TITLE: $p,q,r$ primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational.
QUESTION [8 upvotes]: I want to prove that for $p,q,r$ different primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational. 
Is the following proof correct?
If $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is rational, then  $(\sqrt{p}+\sqrt{q}+\sqrt{r})^2$ is rational, thus $p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}$ is rational, therefore $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational.
If $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational, then $(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})^2$ is rational, therefore $pq+qr+pr+\sqrt{p^2qr}+\sqrt{pq^2r}+\sqrt{pqr^2}$ is rational, therefore $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$ is rational.
Now suppose $p<q<r$. If $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$  and $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ are rational, then $$p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}-p(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})$$
is rational, therefore $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational. 
If $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational, then $((q-p)\sqrt{pr}+(r-p)\sqrt{pq})^2$ is rational, thus $(q-p)^2pr+2(q-p)(r-p)\sqrt{p^2qr}+(r-p)^2pq$ is rational, thus $\sqrt{qr}$ is rational.
But $q,r$ are distinct primes, thus $qr$ can't be a square. Thus $\sqrt{qr}$ is irrational. Contradiction. 
Also, is there an easier proof?

REPLY [4 votes]: A tricky proof may be the following: assume that $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is a rational number $\frac{a}{b}$ with $\gcd(a,b)=1$. Then for every prime $s>b$ we have that the following relation about Legendre symbols:
$$ \left(\frac{p}{s}\right)=\left(\frac{q}{s}\right)=1 $$
implies $\left(\frac{r}{s}\right)=1$. Let $\eta_r$ be the least quadratic non-residue $\!\!\pmod{r}$. 
The Chinese remainder theorem and the Dirichlet's theorem give that there are an infinite number of primes $s$ for which $s\equiv 1\pmod{4pq}$ and $s\equiv \eta_r\pmod{r}$, but quadratic reciprocity then gives:
$$ \left(\frac{p}{s}\right)=\left(\frac{q}{s}\right)=1,\quad \left(\frac{r}{s}\right)=\left(\frac{\eta_r}{r}\right)=-1, $$
contradiction.