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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.

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    $\begingroup$ Note that the answers given use that $x+y+z\neq 0$ because they divide through. This is true because $x,y,z\gt 0$. Neither of the answers so far given explicitly mention this. They are not wrong because of this, but it is an important thing to notice, and good to get into the habit of mentioning it, because one day it will not be true and things will go wrong. $\endgroup$ Oct 8, 2020 at 4:59
  • $\begingroup$ @Mark Bennet I did mention it in my answer ;) it would be a huge overlook on my part to go to the next step otherwise :) $\endgroup$ Oct 8, 2020 at 9:44

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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 $2$.

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  • $\begingroup$ If your first identity is well-known, does it have a name? Where can I find identities like this? $\endgroup$
    – asmaier
    Jun 18, 2021 at 18:17
  • $\begingroup$ Even if it does have a name, I am not aware of it. If you want it in a list of identities, it should be there among the basic factorisation identities in any middle school algebra book, like $$a^2-b^2 = (a+b)(a-b) \\ a^3+b^3 = (a+b)(a^2-ab+b^2)$$ whenever you learn methods like vanishing method and symmetric factorisation (you might have other names for that). The only factorisation identity among the elementary ones which I know has a name is Sophie Germain's identity. $\endgroup$ Jun 20, 2021 at 14:14
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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, $$(x,y,z) = (z+2, ~z+1, z)$$

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  • $\begingroup$ Thanks for the link, I was unaware of the existence of such factorisation $\endgroup$ Oct 8, 2020 at 17:23

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