Cyclic Cube Root

Let $x = \sqrt[3]{a-b+1}, y = \sqrt[3]{b-c+1}, z = \sqrt[3]{c-a+1}$ . Observe that $x,y,z$ are real numbers such that $x+y+z=3$ , $x^3+y^3+z^3=3$ and each of $x^3, y^3, z^3$ are integers. Now I am not sure how to solve this without first proving that $x,y,z$ must be integers. It took me way too long to prove this, and I'll post my proof of it below if anyone wants to see it.

But for now, let's assume that $x,y,z$ are integers. Then: $(x+y+z)^3 - (x^3 + y^3 + z^3) = 3^3 - 3 = 24$ $3x^2 y + 3x^2 + 3y^2 x + 3y^2 z + 3z^2 x + 3z^2 y + 6xyz = 24$

Luckily, the left-hand side factors nicely into $3(x+y)(y+z)(x+z) = 24$ . Now use the fact that $x+y+z=3$ and we get: $(x+y)(3-x)(3-y) = 8$ If we do a little case checking, we see that the only solutions are $(x,y) = (1,1),(4,4),(4,-5),(-5,4)$ .

Therefore, there are essentially two possibilities: $\{x,y,z\} = \{1,1,1\}$ and $\{x,y,z\} = \{4,4,-5\}$ . Now it’s fairly simple to show that $a = 671 + \frac{x^3 - z^3}{3}, b = 671 + \frac{y^3 - x^3}{3}, c = 671 + \frac{z^3 - y^3}{3}$ Therefore, in the first case, we get $\{a,b,c\} = \{671,671,671\}$ , and in the second case we get $\{a,b,c\} = \{734,671,608\}$ . The smallest of these is 608. Therefore, $min\{min\{a,b,c\}\} = \boxed{608}$ .