Solution
= Solution
Since $N=3p$, its nontrivial difference-of-squares representation is
$$
3p=\left(\frac{p+3}{2}\right)^2-\left(\frac{p-3}{2}\right)^2.
$$
As $p$ is prime, every earlier square difference would give another nontrivial factorization, which is impossible. Thus the first successful value is $r=(p+3)/2$, and
$$
\boxed{k=\frac{p+3}{2}-\lfloor\sqrt{3p}\rfloor.}
$$