Proof: Since $(\sqrt{p+r}-\sqrt{p})^{n}(\sqrt{p+r}-$ $\sqrt{p})^{-n}=1$, and by reference [1] we know
$$(\sqrt{p+r}-\sqrt{p})^{n}=\sqrt{m+r^{n}}-\sqrt{m} \text{. }$$

Therefore, $(\sqrt{p+r}-\sqrt{p})^{-n}$
$$=\frac{1}{\sqrt{m+r^{n}}-\sqrt{m}}=\frac{\sqrt{m+r^{n}}+\sqrt{m}}{r^{n}} \text{. }$$

By the extended theorem 3, we easily obtain the following corollary:
For any natural numbers $n$, $p$, $r$, there exists a natural number $m$ such that
$$(\sqrt{p+r}+\sqrt{p})^{n}=\sqrt{m+r^{n}}+\sqrt{m} \text{. }$$