# Elementary proof that finite sums of square roots of primes is irrational

It is relatively easy to show that if $p_1$, $p_2$ and $p_3$ are distinct primes then $\sqrt{p_1}+\sqrt{p_2}$ and $\sqrt{p_1}+\sqrt{p_2}+\sqrt{p_3}$ are irrational, but the only proof I can find that $\sqrt{p_1}+\sqrt{p_2}+...+\sqrt{p_n}$ is irrational for distinct primes $p_1$, $p_2$, ... , $p_n$ requires we consider finite field extensions of $\mathbb{Q}$.

Is there an elementary proof that $\sqrt{p_1}+\sqrt{p_2}+...+\sqrt{p_n}$ is irrational exist?

(By elementary, I mean only using arithmetic and the fact that $\sqrt{m}$ is irrational if $m$ is not a square number.)

The cases $n=1$, $n=2$, $n=3$ can be found at in the MSE question sum of square root of primes 2 and I am hoping for a similar proof for larger $n$.

• More is true: math.stackexchange.com/questions/30687/….
– lhf
Nov 11, 2015 at 0:46
• mathematical induction... Nov 19, 2015 at 5:57
• That does seem the obvious approach, but it I can't seem to get it to work Nov 19, 2015 at 9:16

I do not know if this counts as an elementary proof, but one may exploit the fact that the Legendre symbols $\left(\frac{p_1}{q}\right),\ldots,\left(\frac{p_n}{q}\right)$ are somewhat independent.
In more rigorous terms: let us assume that $p_1,\ldots,p_n$ are distinct primes. Let $\eta$ be the least quadratic non-residue $\!\!\pmod{p_n}$. By Dirichlet's theorem, there are infinite primes $q$ such that
$$q\equiv 1\!\!\!\pmod{4},\quad q\equiv 1\!\!\!\pmod{p_1},\quad \ldots,\quad q\equiv 1\!\!\!\pmod{p_{n-1}}$$ and $q\equiv\eta\!\pmod{p_n}$. By quadratic reciprocity, $p_1,\ldots,p_{n-1}$ are quadratic residues $\!\!\pmod{q}$ and $p_n$ is a non-quadratic residue. Let $f(x)$ be the minimal polynomial of $$\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n}$$ over $\mathbb{Q}$. Our claim is that $\deg f>1$ since $f(x)$ does not completely factor over $\mathbb{F}_q$.
Indeed $\pm\sqrt{p_1},\ldots,\pm\sqrt{p_{n-1}}$ can be regarded as elements of $\mathbb{F}_q$, but by construction there is no element $a\in\mathbb{F}_q$ such that $a^2=p_n$, so the degree of the splitting field of $f(x)$ over $\mathbb{F}_q$ is at least two.
In particular $\sqrt{p_1}+\ldots+\sqrt{p_n}$ is not a rational number, neither it is any combination of the form $\pm\sqrt{p_1}\pm\sqrt{p_2}\pm\ldots\pm\sqrt{p_n}$. $\square$
Along the same lines we can prove that these numbers are algebraic conjugates over $\mathbb{Q}$, hence $\sqrt{p_1}+\ldots+\sqrt{p_n}$ is a constructible algebraic number over $\mathbb{Q}$ with degree $2^n$.