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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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    $\begingroup$ 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? $\endgroup$ – Steven Stadnicki Aug 14 '19 at 14:47
  • $\begingroup$ Do you want to find all values for x,y such that x=y and the equation is satisfied? $\endgroup$ – NoChance Aug 14 '19 at 14:59
  • $\begingroup$ Thanks for the members who commented on my solution. The requirement is a bit vague to me! $\endgroup$ – NoChance Aug 14 '19 at 15:22
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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 $x = 180, y = 45$.

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

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    $\begingroup$ Wouldn't $$(\sqrt{x}-\sqrt{20})(\sqrt{y}-\sqrt{20}) = 20 \neq 2\sqrt{5}$$ $\endgroup$ – InterstellarProbe Aug 14 '19 at 14:58
  • $\begingroup$ @InterstellarProbe: indeed. I just came to that conclusion while checking why my alternate solution didn't match $\endgroup$ – robjohn Aug 14 '19 at 15:04
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$x=\dfrac{20y}{(y+20-4\sqrt{5y}}$

$y=5c^2$

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

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)

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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.

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