How many primes $p$ are there such that $2p^{3} + 206$ is a perfect square?
My approach: Let the square be $k^{2}$, then
$2p^{3} + 206 = k^{2}$ $2p^{3}=k^{2} - 206$
$2p^{3}=(k+√206)(k-√206)$
Now, let $(k+√206)=p^{3}; (k-√206)=2$ and
$(k-√206)=p^{3}; (k+√206)=2$
But I couldn't solve it further. Are there no such primes? Am I on the right track? Please help.
I got $19$ by hit-and-trial. Is there any analytic way?