Linear independence of the square roots over Q

Question

Does there exist a real number $a$ such that the numbers $\sqrt{n^2 + a^2}$ (for all natural $n$) are linearly independent over the field of rational numbers? It is evident that $a$ cannot be rational. Is it possible to prove independence for $a=\pi$?

Answer 1

Suppose that $S_c:=\{\sqrt{n^2+c^2}:n\in\mathbb{N}\}$ is linearly dependent over $\mathbb{Q}$. This means that there is a finite list of rational numbers $a_1,\ldots,a_r$ so that $$ \sum_{n=1}^r a_n\sqrt{n^2+c^2} = 0.$$ Hence the set of $c$ values such that $S_c$ is $\mathbb{Q}$-linearly dependent is smaller than the set of finite sequences of rational numbers. Since the latter set is countable, so is the set of such $c$. Hence the set $S_c$ is $\mathbb{Q}$-linearly independent for almost all $c\in\mathbb{R}$. Of course, this Cantor argument is useless for proving anything about any particular $c$, such as $c=\pi$.

Answer 2

If $\pi$ is transcendental, the $\sqrt{n^2+\pi^2}$ are linearly independant over ${\mathbb Q}$: take a linear combination and notice that $\sqrt{n^2+\pi^2}$ is the only member of the family that is not smooth at $\pi=in$ (this proof would also show independance over ${\mathbb Q}(\pi)$).

If one wants to be more precise, one can argue in this way. Let $C$ be the algebraic curve over ${\mathbb Q}$ defined by equations $Y_n^2=X^2+n^2$ (for $n$ in a finite set $I$), then the point $\pi$, $\sqrt{\pi^2+n^2}$, $n\in I$ is a generic point of this curve as $\pi$ is transcendental. So, if a function vanishes at $\pi$, it is identically $0$. One can also use down to earth arguments by taking the product of the $\sum \pm a_n\sqrt{X^2+n^2}$ to get a polynomial in $X$ with rational coefficients: if this polynomial is $0$ at $\pi$ then it is identically $0$ and one of the factors is identically $0$, etc.