Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\in\Bbb Z$ 
Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers.

I tried to do it by comparing the two expressions but I can't proceed.

Let $f(x)=(x+b)(x+c)$ with $b,c\in\Bbb Z$.
Wlog $b\le c$. 
For $-c\le x\le b$, we have $f(x)\le 0$, hence $f(x)-1<0$.
If $x\ge -b+2$, we have $f(x)-1\ge 2\cdot 2-1>0$. By symmetry, $f(x)-1>0$ also for $x\le -c-2$.
Hence the only possibly integer roots of $f(x)-1$ are $-b+1$ and $-c-1$.
Moreover, $f(-b+1)-1=f(-c-1)=c-b$, hence $f(x)-1$ has integer roots iff $b=c$ and in that case these roots are $-b\pm1$ and hence differ by $2$. 
So if one of the integer roots is $-1991$, then the other, $-a$, must be $-1989$ or $-1993$, i.e., $$a\in\{1989,1993\}.$$
On the other hand indeed, $(x+1989)(x+1991)+1=(x+1990)(x+1990)$ and $(x+1993)(x+1991)=(x+1992)(x+1992)$.
