Proof: Let integers $a, b, c, d$ satisfy $n^{2}a$ and $c > a$, leading to $q > p$ and $v > u$. Since $p, q, u, v$ are all integers, we have $q \geqslant p+1, v \geqslant u+1$. Therefore, we get (note $a = p u > n^{2}$)
$$\begin{array}{l} d = q v \geqslant (p+1)(u+1) = p u + (p+u) + 1 \\ > n^{2} + 2 \sqrt{p u} + 1 > n^{2} + 2 n + 1 = (n+1)^{2} \end{array}$$

Contradiction.