I need to solve $$x(x-1)(x-2) = 2y$$ to find $x$ given $y$. It is known that both $x$ and $y$ are positive integers. Is there an easier way than using cube root formulae?
Edit 1: People have requested context. Here is the actual problem. There is a set $S_1$ containing $x$ distinct elements. Then another set $S_2$ is constructed from all possible triplets of elements in $S_1$, under the following additional conditions:
- All elements in each triplet must be distinct
- Permutation of the last two elements in the triplet is considered the same element. For example, triplet (A,B,C) is considered identical to the triplet (A,C,B) but distinct from (B,A,C).
I know $y$, the size of the set $S_2$, and I would like to find $x$, the size of the original set. It is not hard to figure out that $x$ and $y$ are related via the equation above. I need a function that solves the above equation for $x$ given a valid $y$. By valid I mean that I will only ever use the values of $y$ that have originally been constructed from some positive integer $x$ using the above equation.
I hope this helps with context requirements
Comment expanded to answer per request.
For any $x > 2$, we have
$$(x-1)^3 = (x-1)(x^2-2x+1) > x(x-1)(x-2) > (x-2)^3$$ This means for any valid $y$ (i.e one satsifies $2y = x(x-1)(x-2)$ for some $x$),
$$(x-1)^3 > 2y > (x-2)^3 \implies (x-1) > \sqrt[3]{2y} > (x-2)$$ Since $x$ is supposed to be an integer, this leads to $$x = 1 + \left\lceil \sqrt[3]{2y} \right\rceil = 2 + \left\lfloor \sqrt[3]{2y} \right\rfloor$$