Given, (x+a)(x+1991)+1 ‌∴(x+a)(x+1991)−1=0 ‌⇒(x+a)(x+1991)=−1 Multiplication of two integers is −1, this implies that either (x+a) is +1 and (x+1991) is −1 or (x+a) is −1 and (x+1991) is +1. Let (x+a)=1, then x+1991=−1 ‌⇒‌‌x=−1−1991 ‌x=−1992∘ ‌∴‌‌a=1−x=1−(−1992)=1+1992=1993 ‌‌ Let ‌(x+a)=−1‌, then ‌x+1991=+1 ‌⇒‌‌x=1−1991 ‌=−1990 ‌∴‌‌a=−1−x=−1−(−1990) ‌=−1+1990=1989 ‌ So, a can be 1993 or 1989.