Prove that that there are no positive integers x and y such that: 1/x^2 + 1/xy + 1/y^2 = 1.

We'll begin from the assumption that x, y>0 and
x,y`in` Z

Now, we'll calculate the least common denominator
of the fractions:

LCD =
x^2*y^2

Now, we'll multiply each fraction by the needed
value, in order to get x^2*y^2 at denominator.

y^2/x^2*y^2
+ x*y/x^2*y^2 + x^2/x^2*y^2 = 1

(y^2 + xy + x^2)/x^2*y^2 =
1

We'll cross multiply and we'll
get:

y^2 + xy + x^2 =
x^2*y^2

We'll multiply (x-y) both
sides:

(x-y)(y^2 + xy + x^2) =
(x-y)*x^2*y^2

We'll get to the left a difference of
cubes:

x^3 - y^3 = x^3*y^2 -
x^2*y^3

Suppose that
x=y=1

1-1=1-1=0

Suppose
x=y=2

8-8 = 8*4 - 4*8 =
0

Suppose x = 2 and y = 3

x^3
- y^3 = 8-27 = -19

x^3*y^2 - x^2*y^3 = 8*9 - 4*27 =
-36

We notice that if the values of x and y
are equal positive integres, the given relation represents an identity, therefore 1/x^2
+ 1/xy + 1/y^2 = 1, while if any x `!=` y>0, x,y`in` Z, then the given expression
is not an identity.