How to solve the diophantine equation $n^3-n-1=k^2-k+1$?

Question

My first idea were some factorization-based solution. For example, adding 1 to both sides, and then:

$$n^3-n-1=k^2-k+1$$

$$n^3-n=k^2-k+2$$

$$(n-1)n(n+1)=k^2-k+2$$

...but I don't have idea, what to do with the right side. Any hint?

Answer 1

Hint: Note that: $n^3 - n \equiv 0 \pmod 3$, thus

$4(k^2-k+1) = 3(k-1)^2 + (k+1)^2 \equiv 4(n^3-n-1) \equiv -1 \pmod 3$

But, $(k+1)^2 \equiv 0,1 \pmod 3$