We find all the values of k when x³+x+k is reducible
and subtract from 1000.
If x³+x+k is reducible over the polynomials
with integer coefficients, then it is factorable as
the product of a linear binomial and a quadratic
trinomial both with leading coefficients 1. That
is, integers A,B,C exist such that:
x³+x+k = (x+A)(x²+Bx+C)
Multiplying the right side out we get
(x+A)(x²+Bx+C) = x³+(A+B)x²+(AB+C)x+AC
So we have the identity:
x³+(A+B)x²+(AB+C)x+AC ≡ x³+x+k
So we can equate coefficients:
(1) A+B= 0
(2) AB+C = 1
(3) AC = k
So from (1), B = -A, and substituting for B in (2),
A(-A)+C = 1
-A²+C = 1
(4) C = A²+1
So the factorization
x³+x+k = (x+A)(x²+Bx+C)
becomes
x³+x+k = (x+A)(x²-Ax+A²+1)
Substituting for C from (4), in (3),
A(A²+1) = k
A³+A = k
Since 1 ≤ k ≤ 1000
1 ≤ A³+A ≤ 1000
A³+A is a strictly increasing function, therefore:
The minimum value of A is 1, when k = A³+A = 1³+1 = 1+1 = 2, and
The maximum value of A is 9, when k = A³+A = 9³+9 = 729+9 = 738.
(For when A = 10, k = A³+A = 10³+10 = 1000+10 = 1010, which is
over 1000.)
So there are 9 values of A, 1 through 9, and therefore 9
factorizations which are:
For k = 1, (x³+x+2) =(x+1)(x²-x+2)
For k = 2, (x³+x+10) =(x+2)(x²-2x+5)
For k = 3, (x³+x+30) =(x+3)(x²-3x+10)
For k = 4, (x³+x+68) =(x+4)(x²-4x+17)
For k = 5, (x³+x+130) =(x+5)(x²-5x+26)
For k = 6, (x³+x+222) =(x+6)(x²-6x+37)
For k = 7, (x³+x+350) =(x+7)(x²-7x+50)
For k = 8, (x³+x+520) =(x+8)(x²-8x+65)
For k = 9, (x³+x+738) =(x+9)(x²-9x+82)
Since for only 9 positive integers 1 ≤ k ≤ 1000, the
polynomial f_k(x)=x³+x+k is reducible, then for the other
1000-9 or 991 positive integers 1 ≤ k ≤ 1000, the
polynomial f_k(x)=x³+x+k is irreducible.
Answer: 991
Edwin
