Find all primes such that,
$\large \color{black}{\begin{align}
y^2-4\pmod x\equiv 0\quad \text{and}\hspace{.33em}\~\
x^2-1\pmod y\equiv 0 \quad \text{where} \quad \{x,y\}\in\mathbb{P}\hspace{.33em}\~\
\end{align}}$

Question

Find all primes such that,
$\large \color{black}{\begin{align} 
y^2-4\pmod x\equiv 0\quad    \text{and}\hspace{.33em}\~\
x^2-1\pmod y\equiv 0 \quad \text{where} \quad \{x,y\}\in\mathbb{P}\hspace{.33em}\~\
\end{align}}$

anonymous · Answer

well the problem wants all of prime pairs like $(x,y)$ such that

$x|y^2-4$ and $y|x^2-1$

mathmate · Answer

Here are exhaustive brute force results for 2<x,y<1000, but have no idea where to go from here (5,3),(15,7)

anonymous · Answer

first suppose that $x \le y$, since $y|(x-1)(x+1)$  and $y>x-1$ it follows that $y|x+1$ and so $y=x+1$, there is only one pair of consecutive primes, so we get from here one solution and that is $(x,y)=(2,3)$

anonymous · Answer

now going with $x>y$ 
Oh, I'm late, I must go somewhere, I'll finish this later ;-))

mathmath333 · Answer

@mathmate 
$\large \color{black}{
(5,3) \normalsize\text{ is a valid pair  but not } \hspace{.33em}\~\
 (15,7) \normalsize\text{ as } \large 15 \normalsize\text{ is composite. }\hspace{.33em}\~\
}$

mathmath333 · Answer

mukushla's start is nice but,


$\large \color{black}{\begin{align} 
3^2-4\pmod 2\cancel{\equiv}  0\hspace{.33em}\~\
\end{align}}$ for the pair of $\large (2,3)$

mathmate · Answer

Oops! Thank you @mathmath333