# 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
2 years, 4 months 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.

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago

Thanks! But the numbers can be negative. ?

- 2 years, 4 months ago

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

- 2 years, 4 months ago