Prove the given equation has no solution

[Albert1]

$x,y\in N$
prove the equation :$\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}=1$
has no solution

 

[soroban]

Hello, Albert!

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

Spoiler

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.

 

[Albert1]

Hello, Albert!

Spoiler

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 !

 

[Albert1]

$x,y\in N$
prove the equation :$\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}=1---(1)$
has no solution

Spoiler

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

 

[kaliprasad]

Spoiler

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

Spoiler

$x= y$
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