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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?

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There is an simplier proof:

Suppose that $\sqrt{p}+\sqrt{q}+\sqrt{r}=w$, where $w$ is rational. Then$\sqrt{p}+\sqrt{q}=-\sqrt{r}+w$, so:

$$(\sqrt{p}+\sqrt{q})^2=(-\sqrt{r}+w)^2$$

Therefore $\sqrt{pq}+w\sqrt{r}$ is rational. But it isn't (the same argument). Suppose that $\sqrt{pq}-w\sqrt{r}=s$ where $s$ is rational, then:

$$\sqrt{pq}=s+w\sqrt{r}$$ $$(\sqrt{pq})^2=(s+w\sqrt{r})^2$$ So $sw\sqrt{r}$ is rational (we also should check case when $s=0$ or $w=0$, but it is simple).

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  • $\begingroup$ Nice argument. Do you know whether my proof would be correct as well? $\endgroup$ – wythagoras Aug 4 '15 at 19:52
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Your proof is actually very close to valid. But in the end, you require that $p, q, r$ be distinct primes, and that is a stronger condition than in the problem as posed.

However, the proof you give is very easy to make complete. For example, if $p = r \neq q$ then we have to show $2\sqrt{p} + \sqrt{q}$ is irrational, and for that the squaring proof works easily. And if $p=q=r$ we have to show that $3\sqrt{p}$ is irrational, again trivial.

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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.

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because $(\sqrt{p}+\sqrt{q})(\sqrt{p}-\sqrt{q}) \in \mathbb{Q}$ we have $$ \sqrt{p} + \sqrt{q} \in \mathbb{Q}(\sqrt{r}) \Rightarrow \sqrt{p} - \sqrt{q} \in \mathbb{Q}(\sqrt{r}) \Rightarrow \sqrt{p} \in \mathbb{Q}(\sqrt{r}) $$ so for some $a,b \in \mathbb{Q}$ with $ab\ne 0$ $$ \sqrt{p} = a+ b \sqrt{r} $$ squaring both sides shows this cannot be the case, so the assumption $\sqrt{p} + \sqrt{q} \in \mathbb{Q}(\sqrt{r})$ must be at fault. in particular we cannot have: $$ \sqrt{p}+\sqrt{q}+\sqrt{r} \in \mathbb{Q} $$

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  • $\begingroup$ Of course the first line trick could also be repeated: If $\sqrt p-b\sqrt r=a\in\Bbb Q$, then from $(\sqrt p-b\sqrt r)(\sqrt p+ b\sqrt r)=p-b^2r\in \Bbb Q$, hence $\sqrt p+ b\sqrt r\in\Bbb Q$ and so $\sqrt p\in\Bbb Q$ $\endgroup$ – Hagen von Eitzen May 28 at 13:03
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Suppose $\sqrt{p}+\sqrt{q}+\sqrt{r} = a \to (a-\sqrt{p})^2 = (\sqrt{q}+\sqrt{r})^2\to a^2-2a\sqrt{p} +p =q+r+2\sqrt{qr}\to a^2-q+p-r=2a\sqrt{p}+2\sqrt{qr}\to (a^2-q+p-r)^2=4a^2p+4qr+8a\sqrt{pqr}\to \sqrt{pqr} $ is a rational number, and this is not possible.

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