## Monday, August 17, 2009

### KBB2 Problem 7

Prove that for any $n > 1$, the polynomial

$P(x) = 2008x^n + 7x + 56$

cannot be factored into two non-trivial integer polynomials.

(A polynomial is non-trivial if its highest degree is greater than one)

#### Solution

Let $P(x) = A(x)B(x)$ where

$A(x) = a_mx^m + a_{m-1}x^{m-1} + \cdots + a_1x + a_0$

$B(x) = b_nx^n + b_{n-1}x^{n-1} + \cdots + b_1x + b_0$

From the constant, we have $a_0 b_0 = 56$, so one of them is divisible by 7. Without loss of generality, we may assume $a_0$. But since 56 is not divisible by 49, then $b_0$ is not divisible by 7.

From the $x$ term, we have $a_1b_0 + a_0b_1 = 7$. Because $a_0b_1$ is divisible by 7, then $a_1b_0$ is divisible by 7, which means $a_1$ is divisible by 7.

We can continue this argument inductively to prove that $a_2,a_3, \cdots, a_m$ is divisible by 7. But since $a_mb_n = 2008$ is not divisible by 7, we arrive at a contradiction.