Algebra (Olympiad)hardFree

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,zx, y, z such that
1(x−y)2+1(y−z)2+1(z−x)2=2014 \frac{1}{(x - y)^2} + \frac{1}{(y - z)^2} + \frac{1}{(z - x)^2} = 2014\,
Solution
Answer: 0
Step 1: Let u=1x−yu = \dfrac{1}{x - y}, v=1y−zv = \dfrac{1}{y - z}, w=1z−xw = \dfrac{1}{z - x}. The left side is u2+v2+w2u^2 + v^2 + w^2. Step 2: Compute the sum of pairwise products.
uv+vw+wu=(z−x)+(x−y)+(y−z)(x−y)(y−z)(z−x)=0,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(z - x) + (x - y) + (y - z) = 0. Step 3: Hence
u2+v2+w2=(u+v+w)2−2(uv+vw+wu)=(u+v+w)2.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(u + v + w)^2 = 2014, so
1x−y+1y−z+1z−x=±2014.\frac{1}{x - y} + \frac{1}{y - z} + \frac{1}{z - x} = \pm\sqrt{2014}.
 Step 5: If x,y,zx, y, z were all rational, the left side would be rational. But 2014\sqrt{2014} is irrational (20142014 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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