Is $f(n)=n$ the only polynomial bijection from $\mathbb{N}$ to $\mathbb{N}$? Is $f(n)=n$ the only polynomial bijection from $\mathbb{N}$ to $\mathbb{N}$ ?
I tried a lot to find another example, but failed. 
If indeed there isn't another, how can I prove such uniqueness?
 A: *

*Its leading coefficient must be positive (or else it would eventually be negative).

*If its degree is larger than $1$, then it would eventually be increasing at a rate that prevents surjection.


So this reduces the problem to finding $a$ and $b$ such that $n \mapsto an+b$ is a bijection.
A: Since a polynomial can only change direction (increasing versus decreasing) finitely many times, any polynomial bijection $\mathbb{N}\rightarrow\mathbb{N}$ must eventually be the identity. In particular, any two polynomial bijections $\mathbb{N}\rightarrow\mathbb{N}$ are infinitely often equal. But any two polynomials which are infinitely often equal are actually equal.

A bit more explanation:


*

*Note that if $f:\mathbb{N}\rightarrow\mathbb{N}$ is a bijection (polynomial or not) and $f(a)>a$, then for some $b>a$ we have $f(b)<f(a)$ - this is a quick application of the pigeonhole principle.

*So if $f:\mathbb{N}\rightarrow\mathbb{N}$ is a bijection such that $f(a)>a$ for infinitely many $a$, then we can find infinitely many pairs $c,d$ with $c<d$ but $f(c)>f(d)$. In particular, this means that $f$ changes direction infinitely many times.

*Finally, we turn to the specific properties of polynomials. Both claims - that no polynomial changes direction infinitely often, and that two polynomials which agree infinitely often are equal - are proved similarly. HINT: think about the fact that any nonconstant polynomial only has finitely many zeroes ...
