2 Answers | Add Yours
1/x^2 + 1/xy + 1/y^2 = 1 Multiply both sides by x^2y^2 to get rid of fractions
y^2 + xy + x^2 = x^2y^2
y^2 + 2xy + x^2 = x^2y^2 + xy
(x+y)^2 = xy(xy + 1)
If x and y are integers then xy = n is an integer.
n(n+1) = n^2 + n which is never a perfect square. The distance between n^2 and (n+1)^2 is 2n +1. so n^2 < n(n+1) < (n+1)^2 for all integers n > 0. So it is impossible for (x + y)^2 = xy(xy+1) for any integers x, y > 0.
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
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.
We’ve answered 319,200 questions. We can answer yours, too.Ask a question