Proving that $\sqrt{1 + x^2}$ is not a polynomial function 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? 
 A: Suppose it is a polynomial, then $$\sqrt{1+x^2}= ax^n+...+b$$
so $$1+x^2= (ax^n+...+b)^2 = a^2x^{2n}+...$$
Since polynomial are equal iff they have same degree we have $2=2n$ so $n=1$ and now we have 
 $$ 1+x^2 = (ax+b)^2 = a^2x^2+2abx+b^2 \implies ab = 0$$
a contradiction, since $|a|=|b|= 1$.
A: Suppose $f$ is a polynomial,namely  $$f(x)=\sqrt{1+x^2}={a_0+a_1x+\cdots+a_nx^n}=p(x)$$ for some  $n$. Then $$\frac{f(x)}{x}=\frac{p(x)}{x}$$ for all non zero $x$. Hence $$\lim_{x \rightarrow \infty} \frac{p(x)}{x}=\lim_{x \rightarrow \infty}\frac{f(x)}{x}=\lim_{x \rightarrow \infty} \frac{\sqrt{1+x^2}}{x}=1$$
which means $p(x)$ and $x$ are polynomials of same degree, a contradiction!
A: Alt. hint: following up on OP's idea to use derivatives, note that $\,f(x) \cdot f'(x) = x\,$, but the only polynomials which satisfy that identity are $\,f(x)=\pm x\,$ (why?).
Alternatively, show that there exists no polynomial $\,f\,$ such that $\,f^2(x)=x^2+1\,$.
A: Another demonstration:
A polynomial of degree one or higher when squared, should result in a polynomial that has all its roots double or more than doubles.
$f ^ 2 = x ^ 2 + 1 = (x + i) (x-i)$ 
That has two roots but they are not equal. And by the fundamental theorem of algebra, that factorization is unique.
Therefore f can not be a polynomial.
A: The limit of the function $f(x)$ when $x$ tends to infinity is $x$.
$$\lim_{x\to\infty} f(x) = x$$
Therefore, either the function $f(x)$ is a first degree polynomial or it is not a polynomial.
Also, $f ^ 2$ must be $x ^ 2 + 1$.
The polynomials of the first degree squared are either simply $x ^ 2$ or have a component $x$ other than zero:
$$ (x+a)^2 = x^2 + 2ax +a^2 $$
Therefore $f (x)$ is not a first degree polynomial ($f^2$ it has no $x$ component, and is not $ x^2 $).
And since it can not have more degrees, $f (x)$ is not a polynomial.
A: 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.
