quadratic sequence Perfect Square values

Question

For quadratic sequence

$$f(n) = n^2 + n + 2534125$$

I need to find a few values of $n$ where $f(n)$ becomes a perfect square, I followed several examples but couldn't find a solution. By trial and error, $n=99,1096$ are the first two values, and I am stuck. This is what I have done so far:

Completing the square :

$$f(n) = (n +1591)^2 -3181n + 2844 = X^2$$

any help is greatly appreciated.

Answer 1

Let $$m^2=n^2+n+2534125$$where $m\in\mathbb{N}$.$$(2m)^2=4m^2=4n^2+4n+10136500=(2n+1)^2+10136499$$ $$(2m+2n+1)(2m-2n-1)=(2m)^2-(2n+1)^2=10136499$$ $$m,n\in\mathbb{N}\Rightarrow2m+2n+1,2m-2n-1\in\mathbb{N}$$ $$2m+2n+1>2m-2n-1$$ $$(2m+2n+1,2m-2n-1)=(10136499,1),(3378833,3),(10167,997),(3389,2991)$$ $$(m,n)=(2534125,2534124),(844709,844707),(2791,2292),(1595,99)$$ $\therefore n$ can be $2534124$, $844707$, $2992$ or $99$.