A walk in the park.

Level 2

Given that $x$ and $y$ are positive integers, find $\sqrt[3]{\dfrac {x^2+y^2+6}{xy}} \in \mathbb Z$ .

The answer is 2.

1 solution

Chew-Seong Cheong
Dec 22, 2017

Let the number given be $N$ . Then $N = \sqrt[3]{\dfrac {x^2+y^2+6}{xy}} = \sqrt[3]{\dfrac xy+\dfrac yx + \dfrac 6{xy}}$ . For $N$ to be a positive integer $\dfrac xy$ , $\dfrac yx$ and $\dfrac 6{xy}$ must all be positive integers. The only possible case is $x=y$ and $N = \sqrt[3]{1+1+\dfrac 6{xy}} = \sqrt[3]{2+\dfrac 6{x^2}}$ . The only possible case is $x=y=1$ , then $N=\sqrt[3]{8} = \boxed{2}$ .

We don't necessarily require that $\dfrac{x}{y}, \dfrac{y}{x}$ and $\dfrac{6}{xy}$ all be integers. We could have $(x,y)$ being either $(1,7)$ or $(7,1)$ , both of which would still yield $N = 2$ .

Brian Charlesworth - 3 years, 5 months ago

Thanks, pal.

Chew-Seong Cheong - 3 years, 5 months ago

If that isn't the right solution, then what is? I want to know!

Thành Đạt Lê - 3 years, 5 months ago

@Thành Đạt Lê – I am trying to work it out.

Chew-Seong Cheong - 3 years, 5 months ago

@Chew-Seong Cheong – Oh, okay! I'll wait.

Thành Đạt Lê - 3 years, 5 months ago

A walk in the park.

The answer is 2.

1 solution

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