What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots? What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots?
$(x + a)(x + 1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$
This is of the form $ax^2 + bx + c$. Applying the quadratic formula $\left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\right)$, we get, that the rooots must equal:
$\frac{-(1991 + a) \pm \sqrt{(1991 + a)^2 - 4(1991a + 1)}}{2} = \frac{-(1991 + a) \pm \sqrt{1991^2 + a^2 + 2\times1991a - 4\times1991a -4}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a)^2 - 2^2}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a - 2)(1991-a+2)}}{2}$
For the formula to yield an integer, the discriminant must be a perfect square or $0$. Clearly, the discriminant will be zero if $a = 1993$ or $a = 1989$. The only thing that remains to be shown is that these are indeed the only possible solutions. 
I couldn't think of any way to do so. Are there any alternate solutions, perhaps some that involve more number theory and less algebra?

You are right, $m^2-4$ cannot be a square for $m>2$. Hint: $m^2 - (m-1)^2 = 2m-1$.
