There exist infinitely many primes of the form $p = \lfloor\sqrt{n} + \sqrt{n + 1}\rfloor$.

Question

Is it difficult to prove this existence problem? There are infinitely many $n \in \Bbb{N}$ such that :

$$ p = \lfloor \sqrt{n} + \sqrt{n+1} \rfloor $$ is a prime number.

Attempt:

Every odd prime $p$ can be written $p = \lfloor{\dfrac{p}{2}} \rfloor + \lfloor \dfrac{p}{2} + 1\rfloor = \lfloor\sqrt{(\dfrac{p}{2})^2} \rfloor +\lfloor \sqrt{(\dfrac{p}{2} + 1)^2} \rfloor$

Answer 1

For any odd prime $p=2k+1$, let $n=(k+1)^2-1$.

Answer 2

Letting $\sqrt{n} + \epsilon = \sqrt{n+1}$ for some $\epsilon > 0$, squaring gives $n + 2\epsilon \sqrt{n} + \epsilon^{2} = n + 1$ therefore the equation $\epsilon^{2} + 2\epsilon \sqrt{n} - 1 = 0$ has solution $\epsilon = \sqrt{n} - \sqrt{n - 1}$

For $n > 1$ this difference will always be less than 1, thus you arrive at the result that $\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2k-1$ when $n+1 = k^2$