the absolute minimum possible value for
a+b+C+d is 4, Ady showed an example
for a+b+c+d=6, so now all that is
left is to determine if a solution exists
for 4 or 5.
a+b+c+d=4
since all are nonzero this requires
a,b,c,d each to be either +1
thus d can be either 1 or 1
the product of the three integer zeros
is equal to d, however neither 1 or 1
can be factored into 3 distinct integer
factors. Thus this absolute minimum is
not possible.
a+b+c+d=5
since all are nonzero this requires
3 of them to be equal to +1 and one to be
equal to +2. Thus d is either +1 or +2
we already ruled out +1 above, so that leaves.
The only way to factor 2 into 3 integers is (1,1,2)
in order for them to be distinct the two 1's must be
of opposite signs, thus giving us either (1,1,2)
or (2,1,1). (1,1,2) was already covered by
Ady thus that just leaves (2,1,1) which gives us
(x2)(x1)(x+1)=x^32x^2+x2 which does not
give us the desired absolute sum of 5.
Thus it is not possible to achieve an asolute
sum of 5 either. Thus 6 must be minimum sought.

Posted by Daniel
on 20130925 14:02:20 