Question 1062344
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How many natural numbers x and y are there such that {{{1/sqrt(x)+1/sqrt(y) = 1/sqrt(20)}}}
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<U>Answer</U>.  How many ? - Zero.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In other words, &nbsp;the given equation has no solutions in natural numbers. 


<U>Proof</U>


<pre>
Let "x" and "y" be the solution. Then   (after multiplication both sides by {{{sqrt(20)*sqrt(x)*sqrt(y)}}} ) you will get


{{{sqrt(20)*(sqrt(x) + sqrt(y))}}} = {{{sqrt(xy)}}}   --->       ( Divide both sides by {{{sqrt(x)}}} )

{{{sqrt(20)*(1 + sqrt(y/x))}}} = {{{sqrt(y/x)}}},   or         ( after introducing the new variable t = {{{sqrt(y/x)}}} )

{{{sqrt(20)*(1 + t)}}} = t   --->   {{{sqrt(20) + sqrt(20)*t}}} = t   --->  {{{sqrt(20)}}} = {{{(1-sqrt(20))*t}}}  --->  t = {{{-sqrt(20)/(sqrt(20)-1)}}}  --->  t = {{{-(sqrt(20)*(sqrt(20)+1))/(20-1)}}}  --->  t = {{{-(20+sqrt(20))/19}}}.


Then from one side, {{{t^2}}} = {{{y/x}}} is a rational number.
From the other side,  {{{(-(20+sqrt(20))/19)^2}}} is an irrational number.

Contradiction.

The contradiction proves the original statement.

The proof is completed.
</pre>


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<U>Comment from student</U>: Thanks for sparing time for me. But one step of your answer is wrong. {{{sqrt(xy)/sqrt(x)=sqrt(y) }}} 
But you accidentally wrote it as {{{sqrt(y/x)}}} That was the part you missed. Anyway thank you.
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<U>My response</U>.  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 <U>perfect</U>.