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

Determine all pairs (r,s)(r,s) of integers which satisfy: r3+s3+3rs=53 r^3+s^3+3rs=53 .

#Algebra

Note by Tomás Carvalho
5 years, 2 months ago

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Comments

r3+s3+1=53+1−3rs\large r^3+s^3+\color{#D61F06}1 = 53+\color{#D61F06}1-3rs

  ⟹  r3+s3+1≥3rs  ⟹  rs≤546=9\large \implies r^3+s^3+1\ge 3rs \implies rs\le\frac{54}{6} = 9

By WLOG we assume r≥sr\ge s & we get s2≤rs≤9  ⟹  ∣s∣≤3  ⟹  −3≤s≤3s^2 \le rs \le 9 \implies |s|\le3 \implies -3\le s\le3

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

It is a symmetric equation in r,sr,s .

r2+s2≥2rs  ⟹  r2−rs+s2≥rs  ⟹  r3+s3≥rs(r+s)  ⟹  53≥rs(r+s+3)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,sr,s is -ve , so let s=−ms=-m

−rm(r−m+3)  ⟹  Since r is +ve ,r+3≥3  ⟹  r+3≥m Since maxs is 3-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≥0  ⟹  −rm(r−m+3)≤0≠53r-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)(1,2,3) we get (2,3),(3,2)(2,3),(3,2) are the only solutions.

Aditya Narayan Sharma - 5 years, 2 months ago

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

Tomás Carvalho - 5 years, 2 months ago

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

Aditya Narayan Sharma - 5 years, 2 months ago

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

Tomás Carvalho - 5 years, 2 months ago

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

Aditya Narayan Sharma - 5 years, 2 months ago
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