Let $\pi(x)$ be the prime counting function. The problem is,
Find all integral values of $x,y$ such that,
$$\left(\pi(x+y)\right)^2=4\pi(x)\pi(y)$$
I have no idea as to where to begin with. I think that probably there will be inequality after some sufficiently large $x$ and $y$ but I can't prove that.
In fact, it appears that for all sufficiently large $x$ and $y$ we will have $4\pi(x)\pi(y)>(\pi(x+y))^2$ but I can't find the lower bound.
Any help will be appreciated.
For every prime $p$ we have $$ \pi(p+k) = \pi(p + k + \ell), $$ where both $k$ and $\ell$ are positive, and $k + \ell < n_p$, and $n_p$ is defined as $$ \pi(p) + 1 = \pi(p+n_p) = \pi(p') $$
We look for solutions $$ \pi^2(p + k + \ell) = 4 \pi(p + k) \pi(\ell), $$
such that $$ \pi(p + k + \ell) = \pi(p + k) = 4 \pi(\ell) $$
which is easier to solve.
Case $\ell = 2$. $$ \ell=2 \Rightarrow \pi(\ell) = 1 : \pi(p) = 4 \Rightarrow p = 7, n_7 = 4, \ell < 4. $$ So these solutions are $$ (x,y) \in \big\{ (2,7), (2,8), (7,2), (8,2) \big\}. $$
Case $\ell = 3$. $$ \ell=3 \Rightarrow \pi(\ell) = 2 : \pi(p) = 8 \Rightarrow p = 19, n_19 = 4, \ell < 4. $$ So these solutions are $$ (x,y) \in \big\{ (3,19), (19,3) \big\}. $$
Higher solutions do not seem to appear, as $\pi(p)$ grows faster then $n_p$...