Integral solutions to $1\times2+2\times3+\cdots+m\times(m+1)=n\times(n+1)$.

Question

I just stumbled across the identity $$1\times2+2\times3+\cdots+34\times35=119\times120,$$ which made me wonder about integral solutions to $$\sum_{k=1}^mk\times(k+1)=n\times(n+1).$$ The left hand side can be rewritten as $$\sum_{k=1}^mk^2+\sum_{k=1}^mk=\frac{m(m+1)(2m+1)}{6}+\frac{m(m+1)}{2}=\frac{m(m+1)(m+2)}{3},$$ which shows that the original problem is equivalent to solving $$m(m+1)(m+2)=3n(n+1),$$ over the integers. This looks nicer already, but I'm still unable to determine its integral solutions. The latter equation also defines an elliptic curve over $\Bbb{Q}$, which might help me if I knew how to use Sage or Magma or the likes. Checking some small values of $(m,n)$ shows that the pairs $$(3,4),\qquad(8,15),\qquad(20,55),\qquad(34,119),$$ are solutions, for what it's worth.

Answer 1

OEIS is our friend. The sequence $n=1,4,15,55,119$ is A102349 and, being integer points on the elliptic curve, $$F(m,n)=\frac{m(m+1)(m+2)}{6} = \frac{n(n+1)}{2}$$ should be finite. The $n$ are solutions to numbers that both triangular and tetrahedral (A027568), with the largest as, $$F(34,\,119) = 7140$$

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$\color{green}{Update}$: In fact, in this post we have,

$$\frac{u(u+1)(u+2)(u+3)}{8} = \frac{v(v+1)(v+2)}{6} = \frac{w(w+1)}{2} =7140$$

with $u=2\times\color{blue}7,\;v=2\times\color{blue}{17},$ and $w= \color{blue}{7\times17} =119\,$. So $7140$ is also $3\times$ the pentatope number $2380$.