# Quod Erat Demonstrandum

## 2012/01/04

### 某題-多項式

Filed under: HKALE,Pure Mathematics — johnmayhk @ 11:08 上午

$p(x)=p(x-1)$ for all $x \in \mathbb{R}$

$p(x)=p(0)$ for all $x \in \mathbb{R}$

$p(x)=p(x-1)$ for all $x \in \mathbb{R}$

$p(n)=p(n-1)=p(n-2)=\dots =p(0)$

$p(x)-p(0)=0$

$p(x)-p(0)\equiv 0$

$xq(x-1)=(x-3)q(x)$ for all $x \in \mathbb{R}$

$q(x)\equiv Cx(x-1)(x-2)$.

$xq(x-1)=(x-3)q(x)$ for all $x \in \mathbb{R}$

$q(x)=p(x)x(x-1)(x-2)$

$xq(x-1)=(x-3)q(x)$ for all $x \in \mathbb{R}$

$x(x-1)(x-2)(x-3)p(x-1) \equiv (x-3)x(x-1)(x-2)p(x)$

$p(x-1) \equiv p(x)$