what is the smallest value that x can have?

2008-07-31 1:57 am
consider the solutions (x,y,z,u) of the system of equations
x+y = 3 (z+u)
x+z = 4 (y+u)
x+u = 5 (y+z)
where x,y,z,u are positive integers. what is the smallest value that x can have?

回答 (1)

2008-07-31 5:53 am
✔ 最佳答案
Rewrite the system of equations as
3u - y + 3z = x
4u + 4y - z = x
-u + 5y + 5z = x.
Note that the determinants
  |3 -1 3|
Δ = |4 4 -1|
  |-1 5 5|
 = [(3)(4)(5) + (-1)(-1)(-1) + (3)(4)(5)] - [(3)(4)(-1) + (3)(-1)(5) + (-1)(4)(5)]
 = 166,
  |x -1 3|
Δu = |x 4 -1|
  |x 5 5|
 = x{[(1)(4)(5) + (-1)(-1)(1) + (3)(1)(5)] - [(3)(4)(1) + (1)(-1)(5) + (-1)(1)(5)]}
 = 34x,
  |3 x 3|
Δy = |4 x -1|
  |-1 x 5|
 = x{[(3)(1)(5) + (1)(-1)(-1) + (3)(4)(1)] - [(3)(1)(-1) + (3)(-1)(1) + (1)(4)(5)]}
 = 14x,
  |3 -1 x|
Δz = |4 4 x|
  |-1 5 x|
 = x{[(3)(4)(1) + (-1)(1)(-1) + (1)(4)(5)] - [(1)(4)(-1) + (3)(1)(5) + (-1)(4)(1)]}
 = 26x.
For any fixed x, by Cramer's rule, the solutions are hence given by
(u, y, z) = (17x/83, 7x/83, 13x/83).
For the system to have positive integral solutions, x must be a multiple of 83.
Hence the smallest value of x is 83.


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