(1) The digit \(a\cdot n\) changes by \(a_i\) and \(b\cdot n\) by \(\bar b_i\) modulo \(q\), so \(\Delta _i\equiv qa_i+\bar b_i-q\omega =\frac13q^2+\frac13qw_i+\bar b_i-q\omega \) modulo \(q^2\), with \(\omega \in \{ 0,1\} \) the wrap of the third digit (a wrap of \(a\cdot n\) changes \(\Delta _i\) by \(q^2\)). (2) Along a cycle the third digit takes the values \(b\cdot n+t\bar b_i\bmod q\), \(t=0,\dots ,q-1\), and returns after a total increase \(q\bar b_i\), so it wraps exactly \(\bar b_i\) times and does not wrap \(q-\bar b_i\ge 1\) times; this also shows that both cases of (1) occur if \(\bar b_i\ne 0\). Under the range condition the representatives \(\frac13q^2+\varepsilon _i\) and \(\frac13q^2+\varepsilon _i-q\) of (1) lie in the window \((-\frac16q^2,\frac56q^2]\), so the excesses are \(\varepsilon _i\) and \(\varepsilon _i-q\), and those of the cycle sum to \(q\varepsilon _i-q\bar b_i=\frac13q^2w_i\). (Without it the identity can fail: for \(q=5\), \(w=(-8,4,4)\) and \(b=(0,1,-1)\) the excesses of a cycle in the direction \(\mathbf{e}_x\) sum to \(\frac{175}3\), not \(-\frac{200}3\).) (3) Then \(-\frac53q\le \varepsilon _i-q{\lt}\varepsilon _i\le \frac53q\), and \(\frac53q\le \frac16q^2\) for \(q\ge 10\), so every excess lies in \([-\frac16q^2,\frac13q^2]\) and the largest one is \(\max _i\varepsilon _i\) (attained, since the third digit does not wrap on some step of each cycle, by (2)). (4) As \(w_x+w_y+w_z=0\), \(\sum _i\varepsilon _i=\sum _i\bar b_i\equiv \sum _ib_i\equiv 0\pmod q\), and \(\bar b\ne 0\): otherwise \(b\equiv 0\) and the box with depth \(3\) on a coordinate, whose minor is a \(2\times 2\) minor of \((a,b)\) (Lemma 2), would be singular. So \(\sum _i\bar b_i\ge q\) and \(\max _i\varepsilon _i\ge \frac13q\). For \(q\equiv 1\pmod3\), \(w_c\equiv 2\pmod3\) and \(\sum _cw_c=0\), so some \(w_c\ge 2\) and \(\varepsilon _c\ge \frac23q\).

If \(g{\gt}\frac13q^2\), every step value \(\Delta _i([n])\) has distance more than \(\frac13q^2\) from \(\frac23q^2+q^2\mathbb {Z}\), so its residue modulo \(q^2\) lies in the open interval \((0,\frac13q^2)\). The three steps \([n]\to [n+\mathbf{e}_x]\to [n+\mathbf{e}_x+\mathbf{e}_y]\to [n+\mathbf1]=[n]\) have step values that sum to \(0\), so their residues sum to a multiple of \(q^2\); but that sum lies in \((0,q^2)\).

(1) The point of cell \(n\) has the digits \(n_c\), \(\lfloor \varphi /q\rfloor \), \(\varphi \bmod q\) in every coordinate. The box of shape \((1,1,1)\) holds one point per cell. A box with depth \(3\) on \(c\) prescribes \(n_c\) and \(\varphi \); the \(q^2\) cells with a given \(n_c\) meet every column exactly once (\(n+t\mathbf1\) for the one \(t\) with \(n_c+t\) given), so these boxes hold one point each if and only if \(\varphi \) is a bijection. A box with depth \(2\) on \(c\), \(1\) on \(c'\) and \(0\) on \(c''\) prescribes \(n_c\), \(n_{c'}\) and \(\lfloor \varphi /q\rfloor \); the \(q\) cells with given \(n_c,n_{c'}\) form a line in the direction \(\mathbf{e}_{c''}\), whose columns form a line of \(\mathbb T\) in the direction \([\mathbf{e}_{c''}]\), and every such line of \(\mathbb T\) arises. These are all box shapes.

(2) Identify \(\mathbb T\) with \((\mathbb {Z}/q\mathbb {Z})^2\) by \([n]\mapsto (n_x-n_z,n_y-n_z)\); the lines in the directions \([\mathbf{e}_x]\), \([\mathbf{e}_y]\), \([\mathbf{e}_z]\) become the horizontal lines, the vertical lines and the lines of slope \(1\) of \((\mathbb {Z}/q\mathbb {Z})^2\). A level set \(\{ \lfloor \varphi /q\rfloor =v\} \) has \(q\) elements and meets each of these lines once, by (1). So it is the graph of a permutation \(\pi \) of \(\mathbb {Z}/q\mathbb {Z}\) (once on every horizontal and every vertical line) such that \(y\mapsto \pi (y)-y\) is a permutation too (once on every line of slope \(1\)). Then \(\sum _y(\pi (y)-y)\equiv 0\), while the sum over all residues is \(\frac q2\not\equiv 0\pmod q\) for even \(q\). In other words, \(y\mapsto \pi (y)-y\) is a complete mapping of \(\mathbb {Z}/q\mathbb {Z}\), and this is the classical argument that a cyclic group of even order has none [ 48 , Thm. 1 and its corollary ] ; more generally, no finite group with a nontrivial cyclic Sylow \(2\)-subgroup has one [ 29 , Thm. 5 ] . In the language of Latin squares, (1) says that each level set is a transversal of the cyclic Latin square \((a,b)\mapsto b-a\), and that the array \(\lfloor \varphi /q\rfloor \) on \((\mathbb {Z}/q\mathbb {Z})^2\) is a Latin square orthogonal to it.

(3) Put \(\eta =\frac13q^2-g\). By Corollary 96, \(0\le \eta \le \frac13q^2\), and by Theorem 50 \(d^2=q^4-2q^2\eta +3\eta ^2\), which decreases in \(\eta \) on this interval and equals \(d^2(\mathcal C_q)\) at \(\eta _0=\frac13q+1\) (\(q\equiv 2\)) resp. \(\eta _0=\frac23q\) (\(q\equiv 1\)). So it suffices to show \(\eta \ge \eta _0\). Suppose \(\eta {\lt}\eta _0\).

Then \(\eta {\lt}\frac1{12}q^2\). For \(q\ge 8\) this holds because \(\eta _0\le \frac1{12}q^2\). For \(q=5\) and \(7\): \(3g=|3U+q^2|\) for an integer \(U\), which is not divisible by \(3\) since \(q^2\equiv 1\pmod3\), so \(3\eta =q^2-3g\equiv 0\) or \(2\pmod3\); hence \(\eta {\lt}\frac83\) gives \(\eta \le 2{\lt}\frac{25}{12}\) at \(q=5\), and \(\eta {\lt}\frac{14}3\) gives \(\eta \le 4{\lt}\frac{49}{12}\) at \(q=7\).

Windows and triangles. Every step value has distance at least \(\frac13q^2-\eta \) from \(\frac23q^2+q^2\mathbb {Z}\), so it has a unique representative \(u\) in the window \([-\eta ,\frac13q^2+\eta ]\) modulo \(q^2\) (narrower than the window of Section 11.1). As \([\mathbf{e}_x]+[\mathbf{e}_y]+[\mathbf{e}_z]=0\) in \(\mathbb T\), for every column and every order of the three directions the three steps around the triangle \([n]\to [n+\mathbf{e}_i]\to [n+\mathbf{e}_i+\mathbf{e}_{i'}]\to [n]\) have values summing to \(0\), so their representatives sum to a multiple of \(q^2\) in \([-3\eta ,q^2+3\eta ]\), i.e. to \(0\) or \(q^2\). In a triangle with sum \(0\) every \(u\) lies in \([-\eta ,2\eta ]\), and in one with sum \(q^2\) every \(u\) lies in \([\frac13q^2-2\eta ,\frac13q^2+\eta ]\); as \(\eta {\lt}\frac1{12}q^2\) these ranges are disjoint, so two triangles that share a step have the same sum. The step \(([n],i)\) shares a triangle with \(([n+\mathbf{e}_i],i')\) for both \(i'\ne i\), and two such moves give \(([n+\mathbf{e}_i+\mathbf{e}_{i'}],i)=([n-\mathbf{e}_{i''}],i)\); and the classes \([\mathbf{e}_{i''}]\), \(i''\ne i\), generate \(\mathbb T\); so all steps are connected through shared triangles, and all triangles have the same sum. Sum \(q^2\) means that every step is close to \(\frac13q^2\), the configuration of the reflected cube \(H\mathbb {Z}^3\); sum \(0\) means that every step is close to \(0\), the configuration of \(\mathbb {Z}^3\).

Sum \(q^2\) (the reflected cube). Put \(\xi =u-\frac13q^2\in [-2\eta ,\eta ]\), the excess; then \(\operatorname {dist}(u,\frac23q^2+q^2\mathbb {Z})=\frac13q^2-\xi \), so \(\eta \) is the largest excess, and every triangle has excess sum \(0\). The class \([\mathbf{e}_i]\) has order \(q\) in \(\mathbb T\), so the steps in the direction \(\mathbf{e}_i\) form \(q\) cycles \(\gamma \) of length \(q\). Along a cycle the step values telescope to \(0\), so \(\sum _{\gamma }u\in q^2\mathbb {Z}\), and the cycle excess \(\Xi _\gamma =\sum _\gamma \xi \in q^2\mathbb {Z}-\frac13q^3\). Summing the excess over the triangles \([n]\to [n+\mathbf{e}_x]\to [n+\mathbf{e}_x+\mathbf{e}_y]\to [n]\) for all \([n]\in \mathbb T\) counts every step once, so the \(3q\) cycle excesses sum to \(0\). Also \(-2q\eta \le \Xi _\gamma \le q\eta \). If \(q\equiv 1\pmod3\), then \(\Xi _\gamma =q^2z_\gamma -\frac13q^2\) with integers \(z_\gamma \) (as \(\frac13q^3\equiv \frac13q^2\)), and the cycle excesses sum to \(0\) only if \(\sum _\gamma z_\gamma =q\). So some \(z_\gamma \ge 1\), \(\Xi _\gamma \ge \frac23q^2\), and \(q\eta \ge \frac23q^2\): \(\eta \ge \frac23q\), a contradiction. If \(q\equiv 2\pmod3\), then \(\Xi _\gamma =q^2z_\gamma -\frac23q^2\) and \(\sum _\gamma z_\gamma =2q\). Every excess lies in \(\mathbb {Z}+\frac23\) (as \(u\in \mathbb {Z}\) and \(\frac13q^2\in \mathbb {Z}+\frac13\)), and so does \(\frac13q\); hence \(\eta {\lt}\frac13q+1\) gives \(\eta \le \frac13q\), and then every \(\Xi _\gamma \in [-\frac23q^2,\frac13q^2]\), i.e. \(\Xi _\gamma \in \{ -\frac23q^2,\frac13q^2\} \). A cycle excess \(\frac13q^2\le q\eta \) forces \(\xi =\eta =\frac13q\) on each step of its cycle, and a cycle excess \(-\frac23q^2\ge -2q\eta \) forces \(\xi =-2\eta =-\frac23q\) on each step. So every step has \(u=\frac13q(q+1)\) or \(\frac13q(q-2)\), a multiple of \(q\), and \(\varphi \) is constant modulo \(q\) on \(\mathbb T\). This contradicts (1): \(\varphi \) takes every value in \(\{ 0,\dots ,q^2-1\} \).

Sum \(0\) (the cube). Now \(u\in [-\eta ,2\eta ]\) and \(\operatorname {dist}(u,\frac23q^2+q^2\mathbb {Z})=u+\frac13q^2\), so \(\min u=-\eta \). The cycle sums \(\sum _\gamma u\) lie in \(q^2\mathbb {Z}\) and add up to \(0\). If one of them is nonzero, one is at most \(-q^2\); it is at least \(-q\eta \), so \(\eta \ge q\). Otherwise every cycle closes: the partial sums \(S_t\) of \(u\) along a cycle are integers with \(S_t\equiv \varphi (c_t)-\varphi (c_0)\pmod{q^2}\), where \(c_t\) are its columns. By (1) the \(q\) values \(\varphi (c_t)\) lie in \(q\) different blocks \([kq,kq+q)\), so the \(q\) values \(\varphi (c_0)+S_t\) meet all \(q\) blocks modulo \(q^2\); an interval of integers of length \(\mathrm{rg}\) meets at most \(\mathrm{rg}/q+2\) of them, so the range of the walk is \(\mathrm{rg}\ge q(q-2)\). Here \(\eta {\gt}0\), since \(\eta =0\) would make every step value \(0\) and \(\mathrm{rg}=0{\lt}q(q-2)\). A closed walk whose up-steps are at most \(2\eta \) and whose down-steps are at least \(-\eta \) needs at least \(\mathrm{rg}/(2\eta )+\mathrm{rg}/\eta \) steps, so \(q\ge 3\, \mathrm{rg}/(2\eta )\) and \(\eta \ge \frac32(q-2)\). Both \(q\) and \(\frac32(q-2)\) are at least \(\frac23q\ge \eta _0\) for \(q\ge 4\), a contradiction.

Equality holds for \(\mathcal C_q\) by Theorem 51(2).