# some positive integer pairs(x,y) satisfy $\frac{1}{ \sqrt{x}}+\frac{1}{ \sqrt{y}}=\frac{1}{ \sqrt{20}}$How many values of $xy$ are there?

by Tyrone   Last Updated August 14, 2019 15:20 PM

There are some positive integer pairs(x,y) that satisfy $$\frac{1}{ \sqrt{x}}+\frac{1}{ \sqrt{y}}=\frac{1}{ \sqrt{20}}$$ How many different possible values of the product x and y are there?

I found one value when $$x=y$$

$$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{20}}$$

which simplifies to

$$\frac{2\sqrt{x}}{x}=\frac{1}{\sqrt{20}}$$

which gives $$x=y=80$$

therefore one possible value of the product is $$80^2$$

but I don't know how to prove this is the only solution or if there are more

suggestions, help, and solutions would all be appreciated.

taken from the 2019 IWYMIC/SAIMC(Question 7) https://chiuchang.org/imc/wp-content/uploads/sites/2/2019/08/SAIMC-2019_Keystage-3_Individual_Final.x17381.pdf

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Assume firstly that $$xy\neq 0$$, then

$$2\sqrt{5}\left(\sqrt{x}+\sqrt{y}\right)=\sqrt{xy}$$ or to: $$(\sqrt{x}-\sqrt{20})(\sqrt{y}-\sqrt{20})=2\sqrt{5} \tag{*}$$ or, assuming $$x=5p^2,y=5q^2$$, $$(p-2)(q-2)=2.$$ Since $$2$$ is a prime, then solutions are given by $$(0,1),(1,0),(3,4)$$ and $$(4,3)$$. Now show that $$x=5p^2,y=5q^2$$ is enought to fulfill $$(*)$$.

Kevin
August 14, 2019 14:49 PM

I assume you want pairs of x and y that are equal and satisfy the equation.

Set $$k= \sqrt{20}$$ $$\frac{1}{ \sqrt{x}}+\frac{1}{ \sqrt{y}}=\frac{1}{ \sqrt{20}}$$

to get:

$$\frac{1}{ \sqrt{x}}+\frac{1}{ \sqrt{y}}=k$$

re-arrange to get:

$$y=\frac{1}{(k-\frac{1}{\sqrt(x)})^{2}}$$

When $$x=y$$ the above equation becomes:

$$x (k-\frac{1}{\sqrt(x)})^{2}=1$$

That is: $$(k\sqrt(x)-1)^2=1$$

Taking the square root of both sides, We have 2 cases:

$$k\sqrt(x)-1 =1$$ This implies that $$x=0$$ This may be rejected.

The other case is:

$$k\sqrt(x)-1 =1$$

$$k\sqrt(x) =2$$

That is:

$$k^2(x)=4$$ and $$x=4/k^2$$ $$x=80$$

I don't see other cases!

Desmos Graph

NoChance
August 14, 2019 15:14 PM

First, we have $$\frac{1}{\sqrt x} - \frac{1}{\sqrt{20}} = -\frac{1}{\sqrt y}$$, so $$\frac{20 + x - 2\sqrt{5 x}}{20x} = \frac{1}{y}$$. As $$x$$ and $$y$$ are rational, so is $$\sqrt{5x}$$, thus $$x = 5\cdot a^2$$ for some integer $$a$$. Similarly $$y = 5\cdot b^2$$. Substituting it into original equation, we get $$\frac{1}{a} + \frac{1}{b} = \frac{1}{2}$$ $$2a + 2b = ab$$ $$(a - 2)(b - 2) = 4$$

As $$4 = 1 \cdot 4 = 2 \cdot 2 = 4 \cdot 1$$ are the only decompositions of $$4$$ to product of positive integers, we have variants $$a - 2 = 1, b - 2 = 4$$, $$a - 2 = 2, b - 2 = 2$$, $$a - 2 = 4, b - 2 = 1$$. They correspond to $$x = 45, y = 180$$, $$x = y = 80$$ and $$x = 180, y = 45$$.

mihaild
August 14, 2019 15:16 PM