Consider three positive integers x < y < z in arithmetic sequence, and determine analytically all possible solutions of each of the following equations:

(I) x² + y² = z² - 135

(II) x² + 3y² = z² - 105

(III) x² +y² = z² - xyz

Since x,y,z are in arithmetic sequence, there exists a positive number d so that y = x + d and z = y + d = x + 2d
Therefore y² = x² + 2dx + d² and z² = x² + 4dx +4d²

(I) x² + y² = z² - 135
x² + x² + 2dx + d² = x² + 4dx + 4d² - 135
x² - 2dx + 135 - 3d² = 0
x[1,2] = 1/2 * ( 2d ± √(4d² - 4(135 - 3d²)) ) =
=d ± √(d² - 135 + 3d²) = d ± √(4d² - 135)
For this to be a viable answer, two conditions must exist: The stuff in the square root needs to be bigger or equal to 0, and x needs to be bigger or equal to 0.
4d² - 135 ≥ 0
d² ≥ 33.75
d ≥ 5.81
x = d ± √(4d² - 135) ≥ 0
If you choose the plus, then it's gotta be positive, so we'll check the minus.
d - √(4d² - 135) ≥ 0
d ≥ √(4d² - 135)
d² ≥ 4d² - 135
d² ≤ 45
d ≤ 6.708
Therefore for all ds for which 5.81 ≤ d ≤ 6.708 , x = d - √(4d² - 135) is an answer (and of course y and z can easily be found from that).
And for all ds for which d ≥ 5.81 , x = d +√(4d² - 135) is an answer.

(II) x² + 3y² = z² - 105
x² + 3x² + 6dx + 3d² = x² + 4dx + 4d² - 105
3x² + 2dx + 105 - d² = 0
x[1,2] = 1/6 * ( -2d ± √(4d² - 12(105 - d²)) ) =
= 1/3 * ( -d ± √(4d² - 315) )
The same conditions need to apply here, except we can already rule out the minus, so we need only calculate the plus. If x ≥ 0, then obviously the root is too.
-d + √(4d² - 315) ≥ 0
√(4d² - 315) ≥ d
4d² - 315 ≥ d²
d² ≥ 105
d ≥ 10.247
Therefore, for all ds for which d ≥ 10.247 , x = 1/3 * ( -d + √(4d² - 315) ) is an answer.

Darn, it's later than I thought. I can post the solution to (III) tomorrow, I need to get some sleep.

...err, this is the right solution, isn't it? Because it seems a tad too easy...

Edited on January 16, 2007, 4:49 pm

Edited on January 16, 2007, 4:51 pm

Posted by TamTam on 2007-01-16 16:41:53