Algebra (Olympiad)hardFree

Algebra (Olympiad): Number Real Values Satisfying Equations

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

Question
The number of real values of x,y,zx, y, z for satisfying the equations
x2−y=z−1,y2−z=x−1,z2−x=y−1.\sqrt{x^2 - y} = z - 1, \qquad \sqrt{y^2 - z} = x - 1, \qquad \sqrt{z^2 - x} = y - 1.
Solution
Answer: 1
Step 1: Each square root is non-negative, so z≥1z \ge 1, x≥1x \ge 1, y≥1y \ge 1. Squaring each equation,
x2−y=z2−2z+1,y2−z=x2−2x+1,z2−x=y2−2y+1.x^2 - y = z^2 - 2z + 1, \qquad y^2 - z = x^2 - 2x + 1, \qquad z^2 - x = y^2 - 2y + 1.
 Step 2: Add the three equations.
(x2+y2+z2)−(x+y+z)=(x2+y2+z2)−2(x+y+z)+3.(x^2 + y^2 + z^2) - (x + y + z) = (x^2 + y^2 + z^2) - 2(x + y + z) + 3.
 Step 3: Simplify.
−(x+y+z)=−2(x+y+z)+3  ⟹  x+y+z=3.-(x + y + z) = -2(x + y + z) + 3 \implies x + y + z = 3.
 Step 4: Rewrite as (x−1)+(y−1)+(z−1)=0(x - 1) + (y - 1) + (z - 1) = 0. Since each of x−1,y−1,z−1x - 1, y - 1, z - 1 is non-negative, the sum is zero only if each is zero, that is x=y=z=1x = y = z = 1. Answer: (x,y,z)=(1,1,1)(x, y, z) = (1, 1, 1).
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