# Linear independence of the square roots over Q

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 $a=\pi$?

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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 $c=\pi$.

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Minor nitpick: your definition of $S_c$ doesn't have $c$ in it. –  Mike Miller Feb 1 '14 at 12:31
It it obvious that the same finite set $\{a_n\}$ can't work for more than one real $c$? –  David Loeffler Feb 1 '14 at 13:06
@DavidLoeffler Well, each finite set will corrrespond to the finitely many $c$'s that satisfy $\sum a_n\sqrt{n^2+c^2}=0$. –  Joe Silverman Feb 1 '14 at 13:32
@Mike Thank you, I'll fix that typo. –  Joe Silverman Feb 1 '14 at 13:32
In particular, all such $c$ would have to be algebraic. –  Lev Borisov Feb 1 '14 at 14:54

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.

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Please tell more about your proof. Where the transcendence is used? –  Anton Feb 1 '14 at 22:49
$\pi$ is transcendent, hence $\mathbb{Q}(\pi)$ is isomorphic to the field $\mathbb{Q}(x)$ of rational functions. To show that the numbers $\sqrt{n^2+\pi^2}$ are $\mathbb{Q}$-linearly independent is therefore equivalent to the statement that the functions $x\mapsto\sqrt{n^2+x^2}$ are $\mathbb{Q}$-linearly independent. –  Jan-Christoph Schlage-Puchta Feb 5 '14 at 21:03