Finding all triplets of primes $(p,q,r)$ such that $pq=r+1$ and $2(p^2+q^2)=r^2+1$

Question

Find the number of triplets $(p,q,r)$ such that $p,q,r$ are prime and they satisfy

$$pq=r+1$$ & $$2(p^2+q^2)=r^2+1$$

Answer 1

HINT :

$$pq=r+1\tag 1$$ $$2(p^2+q^2)=r^2+1\tag2$$ The LHS of $(2)$ is even, so $r$ has to be odd. So, the RHS of $(1)$ is even.

It follows from this that we have either $p=2$ or $q=2$.

Answer 2

Hint:-$r$ must be odd form the second equation.Try to prove why.The proof follows.

Answer 3

It is clear that $r$ must be odd so $r+1$ is even. Hence by symmetry we can do $p=2$ so the system $2q=r+1$ and $2(4+q^2)=r^2+1$ determines uniquely the solutions $(q,r)=(3,5)$

Answer 4

From the first equation, $r+1=pq\ge 4$, hence $r\ge 3$, so $r+1$ is even. This implies $p$ or $q$ is an even prime. As the system of equations is symmetric in $p$ and $q$, we may suppose $p=2$. The system becomes $$\begin{cases}r=2q-1\\8+2q^2=4q^2-4q+2\end{cases}\iff\begin{cases}r=2q-1\\q^2-2q-3=0\end{cases}$$ The second equation has $-1$ and $3$ as roots. Hence the solution is $$(p,q,r)=(2,3,5)\enspace\text{or}\enspace (3,2,5).$$