Solve $n(n+1)(n+2)=6m^3$ in positive integers

Solution 1:

The paper Rational approximation to algebraic numbers of small height: the Diophantine equation $\left|ax^n-by^n\right|=1$ by Michael A. Bennett (J. reine angew. Math. 535 (2001),1–49) shows that the equation in the title has at most one solution in positive integers $x,y$ for given positive integers $a,b,n$ with $n\ge3$.

It follows that $x=y=1$ is the only solution both of $\left|2x^3-y^3\right|=1$ and of $\left|2x^3-3y^3\right|=1$.

From now on, $n$ is the $n$ in the question, not in the paper.

$n$, $n+1$ and $n+2$ have no factors other than $2$ in common. It follows that except for factors of $2$ and one factor of $3$ to cancel the factor of $3$ in the factor $6$, $n$, $n+1$ and $n+2$ must all be perfect cubes. This allows us to exclude non-trivial solutions by case analysis:

If $n+1$ is even, it does not contain the single uncubed factor of $3$, since otherwise $n$ and $n+2$ would both be perfect cubes. It has a single uncubed factor of $2$ to cancel the factor of $2$ in the factor $6$, since $n$ and $n+2$ are odd. Thus $n+1=2x^3$, and one of $n$ and $n+2$ must be a perfect cube $y^3$, and with $\left|2x^3-y^3\right|=1$ it follows that $x=y=1$.

If $n$ and $n+2$ are even, one of them, say, $n$, is divisible by $2$ but not by $4$. It cannot also contain the single uncubed factor of $3$, since otherwise the other two would have to be perfect cubes. Thus $n=2x^3$, and $n+1$ is either $y^3$ or $3y^3$. Again it follows that $x=y=1$.