TITLE: Proving that $\sqrt{1 + x^2}$ is not a polynomial function
QUESTION [12 upvotes]: The given task goes as follows:

Show that $ f: \mathbb{R} \longrightarrow \mathbb{R}$ defined by $f(x) = \sqrt{1 + x^2} $ is not a polynomial function.

I tried this approach - if $f(x)$ is a $n$-degree polynomial function, then the $(n+1)$-st derivative equals to 0 and I was trying to determine the $k$-th derivative of $f(x)$ (and show it differs from 0 for any $k$) but without success. Since $f(x)$ is continuous and defined over whole R domain, I have no idea how to carry on. Any ideas?

REPLY [2 votes]: Assume $\sqrt{x^2+1}= a_nx^n+...a_1x +a_0$ , with an $a_n \not =0$.
Since $\sqrt{x^2+1}$ is an even function, the polynomial has to be even, i.e. only even powers occur.
Set $y^2:=x^2+1$, $y \ge 1$, then
$y = a_{2k}(y^2-1)^{2k} + ........+a_0.$
RHS: A polynomial in $y$ with even powers.
LHS: $y$.
A contradiction.