Can the function $\sum_{n=0}^{N} a_n \sqrt{n^2 + x^2}$ vanish identically on an interval? Let $a_0,a_1,\dots,a_N$ be real numbers not all equal to zero, and consider the function $f:\mathbb{R} \rightarrow \mathbb{R}$ defined by
\begin{equation}
f(x) = \sum_{n=0}^{N} a_n \sqrt{n^2 + x^2}
\end{equation}
Could anyone prove that this function cannot vanich identically in any interval $(a,b)$?
Even though this fact seems quite obvious, I could not be able to find a proof.
Thank you very much in advance for your help.
NOTE. This question was motivated by the post Linear Independence of Square Roots over Q, where it is taken for granted that the function $f$ can only have finitely many zeroes.
Let us note here that if we can prove that it cannot vanish identically on any interval, then we can easily deduce that it has only finitely many zeroes. The argument goes as follows. Let $S$ be the set of all maps $\sigma:\{1,\dots,N\} \rightarrow \{0,1 \}$, and consider the polynomial
\begin{equation}
P(X_0,\dots,X_N)= \prod_{\sigma \in S} \left(a_0 X_0 + \sum_{n=1}^{N} (-1)^{\sigma(n)} a_n X_n \right).
\end{equation}
Since by replacing $X_i$ by $-X_i$, for some $i \geq 1$, we get the same expression on the righthand side of the equation, we see that $X_1,\dots,X_N$ only appear with even powers in the monomials of $P$. On the other hand, by replacing $X_0$ by $-X_0$, we see from the expression on the righthand side that we get $(-1)^{2^N} P=P$, since $|S|=2^N$. So also $X_0$ appears only with even powers in each monomial of $P$. Then by setting
\begin{equation}
R(x)=P\left(x,\sqrt{1+x^2},\sqrt{4+x^2},\dots,\sqrt{N^2+x^2} \right),
\end{equation}
we get a polynomial in $x$ with real coefficients.
Let now define
\begin{equation}
f_\sigma (x) = a_o x + \sum_{n=1}^{N} (-1)^{\sigma(n)} a_n \sqrt{n^2 + x^2},
\end{equation}
and let $Z_\sigma$ be the set of zeroes of $f_\sigma$. If $Z$ is the set of all zeroes of $R$, we have $Z= \cup_{\sigma \in S} Z_{\sigma}$. Now assume that we have proved the statement in the post and that $R$ were the null polynomial. Then we would have $\mathbb{R} = \cup_{\sigma \in S} Z_{\sigma}$, and so by Baire's Theorem some $Z_\sigma$ should have a nonempty interior, a contradiction. We conclude that $R$ is a non-null polynomial, which implies that $Z$, and so each $Z_\sigma$, is finite.
 A: Finally I found the proof I was looking for, even though it is not an "elementary" one since it uses complex analysis.
Since the function we are interetsed is even, we can limit ourselves to define $f$ on $(0,\infty)$ as
\begin{equation}
f(x) = a_0 x + \sum_{n=1}^{N} a_n \sqrt{n^2 + x^2}.
\end{equation}
Now let $z \mapsto \sqrt{z}$ the analytic extension of the square root $x \mapsto \sqrt{x}$ of positive real numbers to the complex domain $\mathbb{C} \backslash \{ x \in \mathbb{R}: x \leq 0 \}$. Each function $z \mapsto \sqrt{n^2 + z^2}$ is then holomorphic on the domain $\mathbb{C} \backslash \{ iy : y \in \mathbb{R}, |y| \geq n \}$, so in particular $f:(0,\infty) \rightarrow \mathbb{R}$ extends to the complex function
\begin{equation}
F(z) = a_0 z + \sum_{n=1}^{N} a_n \sqrt{n^2 + z^2},
\end{equation}
which is holomorphic on the domain $\mathbb{C} \backslash \{ iy : y \in \mathbb{R}, |y| \geq 1 \}$. Now, if all the coefficients $a_1,\dots,a_N$ are equal to zero, our thesis is trivial. If not, let $n$ be the smallest number in $\{1,2,\dots,N\}$ such that $a_n \neq 0$. Fix a real number $y$ such that $n < y < n+1$, and note that for each $m > n$, the function $z \mapsto \sqrt{m^2 + z^2}$ is continuous in $z=iy$ (since this point is contained in its domain of holomorphy). So we have
\begin{equation}
\lim_{\epsilon \rightarrow 0^{+}} [F(\epsilon + iy) - F(-\epsilon + iy)]=
\lim_{\epsilon \rightarrow 0^{+}} a_n\left( \sqrt{n^2+(\epsilon +iy)^2}-\sqrt{n^2+(-\epsilon+iy)^2} \right)=\\
=\lim_{\epsilon \rightarrow 0^{+}} a_n \left( \sqrt{n^2+\epsilon^2 -y^2 +2i\epsilon y}-\sqrt{n^2+\epsilon^2 -y^2 -2i\epsilon y} \right)= 2i a_n (y^2 - n^2) \neq 0.
\end{equation}
So $F$ is not identically zero. By the Identity theorem we then conclude that $f$ cannot vanish identically on any interval $(a,b)$. QED
A: Another (somewhat more general) idea is as follows.

Let $a_0,\ldots,a_n$ be distinct positive real numbers, and $\alpha\in\mathbb{R}\setminus\{0,\ldots,n\}$. If the function $$F(x)=\sum_{k=0}^n c_k(x+a_k)^\alpha$$ (with real numbers $c_0,\ldots,c_n$) satisfies $F^{(k)}(0)=0$ for $0\leqslant k\leqslant n$ (in particular, if it vanishes in some neighborhood of zero), then we have $c_0=\ldots=c_n=0$.

The proof is easy: the quantities $x_k:=a_k^{\alpha-n}c_k$ satisfy a (homogeneous) system of linear equations, whose matrix (being the Vandermonde matrix in $a_0,\ldots,a_n$) is nondegenerate. The question reduces to the above by considering $f(\sqrt{x+c})$, where $a<c<b$ and (we may assume that) $0\leqslant a<b$.
