Different ways to write consecutive integer sum

Relevant wiki: Pell's Equation

Recall the the algebraic identity , $1+2+3+\cdots +N = \frac12 N(N+1)$ .

$\begin{aligned} &&1+2+\cdots+n=(n+1)+(n+2)+\cdots+(n+m)\\ &\Leftrightarrow & 2 [ 1+2+\cdots+n] =1 + 2 + \cdots + (n+1)+(n+2)+\cdots+(n+m) \\ &\Leftrightarrow& n(n+1)=\frac{(n+m)(n+m+1)}{2} \\ &\Leftrightarrow & n^2-(2m-1)n-m(m-1)=0\end{aligned}$

with the quadratic formula we get $n=\frac{2m-1+\sqrt{8m^2+1}}{2}$ we know that $n$ is a positive integer so we must have $8m^2+1=t^2$ for some positive integer $t$ .

Consider a Pell's equation , $t_k^2-8m_k^2=1$ with a fundamental solution $(t_1 , m_1) = (3,1)$ we can construct a family of solution with $t_k + \sqrt{8}m_k = (3+\sqrt{8})^2$ . Put $k=1,2,3,4,\ldots$ and we notice that the next solution of $n>84$ occurs when $k=4$ which gives us $m=204,t=577 \\$ .

$n=\frac{2\cdot204-1+577}{2} = \boxed{492}$ .