For what values of a ∈ Z , the quadratic expression x + a x + 1991 + 1 can be factorised as x + b x +…

Since, he quadratic expression $\left(x+a\right)\left(x+1991\right)+1$ can be factorised as $\left(x+b\right)\left(x+c\right)$, therefore

$\left(x+a\right)\left(x+1991\right)+1=\left(x+b\right)\left(x+c\right)$

This implies that $\left(x+a\right)\left(x+1991\right)+1$ has roots $-b \& -c$ i.e., integer roots. So, discriminant of  $x^2+\left(1991+a\right)x+1991a+1=0$must be perfect square.

${\left(1991+a\right)}^2-4\left(1991a+1\right)=m^2$

$\Rightarrow {\left(1991-a\right)}^2=m^2+4$

$\Rightarrow {\left(1991-a\right)}^2-m^2=4$

$\Rightarrow \left(1991-a-m\right)\left(1991-a+m\right)=4$

Since, both $\left(1991-a-m\right)$and $\left(1991-a+m\right)$ are integers, therefore possible choices are

$\left(1991-a-m\right)\left(1991-a+m\right)=1\times 4$

$\left(1991-a-m\right)\left(1991-a+m\right)=-1\times \left(-4\right)$

$\left(1991-a-m\right)\left(1991-a+m\right)=2\times 2$

$\left(1991-a-m\right)\left(1991-a+m\right)=-2\times \left(-2\right)$

Only possible result is $m=0$, so

$a=1989, 1993$