I need to solve following problem, but don't know to start with, please provide hints and solutions.
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 $b,c$ are integers .
This is equivalent to find $a\in\Bbb{Z}$ such that the equation $L$: $$(x+a)(x+1991)+1=0$$ has $2$ integer roots.
Notice integer roots exist if the discriminant of $L$ is a perfect square(*), that is ($T$), $$(1991+a)^2-4(1)(1991a+1)=t^2$$ for some $t\in\Bbb{N^0}$.
Rearrange ($T$) and we get $$(1991-a)^2-t^2=4$$ $$(1991-a-t)(1991-a+t)=4$$
Since both $1991-a-t$ and $1991-a+t$ are integers, by the pairs $$(1991-a-t,1991-a+t)=(1,4)/(-1,-4)/(2,2)/(-2,-2)$$
among which the only possible result is $t=0$.
Therefore $a=1989$ or $1993$
(*) Quoted from Wiki: "For a quadratic polynomial with rational coefficients, it factors over the rationals if and only if the discriminant is a square."
(**)It is also obvious that if the discriminant is not a square, the roots will be irrational.