Problem No. 45
How many triples ( x , y , z ) of positive integers satisfy the equation
x y z y z x z x y = 3 x y z
The answer is 3.
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1 solution
( 1 , 2 , 3 ) , ( 2 , 3 , 1 ) , ( 3 , 1 , 2 ) are not solutions. Also, you can't assume wlog x ≤ y ≤ z , since the equation is cyclic, not symmetric. The "correct" answer is wrong. The actual answer is 3 . All the solutions are ( 1 , 3 , 2 ) , ( 2 , 1 , 3 ) , ( 3 , 2 , 1 ) .
Correct solution: The equation is cyclic. Two cases: if wlog x = 1 , then ( x , y , z ) = ( 1 , 3 , 2 ) . Otherwise, x , y , z ≥ 2 and 3 x y z ≥ x 4 y 4 z 4 ⟺ 3 ≥ x 3 y 3 z 3 , impossible. Answer: ( x , y , z ) = ( 1 , 3 , 2 ) , ( 2 , 1 , 3 ) , ( 3 , 2 , 1 ) are all the solutions.
Assume WLOG that x ≤ y ≤ z . Case 1 (x=1): y z = 3 y ⇒ y = 3 z = 2 This triple can be arranged in 3! = 6 ways leaving 6 triples. Case 2 (x=2): 2 y z . y z 2 . z 2 y = 6 y z Now y = 2 suggests that z is a multiply of 3 which produces no solutions (as LHS > RHS) and y = 3 produces 8 z . 3 z 2 . z 7 = 1 8 so there are no solution here for the same reason. Now as y gets larger the LHS gets larger faster than the RHS.
Case 3 ( x ≥ 3 ): x y z − 1 . y z x − 1 . z x y − 1 = 3 ⇒ y z − 1 ≤ 1 ⇒ y z ≤ 2 This contradicts the assumption that x ≤ y ≤ z . Hence, there are only 3 triples given by (1,3,2) ,(2,1,3) and (3,1,2) (3 of the 6 arrangements are not valid).