<s>[INST] Solving $\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$ for natural numbers $a,b,c$ Find natural numbers $a<b<c$ such that:
$$\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$$

I found infinitely many solutions when $a=1$, where $b$ and $c$ can be found in the following way:
Calculate, in the simplest form, the continued fractions of the series: $1+\frac{1}{2}$,$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}$, where each term contains an odd number of $2$'s.
Then $b$ is $\frac{1}{2}D$ and $c$ is $\frac{N-1}{2}$ where $D$ and $N$ are the numerator and the denominator of any term of the previous series, respectively.
For example,$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}$, So $b=\frac{1}{2}(12)=6$, and $c=\frac{17-1}{2}=8$.
So, $a=1,b=6,c=8$ is satisfied.

Calculating continued fractions by hand is easy, but tedious!, So is there a general way to calculate $b$ and $c$ given any natural number $a$?

Your help would be appreciated. Thanks!
 [/INST] Not an answer, but rather a long comment.
Hint. Use the well-known formula $$\sum_{i=1}^n i=\frac{n\cdot (n+1)}{2}$$ in order to deduce $$\sum_{k=a}^{b-1} k=\sum_{k=1}^{b-1}k-\sum_{k=1}^{a-1}k=\ldots=\frac{(b-a)(a+b-1)}2$$ Similarly $$\sum_{k=b+1}^c k=\frac{(c-b)(c+b+1)}2$$Thus, your problem comes down to solving the following Diophantine equation $$(b-a)(a+b-1)=(c-b)(c+b+1)\iff2b^2=a^2-a+c^2+c$$ Given some $a\in\mathbb N$, observe that the latter equation is equivalent to $$\begin{align*}8b^2+2=(2a-1)^2+(2c+1)^2&\iff(2a-1)^2-2=8b^2-(2c+1)^2\\&\iff ... \end{align*}$$
I've been playing with Wolfram Alpha for a while and can't find any pattern for the solutions of the respective $a$ values, neither can the OEIS.
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