Determine all integers such that $\displaystyle x^2 + 19x + 92$
is a square.
Let $\displaystyle =k^2$
$\displaystyle x^2+19x+92-k^2=0$
Apply quadratic formula:$\displaystyle x=\frac{-19\pm{\sqrt{19^2-4(92-k^2)}}}{2}$
$\displaystyle x=\frac{-19\pm{\sqrt{4k^2-7}}}{2}$
Determine integers such that $\displaystyle 4k^2-7=m^2$
$\displaystyle 4k^2-m^2=7$
$\displaystyle (2k+m)(2k-m)=7$
As k and m are integers, these 2 factors must be factors of 7
$\displaystyle 2k+m=7$
$\displaystyle 2k-m=1$
$\displaystyle 4k=8\rightarrow{k}=2$
Insert k into the equation to receive your answer
$\displaystyle x=\frac{-19\pm3}{2}$