Solve it!

We wish to find $n$ such that $n^{2} + 19n + 92 = m^{2}$ for some integer $m.$ Completing the square, we then have that

$(n + \frac{19}{2})^{2} - (\frac{19}{2})^{2} + 92 = m^{2} \Longrightarrow (n + \frac{19}{2})^{2} + \frac{7}{4} = m^{2}$

$\Longrightarrow (2m)^{2} - (2n + 19)^{2} = 7 \Longrightarrow (2m + (2n + 19))(2m - (2n + 19)) = 7.$

Since we are looking for integer solutions, we must then have that either

• (i) $2m + (2n + 19) = 7, 2m - (2n + 19) = 1 \Longrightarrow m = 2, n = -8$ ,

• (ii) $2m + (2n + 19) = -7, 2m - (2n + 19) = -1 \Longrightarrow m = -2, n = -11,$

• (iii) $2m + (2n + 19) = 1, 2m - (2n + 19) = 7 \Longrightarrow m = 2, n = -11$ , or

• (iv) $2m + (2n + 19) = -1, 2m - (2n + 19) = -7 \Longrightarrow m = -2, n = -8.$

Thus the only possible values for $n$ are $-8$ and $-11$ , which sum to $\boxed{-19}.$

$n^2+19n+92=k^2$ so the discriminant must be integer and be a perfect square.

$D=19^2-4(1)(92-k^2)=4k^2-7=m^2$

$\Rightarrow 4k^2-m^2=7$ so $(2k-m)(2k+m)=7$

We get four sistem of equations:

$2k-m=1$ with $2k+m=7$ , and $2k-m=7$ with $2k+m=1$ have as solution $k=2$

$2k-m=-1$ with $2k+m=-7$ , and $2k-m=-7$ with $2k+m=-1$ have as solution $k=-2$

Then substitute and solve for $n^2+19n+92=k^2$ and get as solutions $n=-11,-8 \therefore$ the sum is $\boxed{-19}$

There is a shortcut to this problem

Given that x²+19x+92= 0

Sum = -b/a Product= c/a

a=1 b=19 c=92 Sum of roots is -b/a

So the answer is -19