given this equation $$S=\sqrt{x^2+1500x-1472}$$ find $x$, such that $x$ is positive integer, and $S$ is positive integer.
I have tried to solve this, and I get that $x = 36$, but I get that $x$ by trial and test, so
if I try $x = 1$, I got $S = 5.3851\dots$ ($S$ is not a positive integer),
if I try $x = 2$, I got $S = 39.140\dots$ ($S$ is not a positive integer),
if I try $x = 3$, I got $S = 55.108\dots$ ($S$ is not a positive integer), and so on
until I try $x = 36$, I got $S = 232$ ($S$ is a positive integer)
Is there alternative to solve this question?
There is only another solution besides that one: $x=140\,244$, in which case $S=140\,992$.
In order to see why, note that\begin{align}x^2+1500x-1472&=(x+750)^2-1472-750^2\\&=(x+750)^2-563\,972.\end{align}Let $Z=x+750$. Then you are after all pairs $(S,Z)$ of natural numbers such that $S^2=Z^2-563\,972$. But this is the same thing as\begin{align}(Z-S)(Z+S)&=563\,972\\&=2^2\times277\times509\end{align}(decomposition into prime factors). So, for each of the twelve divisors $d$ of $563\,972$, if $d'=\frac{563\,972}d$, you should solve the system$$\left\{\begin{array}{l}Z-S=d\\Z+S=d'.\end{array}\right.$$Its solutions are$$Z=\frac{d+d'}2\quad\text{and}\quad S=\frac{-d+d'}2.$$So, you only need to consider those cases in which $d$ and $d'$ have the same parity and $d'>d$. That's when $d=2$ and when $d=554$. If $d=2$, then $Z=140\,994$ (and therefore $x=140\,244$) and $S=140\,992$; and if $d=554$, then $Z=786$ (and therefore $x=36$) and $S=232$.