# 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?

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

• As a very loose hint, x and y are very constrained; for instance, can you convince yourself that x=60 is impossible? Can you understand why it is? – Steven Stadnicki Aug 14 '19 at 14:47
• Do you want to find all values for x,y such that x=y and the equation is satisfied? – NoChance Aug 14 '19 at 14:59
• Thanks for the members who commented on my solution. The requirement is a bit vague to me! – NoChance Aug 14 '19 at 15:22

First, we have $$\frac{1}{\sqrt x} - \frac{1}{\sqrt{20}} = -\frac{1}{\sqrt y}$$, so $$\frac{20 + x - 4\sqrt{5 x}}{20x} = \frac{1}{y}$$. 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 $$x = 180, y = 45$$.

• shouldn't it be $\frac{20+x-2\sqrt{20x}}{20x}=\frac{1}{y}$? – Tyrone Aug 14 '19 at 17:06
• Yes, thanks. Fixed now. – mihaild Aug 14 '19 at 18:08

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 $$(*)$$.

• Wouldn't $$(\sqrt{x}-\sqrt{20})(\sqrt{y}-\sqrt{20}) = 20 \neq 2\sqrt{5}$$ – InterstellarProbe Aug 14 '19 at 14:58
• @InterstellarProbe: indeed. I just came to that conclusion while checking why my alternate solution didn't match – robjohn Aug 14 '19 at 15:04

$$x=\dfrac{20y}{(y+20-4\sqrt{5y}}$$

$$y=5c^2$$

$$x=\dfrac{100c^2}{5c^2+20-20c}=\dfrac{20c^2}{(c-2)^2}$$

Let $$c-2=d,x=20(d+2)^2/d^2$$

As $$(d,d+2)$$ divides $$d+2-d=2$$

If $$d$$ is odd, $$(d+2,d)=1,d$$ must divide $$20\implies d=1$$

What if $$d$$ is even $$=2e$$ (say)

Letting $$x=\frac{20}{r^2}$$, $$y=\frac{20}{s^2}$$ with $$0 \lt r \lt 1$$, $$0 \lt s \lt 1$$ the equation becomes $$r + s = 1$$ or $$s=1-r$$.

Hence we have $$x= \frac{20}{r^2}$$ and $$y = \frac{20}{(1-r)^2}$$

Now letting $$r = \frac{p}{q}$$ with $$GCD(p,q)=1$$ we get $$x = 20 \frac{q^2}{p^2}$$. Hence $$p^2|20$$ which gives $$p=1$$ or $$p=2$$.

From $$y=20 \frac{q^2}{(q-p)^2}$$ we get

for $$p=1$$: $$y=20 \frac{q^2}{(q-1)^2}$$ hence $$(q-1)^2|20$$ giving $$q=2$$ or $$q=3$$. This in turn gives $$(p,q) = (1,2) \to (x = 20*2^2 = 80, y = 20*2^2 = 80)$$ and $$(p,q) = (1,3) \to (x = 20*3^2 = 180, y = 20*3^2/4 = 45)$$

for $$p=2$$: $$y = 20 \frac{q^2}{(q-2)^2}$$ hence $$(q-2)^2|20$$ giving $$q = 3$$ or $$q=4$$. The latter is ruled out by $$GCD(p,q)=1$$ hence we obtain

$$(p,q) = (2,3) \to (x = 5 q^2 = 45, y = 20 \frac{3^2}{(3-2)^2} = 180)$$

Summarising we have obtained two essentially different solutions $$(x,y) = (80,80)$$ and $$(x,y) = (45,180)$$.

Remark: it remains to show that the rational ansatz for $$r$$ is the most general.