Solution
3x2 -4xy-2xz+2y2+2yz+z2-2x+1=(x-1)^2+(y-x)^2+(z-x+y)^2
No, there is a solution. 3[x-(2/3)y-(1/3)z]2+(2/3)[y+(1/2)z]2 +(1/2)z2 does't equal to -1. You have lost (-2x).
If you fix x and y, you can find an z, so it has infinite solutions. Have you tried with mathematica or wolframalpha.com.
It is a small try from wolframalpha.com, which gives the following answer (see attached)
Integer solution x=1,y=1,z=0
The real solutions may be finite, due to the square root, but I hope it gives an idea to proceed further
In fact, I found that the square root has no positive real solutions. So, the only real solution should be (1,1,0). If I am wrong, let me know
It is a singular quadric with a center [1,1,0] in original coordinates, as Panchatcharam wrote. This is also the unique real solution.
Theoretical framework:
The corresponding matrices A4x4 and B3x3 (enclosed) have determinants det(A) = 0 and det(B) = 1, all eigenvalues of B are positive.
Thus the equation can be transformed in X2/a2+Y2/b2+Z2/c2 = 0.
The transformed equation (numerical values) is enclosed as well.
Dear Leonid,
maybe is this an equation of a tumor. The equation is that of an ellipsoid [reduced in 1 point, its center]. A small change in one parameter, say the linear term -2x replaced by -(2+d)x, would give a non-reduced ellipsoid.
This is not a solution. It is simply another representation of the given equation. The one and only real solution is (1,1,0)
Dear Leonid, your form in squares is excellent and gives immediately the answer.
Dear Panchatcharam, Leonid's word solution confused me as well. Apparently it is used in a different sense than we expected.
Dear Viera,
Yes. It is confusing me. Since he mentioned "finite real solutions", I thought that he is expecting set of all (x,y,z) in R3 which satisfies the equation. So, it may be better to rephrase his question to avoid the ambiguity of the word "solve".
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