Show that if $p$ is the smallest prime factor of $n$ then $x^2-n$ will not be a perfect square for $x>(n+p^2)/2p$ except $x=(n+1)/2$

Question

Don't really understand the solution

How do we know $x+a>n/p$?

Answer 1

Since $a>\frac{n-p^2}{2p}$ and $x>\frac{n+p^2}{2p}$, we have that$$x+a>\frac{n-p^2}{2p}+\frac{n+p^2}{2p}=\frac np.$$