341did you provide a hint or was that part of the problem statement?You can get a = 1989, 1993 in an easier manner
find all integers a such that (x+a)(x+1991) + 1 can be written as (x+b)(x+c) where b and c are integers themselves.
it implies that we have to find all integers a , b and c such that
1991 + a = b + c and 1 + 1991a = bc for some integers b and c.
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2 Answers
341341I find it intriguing that u ask a qn and dont follow up for three days!
Anyway, the solution:
Letting x = -1991, we get (1991-b)(1991-c) =1, so that either b = c = 1990 or, b = c = 1992
Again setting x=-a, we get (a-b)2=1 so that a-b=±1
Noting that a ≠1991, we get a=1989 or 1993
