P(x) is a monic cubic polynomial with all coefficients positive integers. Also for all integers n, P(n) is a multiple of m.
What is the largest value of m?
Let
P(x) = x^{3} + ax^{2} +
bx + c
For integral n > 0:
m divides all P(n) if and only if m divides P(0) (i.e. m divides c), and
m divides all differences between consecutive terms P(i) and P(i+1).
But, if m divides all these first differences then it must also divide their
differences. i.e. the second differences and, by the same argument,
the third differences.
Now, the third differences of a monic cubic are all 6, so m can be no
bigger than 6, and can only be 6 if values of a, b and c can be found
that allow this.
Lots of choices exist for a, b and c. For example: (a, b, c) = (3, 2, 6), as
armando discovered.
Proof: P(n) = n^{3} + 3n^{2}
+ 2n + 6
= n(n+1)(n+2) + 6, both parts of which are divisible by 6.
So 6 is the largest possible value of m.

Posted by Harry
on 20160502 18:36:32 