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

## Squares and square roots – HANOI 2018

Let a=\((\sqrt2+\sqrt3+\sqrt6)(\sqrt2+\sqrt3-\sqrt6)(\sqrt3+\sqrt6-\sqrt2)(\sqrt6+\sqrt2-\sqrt3)\)

b=\((\sqrt2+\sqrt3+\sqrt5)(\sqrt2+\sqrt3-\sqrt5)(\sqrt3+\sqrt5-\sqrt2)(\sqrt5+\sqrt2-\sqrt3)\). The difference a-b belongs to the set

- is [-4,0)
- is {6}
- is [-8,-6]
- cannot be determined from the given information

**Key Concepts**

Algebra

Squares and square roots

Number Theory

## Check the Answer

But try the problem first…

Answer: is [-4,0).

Source

Suggested Reading

HANOI, 2018

Elementary Number Theory by David Burton

## Try with Hints

First hint

(x+y+z)(x+y-z)(x-y+z)(-x+y+z)=2\((x^{2}y{2}+y^{2}z^{2}+z^{2}x^{2})-x^{4}-y^{4}-z^{4}\).

Second Hint

We get a-b=2(2+3)(6-5)-\(6^{2}+5^{2}\)=-1.

Final Step

Then a-b belongs to [-4,0).

## Other useful links

- https://www.cheenta.com/largest-hexagon-in-equilateral-triangle-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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