An old RMO Problem

$(x+a)(x+1991)+1=(x+b)(x+c)$ Put $x=-b$ to get $(a-b)(1991-b)=-1$ and knowing the fact that product of two integers is $-1$ iff they are $1,-1$ . Setting $I. ~a-b=-1, 1991-b=1$ gives $a=1989$ .

Similarly setting $II. ~a-b=1, 1991-b=-1$ gives $a=1993$ . Hence sum $\large =1989+1993=\boxed{\color{#007fff}{3982}}$ .

We are given the expression $(x+a)(x+1991)+1$ .We need to factor it in the form $(x+b)(x+c)$ where b,c are integers.

So we can equate both the equations,

$(x+a)(x+1991) + 1 = (x+b)(x+c)$

or, $x^2 + (1991+a)x + (1991a+1) = x^2 + (b+c)x + bc$

or, $(1991+a)x + (1991a+1) = (b+c)x + bc$ ................................................................ (1)

Comparing the co-efficients of $x$ & $x^0$ we get the following two equations :

$b+c=1991+a$ .................................................. (2)

$bc=1991a+1$ ....................................................(3)

By (3) - (2) ,

$bc-b-c=1990a-1990$

or, $(b-1)(c-1) = 1990a-1989$

since (1990,1989) = 1, so $a= 1989$ which leads to $(b-1)(c-1) = 1989^2$ & thus b & c have integral values 1990 respectively.

Again by (2) + (3) ,

$bc+b+c = 1992a+1992$

or, $(b+1)(c+1) = 1992a+1993$

Again since (1992,1993) = 1 , So $a = 1993$ which leads to $(b+1)(c+1)=1993^2$ & thus b & c have the values 1992 respectively.

The sum of the possible values of a is : $1993+1989 = 3982$ .

Given:(x+a)(x+1991)+1. If the expression is factored as (x+b)(x+c), -b and -c are the roots of the above polynomial. Therefore, (-b+a)(-b+1991)=-1. Either a-b=1,1991-b=-1 or a-b=-1,1991-b=1 (as b and c are integers). Solving, we get a=1989 or a=1993. Required sum=1989+1993=3982.