$(x-y)^2+(y-z)^2+(z-x)^2=6 \to (x,y,z)=(a+1,a,a-1)$ $(a+1)^2=a^2-a+1 \to 3a=0 \to (x,y,z)=(1,0,-1)$ and permutations

We can solve the system on $R$ too: 1) Subtract the first 2 equations, we get: $(x+y+z)(x-y)=0$. If $x=y$ replacing, and subtracting the equations $1$ and $3$, we get $z(x-z)=0$, but in both situations we get contradictions. 2) So $z=-x-y$, replacing in the system, we get $x^2+xy+y^2=1$. On the $R$ we have infinite many solutions (Just consider quadratic equation in $x$), but on $Z$ we get $(1,0,-1)$ and permutations.

Sum the three equations to get that $\sum_{cyc} (x-y)^2=6$ which implies that (WLOG up to permutation) $x=y+1=z+2,$ then solving gives permutations of $(1, 0, -1)$

LLL2019 wrote: Sum the three equations to get that $\sum_{cyc} (x-y)^2=6$ which implies that (WLOG up to permutation) $x=y+1=z+2,$ then solving gives permutations of $(1, 0, -1)$ Yes exactly what you see at the #2.

It is easy to see that $x, y, z$ are distinct(Sketch for proof: Assume $x = y$ and find a contradiction). So WLOG let $x > y > z$, and let $x-y = a, y-z = b$ . Adding all three equations gives us $x^2 + y^2 + z^2 - xy - yz - zx = 3 \implies (x-y)^2 + (y-z)^2 + (z-x)^2 = 6 \implies a^2 + b^2 + (a+b)^2 = 6 \implies a = b = 1$. Now we have $x^2 = (x-1)(x-2) + 1 \implies x = 1 \implies (x, y, z) = (1, 0, -1)$. Therefore the only solutions are $\boxed{(x, y, z) = (1, 0, -1) \text{ and permutations }}$. $\square$

parmenides51 wrote: Solve the following system of equations in integer numbers: $$\begin{cases} x^2 = yz + 1 \\ y^2 = zx + 1 \\ z^2 = xy + 1 \end{cases}$$ $x^2-y^2=(yz+1)-(zx+1)\Rightarrow (x-y)(x+y+z)=0$ and similarly for $x,y,z$. If $x+y+z\ne0$ then $x=y=z$ which is impossible, so $x+y+z=0$. Then: $$x^2+y^2+z^2+2xy+2yz+2zx=(x+y+z)^2=0$$and (by adding all $3$ equations) $$x^2+y^2+z^2=xy+yz+zx+3$$so $x^2+y^2+z^2=2$, which means that $(x,y,z)=(0,\pm1,\pm1)$. Since $x+y+z=0$ we get $(x,y,z)=(0,1,-1)$ and permutations.