The polynomial f(X) has integer coefficients and an integer root. Prove that, for every non negative integer n, the product f(0)f(1)...f(n) is divisible by (n+1)! .

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- Aug 30th 2012, 12:09 PMbrucewayneProving Divisibility
The polynomial f(X) has integer coefficients and an integer root. Prove that, for every non negative integer n, the product f(0)f(1)...f(n) is divisible by (n+1)! .

- Aug 31st 2012, 10:51 AMVlasevRe: Proving Divisibility
This seems like an interesting question. What have you tried so far?

- Aug 31st 2012, 07:18 PMBingkRe: Proving Divisibility
Just some ideas ...

The integer root is very important. You can represent a polynomial as $\displaystyle f(x) = \sum_{i=0}^n a_i x^i$ where the $\displaystyle a_i$ are integers (as stated in the problem). How would you represent the polynomial if you knew one of its roots, say $\displaystyle r$?

What would happen to the product $\displaystyle f(0)f(1) ... f(n)$ if $\displaystyle r \geq 0$ (i.e. $\displaystyle r$ is also nonnegative) and $\displaystyle n \geq r$? what about if $\displaystyle r \geq 0$ and $\displaystyle n < r$? Remember, $\displaystyle r$ is an integer root of the polynomial, and $\displaystyle n$ is nonnegative.

What about similar situations for when $\displaystyle r < 0$ (i.e. what are the possible cases if $\displaystyle r$ is negative?

That should help you get started... - Sep 7th 2012, 11:37 AMbrucewayneRe: Proving Divisibility
I was thinking along the lines of proving by induction, but I'm not really sure.