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Let $f\in\ \mathbb{R}[x]$ be a polynomial with the property that $f(x^2+3x+1)=f^2(x)+3f(x)+1$ and $f(0)=0$. Show that $f=x$.
I tried for $x=-1 => f(-1)=-1$ and for $x=1=>f(1)=1$, but I do not have any idea how to prove for the general case. I think about induction but I'm not really sure about it. Can you help me out?

If you can please try to tell me a way without using sequences.

Ghost
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  • Your request makes no sense. To formulate this problem you need to talk about the natural numbers and functions on them. That is what sequences are. If you can't use sequences, then you can't write your question. –  Jun 06 '18 at 12:35
  • @Kronnected Sorry but I was just saying that "If you can". I see your point, but I want to know if there are any other ways to do it. – Ghost Jun 06 '18 at 12:37

2 Answers2

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You got $f(1)=1$ by putting $x=0$ in the equation.

Putting $x=1$ you get $f(4)=4$. Putting $x=4$ you get $f(29)=29$

Define $x_{n+1}=x_n^2+3x_n+1$ and $x_1=1$. Then $x_n$ is strictly increasing and $f(x_n)=x_n$.

If $f$ has degree $m$, then $f(x)-x=0$ at the $m+1$ distinct points $x_1,x_2,...,x_{m+1}$.

Therefore, $f(x)-x=0$.

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For all such $x_0$,for which $f(x_0)=x_0$, we notice that $f(x_0^2 +3x_0+1)= x_0^2+3x_0+1$.

So, inductively we can generate infinitely many inputs for which $f(x)=x$, even if we know one such input and we already have one, $f(0)=0$.

But, $f(x) - x$ is a polynomial, so it must have finite no. of roots. So, only possibility is $f(x) - x= 0$

Shak
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