Consecutive sets of consecutive numbers which add to the same total

I’m looking at examples of numbers that can be written as the sum of integers from $j$ to $k$ and from $k+1$ to $l$. For example $15$ which can be written as $4+5+6$ or $7+8$. Or $27 = 2+3+4+5+6+7 = 8+9+10$. I have been able to find a few numbers which have two ways to satisfy the above equations. For example,

$$\begin{aligned}105
&= 1+2+\dots +14 = 15+16+\dots+20\\
&= 12+13+\dots+18 = 19+20+\dots+23
\end{aligned}$$

However, I have not been able to find any numbers that can be written as the sum in three ways of consecutive sums. That is, I have not been able to find an $X$ such that,

$$\begin{aligned}X
&= (a+1)+(a+2)+\dots +b = (b+1)+(b+2) +\dots +c\\
&= (d+1)+(d+2)+\dots +e = (e+1)+(e+2) +\dots +f\\
&= (g+1)+(g+2)+\dots +h = (h+1)+(h+2) +\dots +i\\
\end{aligned}$$

Does any such number $X$ exist? If so can you provide an example? If no such number exists can you provide a proof?

Thanks

Solutions Collecting From Web of "Consecutive sets of consecutive numbers which add to the same total"

Following chenyuandong’s answer, you need to find various $x,y$ with the same value of $xy(x^2-y^2)$. A search with Maple (I know, boring) gives
$$x=77,\ y=38\ ;\quad x=78,\ y=55\ ;\quad x=138,\ y=5$$
which leads to
$$\eqalign{
684+\cdots+3686=3687+\cdots+5168&=6561555\cr
2761+\cdots+4554=4555+\cdots+5819&=6561555\cr
8820+\cdots+9534=9535+\cdots+10199&=6561555\ .\cr}$$
There is also a solution with sum $531485955$ and $y<x\le500$ but I don’t have the details. Will post later if I have time 😉

Suppose,

$$X=(a+1)+(a+2)+\dots+b=(b+1)+(b+2)+\dots+c$$

which implies,

$$\Leftrightarrow \frac{b(b+1)}{2}-\frac{a(a+1)}{2}=\frac{c(c+1)}{2}-\frac{b(b+1)}{2}$$

$$2b(b+1)=a(a+1)+c(c+1)$$

or the special Pythagorean triple,

$$(2a-2c)^2+(2a+2c+2)^2=(4b+2)^2$$

where $a,b,c\in\mathbb Z$. We need to find integer solutions of the system,

$$(2c-2a)=s^2-t^2,\quad (2a+2c+2)=2st,\quad (4b+2)=s^2+t^2$$

So,

$$a=\frac{|s^2-t^2-2st|-2}{4},\quad b=\frac{s^2+t^2-2}{4},\quad c=\frac{2st+(s^2-t^2)-2}{4}$$

and just let $s=(2m+1),t=(2n-1)$, where $m\geq n$. Since,

$$X=\bigg(c+\frac{1}{2}\bigg)^2-\bigg(b+\frac{1}{2}\bigg)^2=\bigg(\frac{2st+(s^2-t^2)}{4}\bigg)^2-\bigg(\frac{s^2+t^2}{4}\bigg)^2\\
=\frac{st(t+s)(s-t)}{4}=(2m+1)(2n-1)(m-n+1)mn$$

If the equation below has 3 or more integer solutions then the $z$ is exactly what you want:

$$z=(2m+1)(2n-1)(m-n+1)mn$$

where $m\geq n$. Alternatively,

$$\begin{aligned}
a &= \tfrac{1}{2}\Big(-1+\sqrt{(x^2-2xy-y^2)^2}\Big)\\
b &= \tfrac{1}{2}\big(-1+x^2+y^2\big)\\
c &= \tfrac{1}{2}\big(-1+x^2+2xy-y^2\big)
\end{aligned}$$

where $x>y$ and sign chosen so $a$ is positive. Then,

$$X = \frac{b(b+1)}{2}-\frac{a(a+1)}{2}=\frac{c(c+1)}{2}-\frac{b(b+1)}{2} =\tfrac{1}{2}xy(x^2-y^2)$$

so it suffices to find three pairs of $x,y$ with the same $X$.