Are the numbers $\sqrt{n^2 + q^2}$, $n=0,1,\dots$, linearly dependent over $\mathbb{Q}$? Let $q$ be a non-zero rational number, and consider the set of numbers $\sqrt{n^2 + q^2}$, with $n=0,1,\dots$. Are they linearly dependent over $\mathbb{Q}$? In other terms, can we find some positive integer $N$ and some rational numbers $a_0,\dots,a_N$ not all equal to zero, such that
\begin{equation}
\sum_{n=0}^{N} a_N \sqrt{n^2 + q^2} = 0?
\end{equation}
I found this statement in the post Linear Independence of Square Roots over Q, where the author of the post considers it "evident". For me, not only it is not evident at all, but I have some serious doubt that it is generally true.
What do you think about it?
Thank you very much for your attention in advance.
NOTE. Let us recall, in connection with this problem, that we have the following remarkable result.
Theorem Let $n_1,\dots,n_k$ be square-free integers. Then the numbers $\sqrt{n_1},\dots,\sqrt{n_k}$ are linearly independent over $\mathbb{Q}$ if and onfly if $n_1,\dots,n_k$ are pairwise distinct.
Elementary proofs of this result are given in Linear Independence of Radicals by Iurie Borieco, then a young pluri-medallist at the International Mathematical Olympiads.

Guided by the pythagorean triples $(9,12,15)$ and $(5,12,13)$, we can take $q=12$ and have
$$\sqrt{5^2+12^2}-\frac{13}{15}\sqrt{9^2+12^2}=0.$$
Many pythagorean triples lend themselves to this.
I wonder if any counter-examples are not derived from a pythagorean triple...
