Triples $(p,q,r)$ of primes satisfying the quadratic residue relation $\left(\frac{p}{q}\right)\cdot\left(\frac{q}{r}\right)=\left(\frac{p}{r}\right)$

Question

I stumbled upon this problem while thinking about quadratic residues:

Find all triples $(p,q,r)$ of primes such that $$\genfrac(){}{0}{p}{q}\cdot\genfrac(){}{0}{q}{r} = \genfrac(){}{0}{p}{r}$$

I didn't make much progress beyond some trivial observations.

Answer 1

One can demonstrate infinitely many solutions as follows:-

Let $p$ and $q$ be distinct primes congruent to $3$ modulo $4$. Then $\genfrac(){}{0}{p}{q}$ and $\genfrac(){}{0}{q}{p}$ have opposite signs.

Therefore, for any other odd prime $r$, precisely one of $(p,q,r)$ and $(q,p,r)$ satisfies your rule.