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.
 A: We have that $(x+a)(x+1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$. So if such factorisation exists we must have that the discriminant of the quadratic equation is a square. Now:
$$D = (1991 + a)^2 - 4(1991a + 1) = 1991^2 + 2 \cdot 1991a + a^2 - 4 \cdot 1991a - 4 = (1991 - a)^2 - 4$$
But then $\sqrt{D}, 2, (1991 - a)$ make a Pythagorean triplet, but we know that the only squares whose difference is $4$ are $4$ and $0$, therefore we must have $1991-a = \pm 2 \implies a_1=1989, a_2 = 1993$.
A: 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 $(x+1993)(x+1991)=(x+1992)(x+1992)$.
