determine all natural numbers x, y, z for which the equation 4x^2+45y^2+9z^2-12xy-36yz=25 is valid and where x<y<z
Answers
Given : 4x²+45y²+9z²-12xy-36yz=25
x<y<z
To Find : determine all natural numbers x, y, z
Solution:
4x² +45y²+9z²-12xy-36yz=25
=> 4x² -12xy + 9y² + 36y²+9z² - 36yz=25
=> (2x - 3y)² + ( 6y - 3z)² = 25
as we know (±3)² + (±4) ² = 25 or (±5)² + 0² = 25
2x - 3y = 0 is not possible as x < y
Hence 2x - 3y = - 5 as y > x
x = 2 , y = 3 , x = 5 , y = 5 is not feasible and no further possible
6y - 3z = 0 => 6(3) - 3z = 0 => z = 6
x = 2 , y = 3 and z = 6 is one of the solution
(2x - 3y)² + ( 6y - 3z)² = 25
taking
2x - 3y = - 3 => x = 3 , y = 3 or x = 6 , y = 5 hence no feasible solution
2x - 3y = - 4 => x = 1 , y = 2 is only feasible solution
6y - 3z = ± 3
=> 12 - 3z = ± 3
=> 3z = 12 ± 3 => z = 3 , 5
( x , y , z)
Hence ( 1 , 2 , 3) , ( 1, 2 , 5) and ( 2 , 3 , 6) are the required solution
Lets verify
4x² +45y²+9z²-12xy-36yz=25
( 1 , 2 , 3)
=> 4 + 180 + 81 - 24 - 216 = 25 verified
( 1, 2 , 5)
=> 4 + 180 + 225 - 24 - 360 = 25 verified
( 2 , 3 , 6)
16 + 405 + 324 - 72 - 648 = 25 verified
Hence ( 1 , 2 , 3) , ( 1, 2 , 5) and ( 2 , 3 , 6) are the required solution
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