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.

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 $1991-a = \pm 2 \implies a_1=1989, a_2 = 1993$.
