The odd square-plane conjecture for transversal difference numbers

Let $p$ be an odd prime, and set

$$G_p=(\mathbb Z/p^2\mathbb Z)^2,\qquad H_p=pG_p.$$

For a transversal $T$ of $G_p/H_p$, write $D(T)=T-T$ and let

$$\delta(G_p,H_p)=\min_T |D(T)|.$$

Every normalized transversal has the form $T_f=\{[x]+p[f(x)]:x\in\mathbb F_p^2\}$ for a function $f:\mathbb F_p^2\to\mathbb F_p^2$ with $f(0)=0$. For $u,x\in\mathbb F_p^2$, define

$$c_u(x)=\frac{[x+u]-[x]-[u]}{p},\qquad d_u f(x)=f(x+u)-f(x)+c_u(x),$$

and $A_u(f)=\{d_u f(x):x\in\mathbb F_p^2\}$.

Odd square-plane conjecture. For every odd prime $p$,

$$\delta\bigl((\mathbb Z/p^2\mathbb Z)^2,p(\mathbb Z/p^2\mathbb Z)^2\bigr)=(2p-1)^2.$$

Equivalently, for every function $f:\mathbb F_p^2\to\mathbb F_p^2$,

$$\sum_{u\in\mathbb F_p^2}|A_u(f)|\geq (2p-1)^2.$$

The coordinate box attains $(2p-1)^2$. The paper proves the unconditional lower bound $3p^2-p-1$, leaving a gap of $(p-1)(p-2)$; small-prime computations, probabilistic arguments, and fixed-polynomial evidence support the conjecture, but arbitrary $p$-dependent liftings remain unresolved.