Prove by contradiction that $(x-y)^3+(y-z)^3+(z-x)^3 = 30$ has no integer solutions

Question

By factorizing it I found that $(x-y)(y-z)(z-x) = 10$

Answer 1

Just as Karvens comments:

Let $a=x-y$ and $b=y-z$. Then $-(a+b)=z-x$. Clearly, $a,b,c \in \Bbb Z$. So

$$30=a^3+b^3-(a+b)^3=(a+b)(a^2-ab+b^2)-(a+b)^3=-3ab(a+b)$$

And hence

$$10=-ab(a+b).$$

Therefore $a=\pm 1$ or $a=\pm 2$ or $a=\pm 5$ or $a=\pm 10$. Claerly they cannot be the solution.

Answer 2

Set $a=x-y$, $b=y-z$ and $c=z-x$. Then $a+b+c=0$, and hence, as $abc=10$ two are negative and one positive. By symmetry, assume $$ A=-(x-y)>0, \,\, B=-(y-z)>0, \,\, A+B=z-x>0, $$ and for the integers $A,B,C$ we have $$ (A+B)^3-A^3-B^3=30. $$ Now as $AB(A+B)=10$, then, as $A+B>A,B$, $$ (A,B,A+B)\in \{(1,1,10),(1,2,5), (2,1,5)\}, $$ but none of the triplets works.

Hence no integer solution.