Prove that if x,y,z are positive integers such that $x^3+y^3+z^3=3(x+y+z+xyz)$ then they must be consecutive numbers I was thikning going for contradiction, assuming they're not consecutive but I found it very hard to find a statement that contradicts that condition. Maybe stating that there exist some $m, n \in \Bbb{Z}$ such that $|mn|\ne 1$ and $y=x+m$ and $z=x-n$? Another way would be assuming at least 2 of them are non-consecutive?
I would like to see general ideas on how to proceed.
 A: Use the well-known identity $$x^3+y^3+z^3-3xyz = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right)$$ to have, from your given equation (since the left hand side equals $3(x+y+z)$, $$\begin{aligned} 3(x+y+z) & = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right) \\ \implies 6 &= (x-y)^2+(y-z)^2+(z-x)^2 \ (\because x,y,z>0 \implies  x+y+z\ne 0) \end{aligned}$$
also, $x,y,z$ being positive integers, the squares in the last line $(x-y)^2, (y-z)^2, (z-x)^2$ are perfect square integers, and the only way three perfect squares add up to $6$ is $$6=1+1+4$$
so, two of the differences $|x-y|$ and $|y-z|$ (without loss of generality) must be $1$, naturally the $|x-z|$ must be $2$.
A: First, subtract $3xyz$ from both sides of the equation:
$$x^3+y^3+z^3=3(x+y+z+xyz) \\ 
x^3+y^3+z^3-3xyz = 3(x+y+z)$$
Now,  using this factorisation, observe:
$$
(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=3(x+y+z)$$
Since $x,y$, and $z$ are positive integers which implies $x+y+z\neq 0$, you can divide the equation by $x+y+z$ to get
$$ x^2+y^2+z^2 - xy-yz-zx = 3$$
Now, from the above equation, it's easy to observe that
$$(x-y)^2+(y-z)^2+(z-x)^2=6$$
Now, if two of the differences in the brackets are greater than $1$ then the sum will be greater than $2^2 + 2^2 = 8 > 6$ (contradiction!).
Also, if all are equal to $1$, then the sum will be $3 \neq 6$ (contradiction!).
Hence, one of the differences is $2$ and the other two are equal to $1$ (so that $2^2+1^2+1^2=6$ as desired) which implies $x,y$, and $z$ are consecutive integers.

Why is that? Now goes the explanation:
Since you have already shown that one of the differences is $2$ and the other two are $1$, without loss of generality, you may take
$$
\begin{cases}
x - y = 1 \\ 
y - z = 1 \\ 
x - z = 2
\end{cases} \implies
\begin{cases}
x = y + 1 \\ 
y = z + 1 \\ 
x = z + 2 
\end{cases}
$$
Hence, $$(x,y,z) = (z+2, ~z+1, z)$$
