Expand each term $(x+k-1)(x+k+1)=x^2+2kx+(k^2-1)$ and sum from $k=1$ to $n$:

$n{x}^2+2x(1+2+\dots +n)+{\sum }_{k=1}^n(k^2-1)=4n$

$n{x}^2+n(n+1)x+\frac{n(n+1)(2n+5)}{6}=4n$ (after using $\sum k^2=n(n+1)(2n+1)/6$ and subtracting $n$).

Dividing through by $n$:

$x^2+(n+1)x+\frac{(n-1)(2n+5)}{6}=4$

The roots differ by $2$, so the discriminant must equal $4$. Computing:

$D=(n+1{)}^2-\frac{4(n-1)(2n+5)}{6}+16=\frac{122-2n^2}{6}+\text{const}$

Setting $D=4$ leads to $(n^2-1)/3=16$, hence $n=7$.

The equation becomes $x^2+8x+15=0$, giving roots $-3$ and $-5$. Thus $\alpha =-5$ and $\alpha +2=-3$.

$\therefore n+\alpha =7-5=2$