Find
all integer solutions of:
x2+y2+
z2=2x-3z+5
x2+y2+3x=2
Any remarks?
(x,y,z)
( -1 , +/- 2 , -1 )
( -1 , +/- 2 , -2 )
( -1 , +/- 2 , -1 )
( -1 , +/- 2 , -2 )
(1) x^2+y^2+ z^2 = 2x-3z+5
(2) x^2+y^2+3x = 2
subtract (2) from (1)
z^2 - 3x = 2x-3z+3
z^2 + 3z -(5x+3) = 0
z = [-3 +/- sqrt(9+4(5x+3))]/2
z = -1.5 +/- [sqrt(20x + 21)]/2
If x is -1, then z = {-1 or -2}
For any other values of x, at least those which make
20x + 21 a square, per Eqn (2), y^2 is negative
y^2 = 2 - 3x - x^2
So the above are the only integer values
Edited on September 19, 2020, 11:41 am
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Posted by Larry
on 2020-09-19 11:40:52 |
Solution
