Find $(\ell,m,n)$ such that $\sum\limits_{k=1}^\ell k$, $\sum\limits_{k=\ell+1}^mk$ and $\sum\limits_{k=m+1}^nk$ are in geometric progression Find one triple of integers $(l, m, n)$, where $2 < l < m < n$ such that
$\displaystyle \sum_{k=1}^{l} k, \displaystyle \sum_{k=l+1}^{m} k,\displaystyle \sum_{k=m+1}^{n} k\;$ is a geometric sequence.
My attempted work :
$\displaystyle \sum_{k=1}^{l} k = \frac{l(l+1)}{2}$
$\displaystyle \sum_{k=l+1}^{m} k = \frac{(m-l)(m+l+1)}{2}$
$\displaystyle \sum_{k=m+1}^{n} k = \frac{(n-m)(n+m+1)}{2}$
Please give me some suggestions.
 A: Call the three sums, in order, $A_0, A_1, A_2$; these are in geometric progression iff $A_1 = \lambda A_0$ and $A_2 = \lambda^2 A_0$ for some $\lambda$, or equivalently, iff the quantity $$\Delta := A_0 A_2 - A_1^2$$ vanishes.
A posteriori, it's not easy to see how to solve this by hand: Writing the quantity on the left-hand side as a polynomial in terms of $l, m, n$ does not give something visibly tractable. If one searches with a CAS for solutions with $l, m \leq 10^3$, one can spot a pattern among some of the solutions: For all $l$,
$$m = (2 l + 2) l, \qquad n = (4 l^2 + 6 l + 3 l) l$$
is a solution with $\lambda = (2 l + 1)^2$.
There are other, 'sporadic' solutions $(l, m, n)$ that do not fall into this pattern. The only ones with $n < 10^3$ are:
$$(8, 11, 13), \quad (5, 9, 14), \quad (3, 11, 36), \quad (7, 27, 97) .$$
The first of these is the only one mentioned in this answer with $\lambda < 1$. (These do not exhaust the sporadic solutions, by the way: $(48, 2183, 98283)$ is another.)
All of the solutions mentioned here correspond to factorizations of $\Delta$ of the particular form
$\Delta = l (l + 1) [n + (\mu + 1)](n - \mu)$, which might be a clue to how one could solve this manually.
