Algebra (Olympiad)hardFree

Solve the Symmetric System for All Triples (x,y,z) | IOQM

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

Question
The total number of all ordered triples (x,y,z)(x, y, z) of real numbers such that
x2−yz+xy+zx=82,y2−zx+xy+yz=−18,z2−xy+yz+zx=18.x^2 - yz + xy + zx = 82, \qquad y^2 - zx + xy + yz = -18, \qquad z^2 - xy + yz + zx = 18.
Solution
Answer: 4
Label the equations (i), (ii), (iii) in order. Step 1: Add (ii) and (iii). The cross terms cancel, leaving
y2+z2+2yz=(y+z)2=−18+18=0  ⟹  z=−y....(iv)y^2 + z^2 + 2yz = (y + z)^2 = -18 + 18 = 0 \implies z = -y. \quad \text{...(iv)}
 Step 2: Add (i) and (iii):
x2+z2+2zx=(x+z)2=82+18=100  ⟹  x+z=±10....(v)x^2 + z^2 + 2zx = (x + z)^2 = 82 + 18 = 100 \implies x + z = \pm 10. \quad \text{...(v)}
 Step 3: Add (i) and (ii):
x2+y2+2xy=(x+y)2=82−18=64  ⟹  x+y=±8....(vi)x^2 + y^2 + 2xy = (x + y)^2 = 82 - 18 = 64 \implies x + y = \pm 8. \quad \text{...(vi)}
 Step 4: Using z=−yz = -y from (iv), equation (v) becomes x−y=±10x - y = \pm 10 and (vi) is x+y=±8x + y = \pm 8. Solving the four sign combinations gives
(x,y,z)=(9,−1,1), (1,−9,9), (−1,9,−9), (−9,1,−1).(x, y, z) = (9, -1, 1),\ (1, -9, 9),\ (-1, 9, -9),\ (-9, 1, -1).
 Answer: (x,y,z)=(9,−1,1), (1,−9,9), (−1,9,−9), (−9,1,−1)(x, y, z) = (9, -1, 1),\ (1, -9, 9),\ (-1, 9, -9),\ (-9, 1, -1).
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