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 Common Multiple from a Cubic (Posted on 2016-05-01)
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?

 See The Solution Submitted by Brian Smith Rating: 4.0000 (1 votes)

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 Difference in approach (spoiler) Comment 3 of 3 |

Let  P(x) = x3 + ax2 + 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) = n3 + 3n2 + 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 2016-05-02 18:36:32

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