One solution of the equation
(x-a)(x-b)(x-c)(x-d)=25
is x=7.
If a, b, c, and d are different integers, find the value of a+b+c+d
(In reply to
Puzzle Solution by K Sengupta)
If a,b,c,d were not distinct integers, then P, Q, R and S would not have been distinct, and we would have:
(P, Q, R, S) (- (5,5,-1,-1), (5,5,1,1), (-5,-5,1,1), (-5,-5,-1,-1), (5,-5, 1, -1), (25,1,1,1), (25,-1,-1, 1), (-25, -1, 1, 1), (-25, -1,-1,-1)
Thus, 28-(a+b+c+d) = 8, 12, -8, -12, 0, 28, 24, -24, -28
or, a+b+c+d = 20, 16, 36, 40, 28, 0, 4, 52, 56 ...........(#)
Consequently, for non-distinct quadruplet (a,b,c,d) of integers satisfying the other given conditions, the sum a+b+c+d could have assumed precisely 9 distinct values as given in (#).
Thoughts On A Possible Extension
