    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
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B

2 does not divide z(x + y)

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C

4 divides x + y + z

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D

8 does not divide x + y + z

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Solution

## 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.  Suggest Corrections  1      Similar questions  Related Videos   AM and GM
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