Solving $3 = x^2 + y^2 + z^2 - xy - yz - zx$ for integer $x$, $y$, $z$ I've been thinking about a solution for the following equation for integers $x, y, z$:
$$3 = x^2 + y^2 + z^2 - xy - yz - zx$$
A possible approach would probably be to transform the original equation to the following one:
$$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$$
But how I do solve this?
 A: You have made great Progress to get :
$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$
The Squares $S1,S2,S3$ must be (Case 1) all 3 even or (Case 2) 1 even with 2 odd.
Let $S1 \le S2 \le S3$ , without loss of generality.
Case 1 : This implies that the Squares are (1A) $0,0,6$ (1B) $0,2,4$ (1C) $2,2,2$
We can then see that no Integer Squares give this.
Case 2 : This implies that the Squares are (2A) $0,1,5$ (2B) $0,3,3$ (2C) $1,1,4$ (2D) $1,2,3$
We can then see that no Integer Squares give this , except (2C).
Hence we get Simultaneous Equations to solve.
$x-y=\pm 1$
$y-z=\pm 1$
$z-x=\pm 2$
We have 8 Such Simultaneous Equations which may give Integer Solutions.
Case 2C1 :
$x-y=+ 1$
$y-z=+ 1$
$z-x=+ 2$
No Solution (adding the 3 Equations gives 0 = 4 which will be Inconsistent )
Case 2C2 :
$x-y=+ 1$
$y-z=+ 1$
$z-x=- 2$
This gives $y=x-1$ & $z=x-2$ ( Solution works out )
Case 2C3 :
$x-y=+ 1$
$y-z=- 1$
$z-x=+ 2$
No Solution (adding the 3 Equations gives 0 = 2 which will be Inconsistent )
Case 2C4 :
$x-y=+ 1$
$y-z=- 1$
$z-x=- 2$
No Solution.
Case 2C5 :
$x-y=- 1$
$y-z=+ 1$
$z-x=+ 2$
No Solution.
Case 2C6 :
$x-y=- 1$
$y-z=+ 1$
$z-x=- 2$
No Solution.
Case 2C7 :
$x-y=- 1$
$y-z=- 1$
$z-x=+ 2$
This gives $y=x+1$ & $z=x-2$ ( Solution works out , though it is Equivalent to 2C2 )
Case 2C8 :
$x-y=- 1$
$y-z=- 1$
$z-x=- 2$
No Solution.
Overall Solution Set : $y=x-1$ & $z=x-2$
Examples :
$x=0,y=-1,z=-2$
$x=10,y=9,z=8$
$x=-10,y=-11,z=-12$
$x=1,y=0,z=-1$
A: COMMENT.-(This is not an answer) Each positive solution $(a,b,c)$ gives all permutation of $\{a,b,c\}$ as solution. Taking now $$6=A^2+B^2+C^2$$ given by the O.P. we have for possible values for $A^2=0,1, 4$.
$A=0$ has no solution.
$A=1$ implies $5=B^2+C^2$ has solution $(B,C)=(\pm1,\pm2),(\pm2,\pm1)$
$A=4$ implies $2=B^2+C^2$ has solution $(B,C)=(\pm1,\pm1)$ it follows simple linear systems.
