Question

Let x, y and z be positive integers such that GCD(x, y, z) =1: x<y<z and x²+y²=z². Then which of the following is always true?

• A. 2 does not divide x
• B. 2 does not divide z(x + y)
• C. 4 divides x + y + z
• D. 8 does not divide x + y + z

Answer 1

The correct option is B

2 does not divide z(x + y)

Let x, y and z are 2n, n²- 1 & n² +1.

Let x = 2n, y = n² - 1 and z = n² +1. Option (A) : 2 does not divide x . Clearly 2 divides x. So option (A) is not true. Ex : 8, 15 and 17 are pythogorean triplet. 2 divides 8. Option (B) : 2 does not divide z(x + y) (n² + 1)(2n + n² -1) which is an odd number so, 2 does not divide z(x + y). ∴So, option (B) is true. Option (C) : 4 divides x + y + z Ex : 5, 12, 13 are the pythogorean triplet. Clearly 5 + 12 + 13 = 30, and 4 does not divides 30. ∴So, option (C) is not true. Option (D) : 8 divides x + y + z Ex : 8, 15, 17 are the pythogorean triplet. Clearly 8 + 15 + 17 = 40, and 8 divides 40.

∴So, option (B) is true.