Let x , y and z be all positive integers with x y = 2 1 , x z = 2 8 and y z = 1 2 . Find x 2 + y 2 + z 2 .
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The answer should be 74, not 33.
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Thanks. What had happened to me. Thirty-three must be the answer of previous solution I wrote.
Take the product of the equations to get ( x y ) ( x z ) ( y z ) = 2 1 ∗ 2 8 ∗ 1 2 = 3 2 ∗ 4 2 ∗ 7 2 , or x y z = 3 ∗ 4 ∗ 7 . x y x y z = 2 1 3 ∗ 4 ∗ 7 , z = 4 . Similarly, x = 7 and y = 3 . x 2 + y 2 + z 2 = 3 2 + 4 2 + 7 2 = 5 2 + 7 2 = 7 4
x z x y = z y = 4 3
y = 4 3 z
y z x y = z x = 4 7
x = 4 7 z
x + y + z = 4 7 z + 4 3 z + z = 2 7 z
x z × y z = x y z 2 = 3 3 6
x y x y z 2 = z 2 = 2 1 3 3 6 = 1 6
( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 x y + 2 x z + 2 y z = ( 2 7 z ) 2 = 1 9 6
x 2 + y 2 + z 2 + 2 x y + 2 x z + 2 y z = 1 9 6
x 2 + y 2 + z 2 + 2 ( 2 1 ) + 2 ( 2 8 ) + 2 ( 1 2 ) = 1 9 6
x 2 + y 2 + z 2 + 4 2 + 5 6 + 2 4 = 1 9 6
x 2 + y 2 + z 2 = 7 4
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x y ⋅ x z x 2 y z 1 2 x 2 x 2 ⟹ x ⟹ y ⟹ z = 2 1 ⋅ 2 8 = 2 1 ⋅ 2 8 = 2 1 ⋅ 2 8 = 1 2 2 1 ⋅ 2 8 = 7 2 = 7 = x 2 1 = 3 = x 2 8 = 4
⟹ x 2 + y 2 + z 2 = 4 9 + 9 + 1 6 = 7 4