SOLUTION: How many natural numbers x and y are there such that {{{1/sqrt(x)+1/sqrt(y) = 1/sqrt(20)}}}

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Question 1062344: How many natural numbers x and y are there such that 1%2Fsqrt%28x%29%2B1%2Fsqrt%28y%29+=+1%2Fsqrt%2820%29
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
How many natural numbers x and y are there such that 1%2Fsqrt%28x%29%2B1%2Fsqrt%28y%29+=+1%2Fsqrt%2820%29
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Answer. How many ? - Zero.
              In other words,  the given equation has no solutions in natural numbers.

Proof 

Let "x" and "y" be the solution. Then   (after multiplication both sides by sqrt%2820%29%2Asqrt%28x%29%2Asqrt%28y%29 ) you will get


sqrt%2820%29%2A%28sqrt%28x%29+%2B+sqrt%28y%29%29 = sqrt%28xy%29   --->       ( Divide both sides by sqrt%28x%29 )

sqrt%2820%29%2A%281+%2B+sqrt%28y%2Fx%29%29 = sqrt%28y%2Fx%29,   or         ( after introducing the new variable t = sqrt%28y%2Fx%29 )

sqrt%2820%29%2A%281+%2B+t%29 = t   --->   sqrt%2820%29+%2B+sqrt%2820%29%2At = t   --->  sqrt%2820%29 = %281-sqrt%2820%29%29%2At  --->  t = -sqrt%2820%29%2F%28sqrt%2820%29-1%29  --->  t = -%28sqrt%2820%29%2A%28sqrt%2820%29%2B1%29%29%2F%2820-1%29  --->  t = -%2820%2Bsqrt%2820%29%29%2F19.


Then from one side, t%5E2 = y%2Fx is a rational number.
From the other side,  %28-%2820%2Bsqrt%2820%29%29%2F19%29%5E2 is an irrational number.

Contradiction.

The contradiction proves the original statement.

The proof is completed.


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Comment from student: Thanks for sparing time for me. But one step of your answer is wrong. sqrt%28xy%29%2Fsqrt%28x%29=sqrt%28y%29+
But you accidentally wrote it as sqrt%28y%2Fx%29 That was the part you missed. Anyway thank you.
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My response. Thank you for your feedback. You are right, I was wrong. My mistake.

I just saw it from the solution by Edwin.

Thank you again, and thanks to Edwin for creating right solution, which is perfect.



Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


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%2Fsqrt%28x%29%2B1%2Fsqrt%28y%29+=+1%2Fsqrt%2820%29

1%2Fsqrt%28x%29%2B1%2Fsqrt%28y%29+=+1%2F%282sqrt%285%29%29

Let x=+5a%5E2, and y=5b%5E2

1%2Fsqrt%285a%5E2%29%2B1%2Fsqrt%285b%5E2%29+=+1%2F%282sqrt%285%29%29

1%2F%28a%2Asqrt%285%29%29%2B1%2F%28b%2Asqrt%285%29%29+=+1%2F%282sqrt%285%29%29

Multiply through by sqrt%285%29

1%2Fa%2B1%2Fb+=+1%2F2

2b%2B2a+=+ab

2%28b%2Ba%29=ab

2b%2B2a=ab

2a=ab-2b

2a=b%28a-2%29

%282a%29%2F%28a-2%29=b

b+=+%282a%29%2F%28a-2%29

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

If a=3, b+=+%282%2A3%29%2F%283-2%29=6%2F1=6

If a=4, b+=+%282%2A4%29%2F%284-2%29=8%2F2=4

If a=5, b+=+%282%2A5%29%2F%285-2%29=10%2F3  <--not an integer.

If a=6, b+=+%282%2A6%29%2F%286-2%29=12%2F4=3 

That's all the solutions which are integers,
because %282a%29%2F%28a-2%29 is a decreasing function
from b=3 approaching b=2 asymptotically.

Since x=+5%2Aa%5E2, and y=5%2Ab%5E2

Substituting a=3, b=6 x=+5%2A3%5E2=5%2A9=45, and y=5%2A6%5E2=5%2A36=180

Substituting a=4, b=4 x=+5%2A4%5E2=5%2A16=80, and y=5%2A4%5E2=5%2A16=80

Substituting a=6, b=3 x=+5%2A6%5E2=5%2A36=180, and y=5%2A3%5E2=5%2A9=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.

-------
Checking:

1. x = 45, y = 180

1%2Fsqrt%2845%29%2B1%2Fsqrt%28180%29+=+1%2Fsqrt%2820%29

1%2F%283sqrt%285%29%29%2B1%2F%286sqrt%285%29%29+=+1%2F%282sqrt%285%29%29

2%2F%28+6sqrt%285%29+%29%2B1%2F%28+6sqrt%28180%29+%29+=+1%2F%28+2sqrt%285%29+%29

3%2F%286sqrt%285%29%29+=+1%2F%282sqrt%285%29%29

1%2F%282sqrt%285%29%29+=+1%2F%282sqrt%285%29%29

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

1%2Fsqrt%2880%29%2B1%2Fsqrt%2880%29+=+1%2Fsqrt%2820%29

2%2Fsqrt%2880%29+=+1%2Fsqrt%2820%29

2%2F%284sqrt%285%29%29+=+1%2F%282sqrt%285%29%29

1%2F%282sqrt%285%29%29+=+1%2F%282sqrt%285%29%29

That shows that it is also a solution.

Edwin