In my spare time, I came up with this conjecture:
For any prime $n$, at least one of $2n^2-1$ , $2n^2+1$, or $\sqrt{2n^2-1}$ is prime.
Examples:
\begin{alignat}{2} 2 &\to\quad 2\cdot 2^2-1 &&\quad=7 \\ 3 &\to\quad 2\cdot 3^2-1 &&\quad=17 \\ 5 &\to\quad \sqrt{2\cdot 5^2-1} &&\quad=7 \\ 7 &\to\quad 2\cdot 7^2-1 &&\quad=97 \\ 11 &\to\quad 2\cdot 11^2-1 &&\quad=241 \\ 13 &\to\quad 2\cdot 13^2-1 &&\quad=337 \\ 17 &\to\quad 2\cdot 17^2-1 &&\quad=577 \end{alignat}
Can somebody prove this?
The conjecture is unfortunately not true.
Proof by counterexample. Consider $n=19$.
We have:
$2\times19^2-1=721=7\times103,$
$2\times19^2+1=723=3\times241,$
$\sqrt{2\times19^2-1}\approx 26.8887,$ which is not an integer hence not prime.
None of $2n^2+1, 2n^2-1, \sqrt{2n^2-1}$ for $n=19$ is prime, but $19$ itself is prime. Therefore the conjecture is not true.