Legendre's Conjecture states that there is a prime between every two perfect squares. It's almost certainly true, but remains unproven.
I conjecture that not only is that true, but for every $a\in\mathbb N$, there exists at least one $b\leq 2a$, so that $a^2<a^2+b<(a+1)^2$, such that $a^2+b$ and $a^2+b^2$ are prime. Empirically, this also appears almost certain.
I wanted to share that, but am also curious whether there's any special reason for it and will consider this post answered if anyone can find a counterexample, or give some reasoning on why this is (or is not) likely to be true.
This conjecture is almost certainly true (at least, for all sufficiently large $a$), and definitely out of reach of current technology, just like Legendre's conjecture. In fact, something much stronger is probably true (again, just like Legendre's conjecture): instead of taking $b\in [1,2a)$ you could take $b\in [1,C(\log a)^3]$, for some suitable constant $C>0$. (at least for all but finitely many $a$).
Here's a rough heuristic. The letter $c$ will denote some constant that might change between each occurrence, and depend on previous constants, but the precise value isn't really important.
Fix $a\in \mathbb{N}$. For each $b\in [0,2a)$ the heuristic 'chance' of $a^2+b$ being prime is, by the prime number theorem, $\approx 1/\log(a^2+b)\approx c/\log(a)$. The chance of $a^2+b^2$ being prime is $\approx 1/\log(a^2+b^2)\approx c/\log(a)$. Heuristically we assume these events are independent, so the chance of both being true for any $b\in[1,2a)$ is $\approx c/(\log a)^2$.
Again, for different $b$ these chances are heuristically independent, so we have the chance of all $b\in [1,X]\subset [1,2a)$ failing is
$$\approx (1-c/(\log a)^2)^{X} \approx e^{-cX/(\log a)^2}.$$
If we take $X=C(\log a)^3$ for some suitable constant $C>0$, then this chance is $\leq 1/a^2$, say, and in particular
$$\sum_{a\in \mathbb{N}}\mathbb{P}(\textrm{no suitable }b\in[1,C(\log a)^3])$$
converges, and hence there are only finitely many $a$ which 'fail'.
(This heuristic is of course obviously flawed in that it doesn't take small congruence obstructions into account - e.g. if $a$ is even then all even $b$ automatically fail, but correcting for such obstructions will have a negligible effect on the overally behaviour.)