Let $n\in \mathbb{N}$ be the smallest natural number such that $p < n(n+1)$. Then $p\ge (n-1)n$. Therefore \begin{align*} \left(2 + \sqrt{p+\frac{1}{4}}\right)^2 &\ge \left(2 + \sqrt{n(n-1)+\frac{1}{4}}\right)^2\\ &= \left(2 + \sqrt{(n-\frac{1}{2})^2}\right)^2\\ &=\left(n+\frac{3}{2}\right)^2\\ & \ge (n+1)(n+2) \end{align*} So we have $p < n(n+1) < (n+1)(n+2) < \left(2 + \sqrt{p+\frac{1}{4}}\right)^2$. Which implies $p < n(n+1) < n(n+2) < (n+1)^2 < (n+1)(n+2) < \left(2 + \sqrt{p+\frac{1}{4}}\right)^2$ Then letting $a=n(n+1)$, $b=(n+1)(n+2)$, $c=n(n+2)$ and $d=(n+1)^2$, we have four distinct natual numbers such that $ab=cd$, as desired $\square$