Why is $p$ necessarily greater than $r$ in this number theory problem? From 1998 St. Petersburg City Mathematical Olympiad, presented in Andreescu & Andrica NT: SEP:

Let $n$ be a positive integer. Show that any number greater than $n^4/16$ can be written in at most one way as the product of two of its divisors having difference not exceeding $n$.

The presented solution is this:

Suppose, on the contrary, that there exist $a > c \ge d > b$ with $a-b \le n$ and $ab=cd>n^4/16$. Put $p=a+b, q=a-b, r=c+d,s=c-d.$ Now $$p^2-q^2=4ab=4cd=r^2-s^2>n^4/4.$$ Thus $p^2-r^2=q^2-s^2 \le q^2 \le n^2.$ But $r^2>n^4/4$ (so $r>n^2/2$) and $p>r\dots$

There is more to the solution, but that is irrelevant to my question. Why is $p>r$? It seems that this should be obvious, the way that it is presented. I notice that $p > r \Leftrightarrow p^2-r^2 > 0$, but I cannot prove that this is true. Manipulating the chain inequality $a>c\ge d > b$ has also done nothing for me.

We have $a>c\ge d>b$, and $p=a+b$, $r=c+d$, $q=a-b$, $s=c-d$. These are all positive (as the question deals with only one sign for the divisors for uniqueness of solution, so we take the $+$ve route), although with the one exception of $s$ which could equal zero if $c=d$.
We cannot have $p<r$, since then $p^2-r^2<0$ which contradicts 
$$(p^2-r^2)-(q^2-s^2)=4ab-4cd=0\tag{1}$$
If $p=r$ then $p^2-r^2=0$, which by $(1)$ means $q^2-s^2=0$, and so $q=s$ also. Hence
\begin{align*}
a+b&=c+d\tag{2}\\
a-b&=c-d\tag{3}
\end{align*}
or $a+b-c-d=a-b-c+d=0$. Adding/subtracting $(2)$ and $(3)$ gives either $a=c$ or $b=d$ respectively, contradicting $a>c\ge d>b$, and so we must have $p>r$.
