Solutions in primes of the equation $\,3p^2+q^2=r^2+3$

Question

Let's consider the Diophantine equation $\,3p^2+q^2=r^2+3$.

Actually, I am interested only in the solutions represented by sets $\,(p,q,r)\,$ of prime numbers.

It's easy to prove that if $\,(p,q)\,$ is a pair of twin primes $\,(p\lt q)$, then $\,(p,q)\,$ solves the equation if $\,p\,$ is a Sophie Germain prime and $\,r\,$ represents its safe prime (that is $\,r=2p+1$).

Similarly, if $\,(p,q)\,$ is a pair of twin primes with $\,p\gt q$, then $\,(p,q)\,$ solves the equation if $\,r=2p-1$.

Therefore, here are some solutions of the given equation:

$(3,5,7)\;\;(5,7,11)\;\;(11,13,23)\;\;(29,31,59)\;\;(41,43,83)\;\;...$

$(7,5,13)\;\;(19,17,37)\;\;(31,29,61)\;\;...$

In order to find other classes of solutions, let's force $\,r=p+2q$. This constraint leads to consider only the pairs $\,(p,q)\,$ which satisfy the following condition:

$5(q^2+1)=2((p-q)^2+1)\;\;\;\;\;\;\;\;\;(*)$

A solution satisfing $\,(*)\,$ is $\,(13,5,23)$.

---

Other solutions are:

$(29,17,53)\;\;(41,17,73)\;\;(59,13,103)\;\;(61,17,107)\;\;(71,107,163)\;\;(79,7,137)\;\;(89,53,163)$

The previous solutions have been found observing that the equation can be written

$$3(p^2-1)=r^2-q^2$$

So, $\,72\,$ always divide $r^2-q^2$.

---

I ask to find other, possibly more general, classes of solutions of the given equation.

Answer 1

Generalizing in another way from the solution $(3,5,7)$, there is also a class of solutions in which $(q,r)$ is a pair of twin primes ($q<r$). A parametric solution with $r=q+2$ is:

$$(p,q,r)=(2m+1,3m^2+3m-1,3m^2+3m+1)$$

Values of $m$ which yield solutions in primes include $1,2,9,30$ giving respectively:

$(3,5,7)\;\;(5,17,19)\;\;(19,269,271)\;\;(61,2789,2791)$