Question

Determine the modulus of the sum of the set of integers $n$ for which $n^2+19n+92$ is a square.

Answer 1

Let $n^2+19n+92=m^2,$ m is a non negative integer. Then $n^2+19n+92-m^2=0$
Solving for n, we get
$n=\frac{1}{2}(-19\pm \sqrt{4m^2-7})$
$\therefore 4m^2-7$ is a square i.e., $4m^2-7=p^2$
Where $p\in N$
$\therefore (2m-p)(2m+p)=7$
$\therefore 2m+p$ being positive therefore
$(2m+p)$ is 7 and $2m-p=1$
Hence, $4m=8\Rightarrow m=2$
Thus, we have $n^2+19n+92=4$
$\Rightarrow n^2+19n+88=0$
$\Rightarrow (n+8)(n+11)=0$
$n=-8$ or $-11=-8+-11=-19$

Thus the absolute value is $19$