Show that among the numbers $\sqrt{\frac{pq-2}{3}}$, $\sqrt{\frac{qr-2}{3}}$ and $\sqrt{\frac{pr-2}{ 3}}$ there is at least one irrational one.

Question

question

Let the natural numbers $p$, $q$ and $r$ be greater than $2$. Show that among the numbers $\sqrt{\frac{pq-2}{3}}$, $\sqrt{\frac{qr-2}{3}}$ and $\sqrt{\frac{pr-2}{3}}$ there is at least one irrational one.

idea

We can assume that all numbers are rational. If we show a contradiction to this assumption, then we would solve the problem.

The first thing we can do is to notice that $\frac{pq-2}{3}$, $\frac{qr-2}{3}$ and $\frac{pr-2}{3}$ are perfect squares.

In order for them to be perfect squares $pr-2$, $qr-2$, $pr-2$ are of the perfect square form $*3$.

I don't know what I can do next. I hope you can help me! Thank you in advance

Answer 1

If $\frac{pq-2}3$ is a perfect square, then $3\ |\ (pq-2)$ and it follows

$$ p\equiv1\text{ mod }3\quad\text{or}\quad p\equiv2\text{ mod }3.$$

If $p\equiv1\text{ mod }3$, then we need both $q\equiv2\text{ mod }3$ and $r\equiv2\text{ mod }3$, but then $qr-2$ is not divisible by $3$. Similar contradiction goes for $p\equiv2\text{ mod }3$.

Answer 2

If $\frac{pq-2}3,\frac{qr-2}3.\frac{pr-2}3$ are perfect squares, then $$pq\equiv qr\equiv pr\equiv2\text{ mod }3 $$ which implies $$ (pqr)^2\equiv2\text{ mod }3 \tag1$$ Note $pqr=3n,3n+1$, or $n+2$ for some $n\in\mathbb{N}$ and hence $$ (pqr)^2\equiv0, \text{ or }1 \text{ mod }3. \tag2$$ Now (1) and (2) are against each other. So among the numbers $\sqrt{\frac{pq-2}{3}}$, $\sqrt{\frac{qr-2}{3}}$ and $\sqrt{\frac{pr-2}{3}}$ there is at least one irrational one.

Answer 3

Suppose $\sqrt {\frac{pq-2}{3}} \in Q. $

$\frac{pq-2}{3}=\frac{u^2}{v^2} , gcd(u,v)=1.$

$(pq-2)v^2=3u^2$

$gcd(u,v)=1 \implies v^2|3\implies v=1 \land (3\not| pq)$

$pq=3x^2+2$

$pr=3y^2+2$

$qr=3z^2+2$

$p^2q^2r^2=(3x^2+2)(9z^2y^2+6y^2+6z^2+4)=27x^2y^2z^2+18x^2y^2+18x^2z^2+12x^2+18z^2y^2+12y^2+12z^2+8$

$p^2q^2r^2=27x^2y^2z^2+ 18(x^2z^2+y^2z^2+x^2y^2)+ 12(x^2+y^2+z^2)+8$

$(pqr)^2\equiv 8 \equiv 2 \pmod {3}$

$3\not|pqr \implies (pqr)^2 \equiv 1 \pmod {3}$ by Fermat's Little Theorem. Contradiction.

The assumption that all 3 are rational yields a contradiction.