Let $p$ be a sufficiently large prime number. Show that the number of ordered pairs of positive integers $(a,b,c,d)$ satisfying \[ab+cd=p\]does not exceed $\frac{3p^2}8$.
I did not see how $p$ being prime helps, and simply did \[T_p:=\#\{(a,b,c,d)\in\Bbb N_+^4\mid ab+cd=p\}=\sum_{i=1}^{p-1}\tau(i)\cdot\tau(p-i),\] where $\tau(\cdot)$ counts the number of positive factors.
By using the estimation $\tau(n)\le16\sqrt[3]n$, it suffices to prove \[T_p\le256\sum_{i=1}^{p-1}\sqrt[3]{pi-i^2}\le\frac{3p^2}8.\]
I'm not sure if it is strong enough.
Edit: I've solved it with an easy method.
By Hölder inequality, \begin{align*}\small 256\sum_{i=1}^{p-1}\sqrt[3]i\cdot\sqrt[3]{p-i}\cdot\sqrt[3]1&\small\le256\sqrt[3]{\textstyle\left(\sum\limits_{i=1}^{p-1}i\right)\left(\sum\limits_{i=1}^{p-1}(p-i)\right)\sum\limits_{i=1}^{p-1}1}\\ &\small=256\sqrt[3]{\frac{p^2(p-1)^3}4}, \end{align*} which is less than $\frac{3p^2}8$ for sufficiently large $p$.
Yoichi Motohashi, The binary additive divisor problem, Annales scientifiques de l’É.N.S. 4e série, tome 27, no 5 (1994), p. 529-572, attributes to Ingham the result, $$ D(N)=\sum_{n=1}^{n-1}d(n)d(N-n)=(1+o(1)){6\over\pi^2}\sigma(N)(\log N)^2 $$ The Ingham citation is A.E. INGHAM, Some Asymptotic Formulae in the Theory of Numbers (J . London Math. Soc., Vol. 2, 1927, pp. 202-208). Motohashi may go on to improve on this, but I didn't look any further. The Motohashi paper is available at http://www.numdam.org/item/10.24033/asens.1700.pdf
Of course, in the case where $N$ is prime, we have $\sigma(N)=N+1$.