A complex system

From the equation $x^2 + y = y^2 + z$ , we get

$x^2 - y^2 = z - y,$

which factors as $(x + y)(x - y) = z - y$ . Since $x + y + z = 1$ , we can write this equation as $(1 - z)(x - y) = z - y$ . Similarly, $\begin{aligned} (1 - x)(y - z) &= x - z, \\ (1 - y)(z - x) &= y - x. \end{aligned}$

Taking the product of all three equations, we get

$(1 - z)(1 - x)(1 - y)(x - y)(y - z)(z - x) = (z - y)(x - z)(y - x).$

We are told that $x$ , $y$ , and $z$ are distinct, so we can cancel the factors of $x - y$ , $y - z$ , and $z - x$ , to get

$(1 - x)(1 - y)(1 - z) = -1. \quad (*)$

Also, the equation $(1 - z)(x - y) = z - y$ simplifies to $x - z = xz - yz$ , which factors as

$x - z = z(x - y).$

Similarly,

$\begin{aligned} y - x &= x(y - z), \\ z - y &= y(z - x). \end{aligned}$

Taking the product of all three of these equations, we get

$(x - z)(y - x)(z - y) = xyz(x - y)(y - z)(z - x).$

Again, we can cancel the factors of $x - y$ , $y - z$ , and $z - x$ , to get $xyz = -1$ .

Expanding equation $(*)$ , we get

$1 - (x + y + z) + (xy + xz + yz) - xyz = - 1,$

so $xy + xz + yz = -2$ .

Now, let $k = x^2 + y = y^2 + z = z^2 + x$ . Then

$3k = (x^2 + y^2 + z^2) + (x + y + z).$

Squaring the equation $x + y + z = 1$ , we get

$x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1,$

so $x^2 + y^2 + z^2 = 1 - 2(xy + xz + yz) = 5$ . Hence, $3k = 5 + 1 = 6$ , which means $k = 2$ . Then $x^2 + y = 2$ , so $x - y = x - (2 - x^2) = x^2 + x - 2 = (x - 1)(x + 2)$ . Similarly, $\begin{aligned} y - z &= (y - 1)(y + 2), \\ z - x &= (z - 1)(z + 2). \end{aligned}$

Therefore,

$\begin{aligned} &(x - y)(y - z)(z - x) \\ &= (x - 1)(x + 2)(y - 1)(y + 2)(z - 1)(z + 2) \\ &= (x - 1)(y - 1)(z - 1) \cdot (x + 2)(y + 2)(z + 2) \\ &= [xyz - (xy + xz + yz) + (x + y + z) - 1] \\ &\quad \cdot [xyz + 2(xy + xz + yz) + 4(x + y + z) + 8] \\ &= 7. \end{aligned}$

Note : $(x,y,z)$ must be a cyclic permutation of the numbers

$\left(-2 \cos \dfrac{2 \pi}{7}, -2 \cos \dfrac{4 \pi}{7}, -2 \cos \dfrac{6 \pi}{7}\right).$

We have: $x^2+y=y^2+z$ ,so $z-y=(x-y)(x+y)=(x-y)(1-z)=x-y-xz+yz$ , so $z-x=z(y-x)$ .

Similarly, we obtain $x-y=x(z-y)$ and $y-z=y(x-z)$ .

Therefore $(x-y)(y-z)(z-x)=-xyz(x-y)(y-z)(z-x)$ or $xyz=-1$ .

We have: $x^2+y=y^2+z$ ,so $z-y=(x-y)(x+y)$ .

Similarly, we obtain $y-x=(z-x)(z+x)$ and $x-z=(y-z)(y+z)$ .

Therefore, $(x-y)(y-z)(z-x)$

$=-(x+y)(y+z)(z+x)(x-y)(y-z)(z-x)$ or $(x+y)(y+z)(z+x)=-1$ .

Therefore $(1-x)(1-y)(1-z)=-1$ or $1-xyz-x-y-z+xy+yz+zx=-1$ or $xy+yz+zx=-2$ .

We have: $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=5$ .

Therefore, $(x-y)(y-z)(z-x)$

$=z(x^2-y^2) + y(z^2-x^2) + x(y^2-z^2)$

$=z(z-y)+y(y-x)+x(x-z)$

$=x^2+y^2+z^2-(xy+yz+zx)=7$

$x+y+z=1$ (i)

${x}^{2}+y={y}^{2}+z$ (ii)

${y}^{2}+z={z}^{2}+x$ (iii)

From (i) $z=1-x-y$ , using this in (ii) we get :

${x}^{2}-{y}^{2}+x+2y=1$ (iv)

Using $z=1-x-y$ in (iii) we get :

${x}^{2}+2xy=y$ (v)

Eliminating y from (iv) and (v) we get :

$3{x}^{4}-4{x}^{3}-5{x}^{2}+5x-1=0$ (vi)

Inspecting (i),(ii),(iii) we get one solution as $x=y=z=\frac{1}{3}$

Hence in that polynomial factoring it out we get :

$(3x-1)({x}^{3}-{x}^{2}-2x+1)=0$

Ommiting out $x=\frac{1}{3}$ we get :

${x}^{3}-{x}^{2}-2x+1=0$

Exploiting the symmetry between the equations we establish that :

$x,y,z$ are the roots of the polynomial of :

${p}^{3}-{p}^{2}-2p+1=0$

So we say that :

$(x-y)(y-z)(z-x)=\sqrt{D}$ where D is the discriminant of the cubic.

Working it out we get :

$D=49$ hence :

$(x-y)(y-x)(x-z)=7$