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

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beckden | High School Teacher | (Level 1) Educator

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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.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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