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Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on **Squares and inequality.**

## Squares and inequality – HANOI 2018

Write down all real numbers (x,y) satisfying two conditions \(x^{2018}+y^{2}=2\) and \(x^{2}+y^{2018}=2\).

- is [-1,0)
- is (0,1),(-1,0)
- is (-1,-1),(-1,1),(1,-1),(1,1)
- cannot be determined from the given information

**Key Concepts**

Algebra

Squares and square roots

Inequality

## Check the Answer

But try the problem first…

Answer: is (-1,-1),(-1,1),(1,-1),(1,1).

Source

Suggested Reading

HANOI, 2018

Inequalities (Little Mathematical Library) by Korovkin

## Try with Hints

First hint

If \(x^{2}>1\) then\(x^{2018}>x^{2}>1\) and \(y^{2}<1\) implies that \(y^{2} \gt y^{2018}\) Then \(x^{2018}+y^{2} \gt x^{2}+y^{2018}\) (contradiction) .

Second Hint

Analogically, if \(x^{2} \lt 1\) implies that \(x^{2018}+y^{2} \lt x^{2}+y^{2018}\)(contradiction).

Final Step

Then \(x^{2}=y^{2}=1\).

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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