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If a polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$ for all $x$, and $g(3)=6$, then $g(25)=$?

My try: $x\cdot g(x+1)=(x-3)\cdot g(x)$,

Put $x=3$, we get $g(4)=0$, means $(x-4)$ is a factor of $g(x)$.

Similarly put $x=0$. We get $g(0)=0$, means $x$ is a factor of $g(x)$.

This means $g(x)=x\cdot (x-4)h(x)$, where $h(x)$ is a polynomial.

Then how can I calculate it?

Thanks.

amWhy
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juantheron
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  • Actually, there is no such polynomial $g$... Indeed, $-\mathbb{N}^*$ are all zeroes, so $g=0$, a contradiction. So the question does not make sense. See Andreas' answer. – Julien Feb 19 '13 at 17:35
  • No such polynomial in characteristic $0$. – Julien Feb 19 '13 at 17:43

2 Answers2

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A repeated use of $g(x+1)=\dfrac{x-3}x g(x)$ gives

$$\begin{array} gg(25)&=&\dfrac{21}{24} g(24)\\ &=&\dfrac{22}{24}\dfrac{21}{23} g(23)\\&=&\cdots \\ &=&\dfrac{21}{24}\dfrac{20}{23}\cdots \frac{1}{4} g(3)\end{array}$$

Pedro
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  • @PeterTamaroff So it appears that there is no polynomial $g$ that satisfies the conditions. You might want to edit your answer. – Julien Feb 19 '13 at 17:37
  • @julien Yes, I can see that. What do you suggest? – Pedro Feb 19 '13 at 23:12
  • @PeterTamaroff It's totally up to you. Maybe you could prove that there is no such $g$ in characteristic $0$. – Julien Feb 19 '13 at 23:14
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We are talking polynomials with integer coefficients, right? Let us follow the suggestion of @IshanBanerjee

We have seen that $0$ is a root of $g$. Plug $-1$ for $x$ in $$ x\cdot g(x+1)=(x-3)\cdot g(x) $$ to find $-1$ is also a root. Plug $-2$ to see $-2$ is also a root...

Andreas Caranti
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