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No Distinct Rationals with Sum of 1/(x-y)² Equal to 2014 | IOQM

JEE Maths question with a full step-by-step solution.

Question

The number of pairwise distinct rational numbers

x, y, z

such that

\frac{1}{(x - y)^2} + \frac{1}{(y - z)^2} + \frac{1}{(z - x)^2} = 2014\,

Solution

Answer: 0

Step 1: Let

u = \dfrac{1}{x - y}

v = \dfrac{1}{y - z}

w = \dfrac{1}{z - x}

. The left side is

u^2 + v^2 + w^2

. Step 2: Compute the sum of pairwise products.

uv + vw + wu = \frac{(z - x) + (x - y) + (y - z)}{(x - y)(y - z)(z - x)} = 0,

because the numerator

(z - x) + (x - y) + (y - z) = 0

. Step 3: Hence

u^2 + v^2 + w^2 = (u + v + w)^2 - 2(uv + vw + wu) = (u + v + w)^2.

Step 4: The equation becomes

(u + v + w)^2 = 2014

, so

\frac{1}{x - y} + \frac{1}{y - z} + \frac{1}{z - x} = \pm\sqrt{2014}.

Step 5: If

x, y, z

were all rational, the left side would be rational. But

\sqrt{2014}

is irrational (

2014

is not a perfect square), a contradiction. Hence no such pairwise distinct rationals exist. Answer: No, such rational numbers do not exist.

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