \in [-3,3] - {{0}}\)

It is a symmetric equation in \(r,s\) .

\(r^2+s^2\ge 2rs \implies r^2-rs+s^2\ge rs \implies r^3 + s^3\ge rs(r+s) \implies 53\ge rs(r+s+3)\)

Now suppose exactly one of \(r,s\) is -ve , so let \(s=-m\)

\(-rm(r-m+3)\implies \text{Since r is +ve ,} r+3\ge 3\implies r+3\ge m \text{ Since max{s} is 3} \)

This clearly implies that \(r-m+3\ge 0 \implies -rm(r-m+3)\le 0 \ne 53\) which is a positive quantity.

So checking possiblities with \((1,2,3)\) we get \((2,3),(3,2)\) are the only solutions.

Log in to reply

why do you say that r^3+s^3+1 is bigger or equal to 3rs?

It's by AM-GM . Go through the inequalities section , the brilliant wiki's are a lot useful in this case.

@Aditya Narayan Sharma – Thanks! But the numbers can be negative. ?

@Tomás Carvalho – Well inequalities don't involve negative numbers, and in this case too the range of \(r,s\) is [-3,3] , but only positive cases satisfy the relation so they are only solutions as I hve proved

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:

TopNewest\(\large r^3+s^3+\color{red}1 = 53+\color{red}1-3rs\)

\(\large \implies r^3+s^3+1\ge 3rs \implies rs\le\frac{54}{6} = 9\)

By WLOG we assume \(r\ge s\) & we get \(s^2 \le rs \le 9 \implies |s|\le3 \implies -3\le s\le3\)

Since having \(r\ge s\ge-3\) we bound \((r,s) \in [-3,3] - {{0}}\)

It is a symmetric equation in \(r,s\) .

\(r^2+s^2\ge 2rs \implies r^2-rs+s^2\ge rs \implies r^3 + s^3\ge rs(r+s) \implies 53\ge rs(r+s+3)\)

Now suppose exactly one of \(r,s\) is -ve , so let \(s=-m\)

\(-rm(r-m+3)\implies \text{Since r is +ve ,} r+3\ge 3\implies r+3\ge m \text{ Since max{s} is 3} \)

This clearly implies that \(r-m+3\ge 0 \implies -rm(r-m+3)\le 0 \ne 53\) which is a positive quantity.

So checking possiblities with \((1,2,3)\) we get \((2,3),(3,2)\) are the only solutions.

Log in to reply

why do you say that r^3+s^3+1 is bigger or equal to 3rs?

Log in to reply

It's by AM-GM . Go through the inequalities section , the brilliant wiki's are a lot useful in this case.

Log in to reply

Log in to reply

Log in to reply