Question 1062344
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There are 2 solutions in integers.  [3 if you count 
interchanging the x and y values on one of them as 
a separate solution.]

{{{1/sqrt(x)+1/sqrt(y) = 1/sqrt(20)}}}

{{{1/sqrt(x)+1/sqrt(y) = 1/(2sqrt(5))}}}

Let {{{x= 5a^2}}}, and {{{y=5b^2}}}

{{{1/sqrt(5a^2)+1/sqrt(5b^2) = 1/(2sqrt(5))}}}

{{{1/(a*sqrt(5))+1/(b*sqrt(5)) = 1/(2sqrt(5))}}}

Multiply through by {{{sqrt(5)}}}

{{{1/a+1/b = 1/2}}}

{{{2b+2a = ab}}}

{{{2(b+a)=ab}}}

{{{2b+2a=ab}}}

{{{2a=ab-2b}}}

{{{2a=b(a-2)}}}

{{{(2a)/(a-2)=b}}}

{{{b = (2a)/(a-2)}}}

The smallest integer for "a" that will give
a positive value for b is a=3

If {{{a=3}}}, {{{b = (2*3)/(3-2)=6/1=6}}}

If {{{a=4}}}, {{{b = (2*4)/(4-2)=8/2=4}}}

If {{{a=5}}}, {{{b = (2*5)/(5-2)=10/3}}}  <--not an integer.

If {{{a=6}}}, {{{b = (2*6)/(6-2)=12/4=3}}} 

That's all the solutions which are integers,
because {{{(2a)/(a-2)}}} is a decreasing function
from b=3 approaching b=2 asymptotically.

Since {{{x= 5*a^2}}}, and {{{y=5*b^2}}}

Substituting a=3, b=6 {{{x= 5*3^2=5*9=45}}}, and {{{y=5*6^2=5*36=180}}}

Substituting a=4, b=4 {{{x= 5*4^2=5*16=80}}}, and {{{y=5*4^2=5*16=80}}}

Substituting a=6, b=3 {{{x= 5*6^2=5*36=180}}}, and {{{y=5*3^2=5*9=45}}}

There are 2 solutions, (x,y) = (45,180), (x,y) = (80,80)
There are 3 if you count (x,y) = (180,45) as a separate solution
from the first one.

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Checking:

1. x = 45, y = 180

{{{1/sqrt(45)+1/sqrt(180) = 1/sqrt(20)}}}

{{{1/(3sqrt(5))+1/(6sqrt(5)) = 1/(2sqrt(5))}}}

{{{2/( 6sqrt(5) )+1/( 6sqrt(180) ) = 1/( 2sqrt(5) )}}}

{{{3/(6sqrt(5)) = 1/(2sqrt(5))}}}

{{{1/(2sqrt(5)) = 1/(2sqrt(5))}}}

That shows that it is a solution.
  
x = 80, y = 80

{{{1/sqrt(80)+1/sqrt(80) = 1/sqrt(20)}}}

{{{2/sqrt(80) = 1/sqrt(20)}}}

{{{2/(4sqrt(5)) = 1/(2sqrt(5))}}}

{{{1/(2sqrt(5)) = 1/(2sqrt(5))}}}

That shows that it is also a solution.

Edwin</pre></b></font>