# Does a function exist with the property $f(-n^2+3n+1)=(f(n))^2+1$?

Let $f:\mathbb{R}\to \mathbb{R}$ be a function which fulfills for every $n \in \mathbb{N}$ $$f(-n^2+3n+1)=(f(n))^2+1$$

Is it possible that such a function exists?

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is your function continuous? any thoughts about the problem? where did you face it, the question in the body and in the title are different, is $x\in \mathbb{R}$? I gave a downvote because of the poor formatting (and because the question is not very clear), I will reverse when you fixed it – Dominic Michaelis Mar 26 '13 at 7:08
By $f^2(n)$, do you mean $f(f(n))$, or $f(n)^2$? – Glen O Mar 26 '13 at 7:09
Thank you ,I have edit – math110 Mar 26 '13 at 7:21
@math110 i have edited you question, as you did have two different equalites please check if i took the right one (and i reversed the downvote) – Dominic Michaelis Mar 26 '13 at 7:36
Thank you ,my frend, becasuese my English is very poor. – math110 Mar 26 '13 at 8:13

The main idea here is writing at first some of the equations you get and look if they have common terms. Indeed here the terms for $n=3$ and $n=1$ are very interessting, as in both only occur $f(1)$ and $f(3)$.
\begin{align*} f(3)&=1+f(1)^2 \tag{$i$}\\ f(1)&=1+f(3)^2 \tag{$ii$}\\ \end{align*} As we don't know that much lets try to get an equation only having $f(1)$.
At first we have this equation: $$f(1)=1+f(3)^2$$ No we use $(i)$ to express $f(3)$ in terms of $f(1)$ $$f(1)= 1+(1+f(1)^2)^2=1+1^2+2f(1)^2+f(1)^4$$ is that possible?
Note that $f(1)$ is a solution of $$0= 2-x+2x^2+x^4$$ but this one has no real solution, hence your function can't exist, as for $x \in [0,1]$ $$2-x+2x^2+x^4\geq 2-x> 0$$ and for $x\in [1,\infty)$ we know that $x<x^2$ and hence $$2-x+2x^2+x^4 \geq 2+x^2+x^4>0$$ and for $x\in (-\infty,0]$ $$2-x+2x^2+x^4 \geq 2+2x^2 +x^4 >0$$