Question

If a, b and c are positive integers and x = cy + bz, y = az + cx, z = bx + ay, where x, y and z are not all zero, then number of ordered triplet (a, b, c) satisfying above is

(A) 0

(B) 1

(C) Finitely many

(D) Infinitely many

Answer 1

Answer is (A) 0

The given equations are x - cy - bz = 0, cx - y + az = 0 and bx + ay - z = 0.

It has non-trivial solutions so

⇒ 1(1 - a²) + c(- c - ab) - b(ac + b) = 0

⇒ 1 - 2abc = a² + b² + c² > 0

⇒ abc < 1/2

So, 0 < abc < 1/2.

Clearly, no triplet (a, b, c) of positive integers can satisfy it