MHB Prove the given equation has no solution

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The equation $\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}=1$ is examined for solutions in natural numbers. Participants discuss the impossibility of finding $x$ and $y$ in the natural numbers that satisfy the equation. Key arguments include the behavior of the left-hand side as $x$ and $y$ increase, which approaches zero but never reaches one. The conclusion drawn is that no pairs of natural numbers can satisfy the equation, confirming it has no solution. The discussion emphasizes the mathematical reasoning behind this conclusion.
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$x,y\in N$
prove the equation :$\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}=1$
has no solution
 
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Hello, Albert!

$x,y\in N$
Prove the equation: $\,\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}=1\,$ has no solution.
Multiply by $x^2y^2\!:\;y^2 +xy + x^2 \:=\:x^2y^2$

Add $xy$ to both sides: $\:x^2 + 2xy + y^2 \:=\:x^2y^2+xy$

Factor: $\: (x+y)^2 \;=\;xy(xy+1)$

The left side is a square of an integer.
The right side is the product of two consecutive integers.

This is clearly impossible.
 
soroban said:
Hello, Albert!
Multiply by $x^2y^2\!:\;y^2 +xy + x^2 \:=\:x^2y^2$

Add $xy$ to both sides: $\:x^2 + 2xy + y^2 \:=\:x^2y^2+xy$

Factor: $\: (x+y)^2 \;=\;xy(xy+1)$

The left side is a square of an integer.
The right side is the product of two consecutive integers.

This is clearly impossible.
nice proof !
 
Albert said:
$x,y\in N$
prove the equation :$\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}=1---(1)$
has no solution
let $x>y>0$ then
$1=\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}<\dfrac{1}{y^2}+\dfrac{1}{y^2}+\dfrac{1}{y^2}=\dfrac {3}{y^2}$
$\therefore y^2<3 , or \,\, y=1$, put y=1 to (1)
we get :
$0=\dfrac {1}{x^2}+\dfrac {1}{x}$ it is impossible (since $x,y\in N$)
if $y>x>0$, the proof is the same
 
Last edited:
Albert said:
let $x>y>0$ then
$1=\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}<\dfrac{1}{y^2}+\dfrac{1}{y^2}+\dfrac{1}{y^2}=\dfrac {3}{y^2}$
$\therefore y^2<3 , or \,\, y=1$, put y=1 to (1)
we get :
$0=\dfrac {1}{x^2}+\dfrac {1}{x}$ it is impossible (since $x,y\in N$)
if $y>x>0$, the proof is the same

the part that was missed was

$x= y$
which gives
$1= \frac{1}{x^2} +\frac{1}{xy} + \frac{1}{y^2}= \frac{3}{x^2} $

or
$x^2 = 3$
which is impossible as it gives irrational x
so x = y is not possible
 
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