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Determine all pairs (r,s) of integers which satisfy

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

Note by Tomás Carvalho
5 months, 3 weeks ago

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

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

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. Aditya Sharma · 5 months, 3 weeks ago

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@Aditya Sharma why do you say that r^3+s^3+1 is bigger or equal to 3rs? Tomás Carvalho · 5 months, 3 weeks ago

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@Tomás Carvalho It's by AM-GM . Go through the inequalities section , the brilliant wiki's are a lot useful in this case. Aditya Sharma · 5 months, 3 weeks ago

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@Aditya Sharma Thanks! But the numbers can be negative. ? Tomás Carvalho · 5 months, 3 weeks ago

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@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 Aditya Sharma · 5 months, 3 weeks ago

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