Finding $n$ elements of $\mathbb{Z}_n\times\mathbb{Z}_n$ such that their differences are all different

Question

Let $n\geq 3$ be an integer and consider the group $\mathbb{Z}_n\times\mathbb{Z}_n$ under addition.

Question: Does there always exist a choice of $n$ elements $$ (a_1,b_1),\dots,(a_n,b_n)\in\mathbb{Z}_n\times\mathbb{Z}_n $$ in the group such that the set of differences $$ S = \bigl\{(a_i,b_i) - (a_j,b_j)\, \big|\, i,j\in\{1,\dots,n\} \text{ and }i\neq j\bigr\} $$ contains $n(n-1)$ distinct elements?

I've been able to find solutions for $n$ up to 7, but not enough of a pattern emerges for me to be able to figure out how to generalize it to all $n\geq3$.

• In the case when $n=3$, we may choose the elements $(0,0),(1,0),(0,1)\in\mathbb{Z}_3\times\mathbb{Z}_3$. To see that this choice has the desired property, construct a table containing $(a_i,b_i) - (a_j,b_j)$ in the $ij$ entry for each pair of indices $i,j\in\{1,\dots,n\}$ with $i\neq j$: \begin{array}{rr|ccc} &&(a_i,b_i)\\ & +& (0,0) & (1,0) & (0,1)\\ \hline -(a_j,b_j)&(0,0) & \cdot & (1,0) & (0,1)\\ &(2,0) & (2,0) & \cdot & (2,1)\\ &(0,2) & (0,2) & (1,2) & \cdot \end{array} It is clear that the off-diagonal entries of this table are all different, and thus $\lvert S\rvert = 6 = 3(3-1)$.

• In the case when $n=4$, we may choose the elements $(0,0),(1,0),(0,1),(3,3)\in\mathbb{Z}_4\times\mathbb{Z}_4$. A similar table may be constructed to show that this choice also has the desired property: \begin{array}{rr|cccc} &&(a_i,b_i)\\ & +& (0,0) & (1,0) & (0,1) & (3,3)\\ \hline -(a_j,b_j)&(0,0) & \cdot & (1,0) & (0,1) & (3,3)\\ &(3,0) & (3,0) & \cdot & (3,1) & (2,3)\\ &(0,3) & (0,3) & (1,3) & \cdot & (3,2)\\ &(1,1) & (1,1) & (2,1) & (1,2) & \cdot \end{array} The off-diagonal entries are all unique.

• For $n=5$, we can choose $(0,0), (2,1), (1,2), (0,2), (2,0)\in\mathbb{Z}_5\times\mathbb{Z}_5$.

• For $n=6$, we can choose $(0,0), (2,1), (1,2), (0,2), (2,0), (5,5)\in\mathbb{Z}_6\times\mathbb{Z}_6$.

• For $n=7$, we can choose $(0,0), (2,1), (1,2), (0,5), (5,0), (1,5),(5,1)\in\mathbb{Z}_7\times\mathbb{Z}_7$.

Answer 1

We shall call a natural number $n$ good provided a group $\Bbb Z_n\times\Bbb Z_n$ has a Sidon set of size $n$. In this joriki proved that each odd prime is good.

We claim that each odd square-free number is good. To show this it suffices to show that a product of any two coprime good numbers $n$ and $m$ is good. It is easy to see that an element $(1,1)$ of a product $\Bbb Z_m\times Z_n$ has order $mn$, so $\Bbb Z_m\times Z_n$ is isomorphic to a cyclic group of order $mn$, that is to $\Bbb Z_{mn}$. It remains to remark that if $A_m$ is a Sidon set in $\Bbb Z_m\times \Bbb Z_m$ and $A_n$ is a Sidon set in $\Bbb Z_n\times\Bbb Z_n$ then $A_m\times A_n$ is a Sidon set in $(\Bbb Z_m\times \Bbb Z_m)\times (\Bbb Z_n\times \Bbb Z_n)\simeq \Bbb Z_{mn}\times\Bbb Z_{mn}$.

Now we claim that the group $G=\Bbb Z_8\times \Bbb Z_8$ has no Sidon set $A$ of size eight. Indeed, suppose to the contrary that such a set $A$ exists. Let $H$ be a subgroup of $G$ consisting of elements with both coordinates even. We claim that there exist elements $a$ and $b$ of $A$ such that $a-b\in H+(1,1)$. Indeed, otherwise $A$ is contained in two cosets of $G/H$, and then $A-A$ is contained in three cosets of $G/H$, which follows that $57=|A-A|\le 48$, a contradiction. Put $A’=A-a$. Then $A$ contains an element $b-a=(c_1,c_2)$ with both coordinates odd. There exists natural odd numbers $d_1$ and $d_2$ such that both $d_1c_1$ and $d_2c_2$ equal $1$ modulo $8$. Then a map of $\Bbb Z_8\times \Bbb Z_8$, $(x_1,x_2)\mapsto (d_1x_1,d_2x_2)$ is an isomorphism of $G$, which maps the set $A’$ into a Sidon set $A’’$ containing $(0,0)$ and $(1,1)$. Let $G_2$ be a subgroup of $G$ consisting of elements of order two, that is $G_2=\{(0,0), (0,4), (4,0), (4,4)\}$. Then $x=-x$ for each $x\in G_2$. There are no distinct elements $a,b\in A’’$ such that $a-b\in G_2$, because otherwise $b-a=-(a-b)=a-b$, which contradicts that $A’’$ is a Sidon set. Therefore each of sixteen cosets of $G/G_2$ contains at most one element of $A’’$, and the elements of cosets $(0,0)+G_2$ and $(1,1)+G_2$ are already fixed. So it remains to check all choices of six among fourteen remaining cosests containing elements of $A’’$ and for each chosen coset four possibilities to choose an element for $A’’$. More subtle arguments can decrease a number of cases to check even more, but already these cases were checked by a short Pascal program below at my old computer in less than a half a minute. It found a Sidon set $\{(1,1), (2,0), (2,1),(0,5),(5,0),(4,2),(4,7)\}$ of size seven but no Sidon set of size eight.

program p3491165;
label
 0;
var
 a,b:array[0..7,0..1]of Byte;
   c:array[0..13,0..1]of Byte;
   d:array[0..7,0..7]of Boolean;
 i,j:Word;
 sum,k,l,d0,d1:Byte;
OFi:Text;
begin
c[0,0]:=0;c[0,1]:=1;
c[1,0]:=0;c[1,1]:=2;
c[2,0]:=0;c[2,1]:=3;
c[3,0]:=1;c[3,1]:=0;
c[4,0]:=1;c[4,1]:=2;
c[5,0]:=1;c[5,1]:=3;
c[6,0]:=2;c[6,1]:=0;
c[7,0]:=2;c[7,1]:=1;
c[8,0]:=2;c[8,1]:=2;
c[9,0]:=2;c[9,1]:=3;
c[10,0]:=3;c[10,1]:=0;
c[11,0]:=3;c[11,1]:=1;
c[12,0]:=3;c[12,1]:=2;
c[13,0]:=3;c[13,1]:=3;
assign(OFi,'3491165.txt');
rewrite(OFi);
b[7,0]:=0;b[7,1]:=0;
b[6,0]:=1;b[6,1]:=1;
for i:=0 to 16363 do begin
 sum:=0;
 for k:=0 to 13 do inc(sum, 1 and (i shr k));
 if sum<>6 then Continue;
 l:=0;
 for k:=0 to 13 do if (1 and (i shr k))=1 then begin
  a[l,0]:=c[k,0];
  a[l,1]:=c[k,1];
  inc(l);
 end;

 for j:=0 to 4095 do begin
  for k:=0 to 5 do begin
   b[k,0]:=a[k,0]+4*((j shr (2*k))and 1);
   b[k,1]:=a[k,1]+4*((j shr (2*k+1))and 1);
  end;
  FillChar(d,SizeOf(d),0);
  for k:=0 to 7 do for l:=0 to 7 do begin
   if k=l then Continue;
   d0:=(8+b[k,0]-b[l,0]) mod 8;
   d1:=(8+b[k,1]-b[l,1]) mod 8;
   if d[d0,d1] then goto 0 else d[d0,d1]:=True;
  end;
  for k:=0 to 7 do writeln(OFi,b[k,0],' ',b[k,1]);
0:
 end;

end;
close(OFi);
end.

Answer 2

A set $S\subset \Bbb Z/n\times \Bbb Z/n$ as in the OP will be called an $n$-set, to have a quick terminology. (It will be useful below to consider $\Bbb Z/n$ as a ring, then the two copies of it as a module over it, so matrix operations may be useful, see the analysis of the special case $\Bbb Z/8$ below. It turns out, there is no $8$-set.)

---

Let us show two lemmas first:

Lemma A: Let $m,M$ be two relatively prime numbers. Let $s$ be an $m$-set, and $S$ be an $M$-set. We identify $\Bbb Z/m$ with $M\Bbb Z/mM$, and
$\Bbb Z/M$ with $m\Bbb Z/mM$, and move $s,S$ inside $\Bbb Z/mM$.

Let $(a_i,b_i)$, $1\le i\le m$, be the elements of $s$, moved inside $\Bbb Z/mM$.

Let $(A_i,B_i)$, $1\le I\le M$, be the elements of $S$, moved inside $\Bbb Z/mM$.

Condider the set with the elements $$ (a_i,b_i)+(A_I,B_I)\ ,\ 1\le i\le m\ , \ 1\le I\le M\ . $$ Then this set is an $mM$-set.

Proof: Assume we have an equality of the shape $$ (a_i,b_i)+(A_I,B_I)-(a_j,b_j)-(A_J,B_J) = (a_k,b_k)+(A_K,B_K)-(a_l,b_l)-(A_L,B_L) \ ,\ 1\le i,j,k,l\le m\ , \ 1\le I,J,K,L\le M\ , $$ involving different indices. We project it back to $\Bbb Z/m$ and $\Bbb Z/M$. It implies $(a_i,b_i)-(a_j,b_j)=(a_k,b_k)-(a_l,b_l)$ and $(A_I,B_I)-(A_J,B_J)=(A_K,B_K)-(A_L,B_L)$. Using the $m$-set, and $M-set$ assumed properties, we get $(i,j)=(k,l)$ and $(I,J)=(K,L)$.

$\square$

Explicitly, if $s=\{\ (a_i,b_i)\ :\ 1\le i\le m\ \}$ and $S=\{\ (A_I,B_I)\ :\ 1\le I\le M\ \}$, then consider the set with elements $M(a_i,b_i)+m(A_I,B_I)$.

---

Lemma B: Let $p$ be a prime. Let $s=\{\ (a_i,b_i)\ :\ 1\le i\le p^m\ \}$ be a $p^m$-set, taken with representatives $a_i,b_i\in \Bbb Z$ in $0,1,\dots,(p^m-1)$, and let $S=\{\ (A_I,B_I)\ :\ 1\le I\le p^M\ \}$ be a $p^M$-set, taken with representatives $A_I,B_I\in \Bbb Z$ in $0,1,\dots,(p^M-1)$.

Consider the set with elements: $$ (a_i,b_i)+p^m(A_I,B_I)\ ,\ 1\le i\le p^m\ ,\ 1\le I\le p^M\ . $$ Then this set is a $p^{m+M}$ set.

Proof: Assume we have an equality of the shape $$ (a_i,b_i)+p^m(A_I,B_I)-(a_j,b_j)-p^m(A_J,B_J) = (a_k,b_k)+p^m(A_K,B_K)-(a_l,b_l)-p^m(A_L,B_L) \ ,\ 1\le i,j,k,l\le p ^m\ , \ 1\le I,J,K,L\le p^M\ , $$ involving different indices. We take this modulo $p^m$, thus getting $$ (a_i,b_i)-(a_j,b_j) = (a_k,b_k)-(a_l,b_l)\ . $$ Since $s$ is a $p^m$-set, we get $(i,j)=(k,l)$. From the remained equality, $$ p^m(A_I,B_I)-p^m(A_J,B_J) = p^m(A_K,B_K)-p^m(A_L,B_L)\ , $$ considered now via $p^m\Bbb Z/p^{m+M}\cong \Bbb Z/p^M$ inside the latter group, $$ (A_I,B_I)-(A_J,B_J) = (A_K,B_K)-(A_L,B_L)\ , $$ we obtain $(I,J)=(K,L)$.

$\square$

---

As noticed in the comments of the OP, joriki provided a simple construction of a $p$-set for an odd prime $p$.

Using it and Lemma B, we obtain constructions of $p^m$-sets, for $p^m$ odd prime power.

And for $p=2$ we there is no solution since for some pairs $(a_1,b_1)$, $(a_2,b_2)$, the two differences are equal, $+1=-1$. So we need explicit solutions for $n=4,8$, if any, using them each power $2^m$ can be realized using Lemma B. The solution for $n=4$ was given in the OP. Using it we have solutions for all even powers of two.

---

Let us analyze the case $n=8$ explicitly. Let $R$ be the ring $(\Bbb Z/8)^2$. There is a canonical map to $\bar R=(\Bbb Z/2)^2$, going modulo two on the components. Elements of $R,\bar R$ will be denoted simpler by the "word" $st$ below, instead of the cartesian product notation $(s,t)$, a tuple with the components $s,t$.

Assume we have a solution $a,b,c,d,e,f,g,h$ in $R$ for our problem. (We try to get structural information from it, so that eventually a search can be done easier.)

Observe the fact (F), that subtracting an arbitrary element $r-R$ from the solution, leads to a solution (possibly the same) $a-r,b-r,\dots,h-r$. An other operation is multiplying with a $2\times 2$-matrix with invertible determinant in $\Bbb R/8$, i.e. performing the operation that maps $st$, written as a column vector $\begin{bmatrix}s\\t\end{bmatrix}$ to... $$ \begin{bmatrix}s\\t\end{bmatrix} \to \underbrace{\begin{bmatrix}A&B\\C&D\end{bmatrix}}_{AD-BC\in{1,3,5,7}} \begin{bmatrix}s\\t\end{bmatrix} \ , $$ (and translating the result to the corresponding word).

Let us denote by $\Bbb D$ the set of all $56=8\cdot 7$ different differences $x-y$, where $x,y$ with $x\ne y$ are among $a,b,c,d,e,f,g,h$. We note that the following elements are not in $\Bbb D$, $00,04,40,44$. The $00$ is excluded because of the condition $x\ne y$, the other are excluded since they have order two. So if $x-y$ is among $04,40,44$, then $x-y=y-x$, so one value would be taken twice. There are exactly four other values excluded from $R$ in $\Bbb D$.

We have in $\bar R$ exactly four elements, $00,01,10,11$. They determine four "blocks" in $R$, a partition of $R$ given by the preimages of the projection $\pi:R\to\bar R$, the four blocks are $\pi^{-1}(00)$, $\pi^{-1}(01)$, $\pi^{-1}(10)$, $\pi^{-1}(11)$, but we will denote them abusively also by $00$, $01$, $10$, $11$. (In case of possible confusion, i will use the word element, or block...)

We make a picture of the four blocks: $$ \begin{array}{|c|c|} \hline 00 & 01\\\hline 10 & 11\\\hline \end{array} $$ Then we map $a,b,c,d,e,f,g$ into the above blocks, count how many of them are in each block. Using the fact (F) above, we can and do assume that the block $00\in \bar R$ contains the maximal number of elements, and that $00\in R$. (If not, take the block with most elements, and let $r$ be one of the elements in this maximal block, and apply (F).) We can further arrange that in the block $01$ there are equally many or more elements than in the block $01$. (Switch components, which is also a matrix operation.)

Recall that in the block $00$ there are $16-4$ possible elements of $\Bbb D$, $00,04,40,44$ being excluded.

We use the variables $a,b,c,\dots$ in the order induced by the block order $00,01,10,11$.

• Assume that in the block $00$ there are $5$ or more elements. This is of course impossible, since $\{\ x-y\ :\ x\ne y\ ,\ x,y\text{ among }a,b,c,d,e\ \}\subset \Bbb D$ already delivers $5\cdot 4> 12$ values. Contradiction.

• Assume that in the block $00$ there are $4$ elements. This is of course impossible, since $\{\ x-y\ :\ x\ne y\ ,\ x,y\text{ among }a,b,c,d\ \}\subset \Bbb D$ already delivers $4\cdot 3= 12$ values. Consider two elements of some other block with at least two elements, take the two differences, which land in the block $00$. We get $14>12$ elements of $\Bbb D$ in the block $00$. Contradiction.

• Assume that in the block $00$ there are $3$ elements, $a,b,c$.

  • Assume there is an other block with $3$ elements, $x,y,z$. Then we can build $2\cdot 3^2=18>16$ differences (like $a-x$ and $x-a$) of two elements from the two maximal blocks, which land in the same block. Contraction.
  • It remains the case of the partition $8=3+2+2+1=3+2+1+2$, so the $11$ block can have either $1$ or $2$ elements. In a picture, the two cases are $$ \begin{array}{|l|l|} \hline 00\ :\ a,b,c & 01\ :\ d,e\\\hline 10 \ :\ f,g& 11\ :\ h\\\hline \end{array} $$ and $$ \begin{array}{|l|l|} \hline 00\ :\ a,b,c & 01\ :\ d,e\\\hline 10 \ :\ f& 11\ :\ g,h\\\hline \end{array} $$ and they are maped in each other by the following matrix operation, $$ \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}$$ where the computations are done modulo two, and the columns of the $2\times 4$ matrices correspond to the four blocks. So it is enough to restrict to one of the cases, let us consider the symmetric case $3+2+2+1$, and in this case $\Bbb D$ has the following distribution of elements: $$ \begin{array}{|l|l|} \hline 00 & 01\\ a-b\ (6) & d-a\ (12)\\ d-e\ (2) & h-f\ (4)\\ f-g\ (2) &\\ \text{Totally $6+2+2=10$ elements} & \text{Totally $12+4=16$ elements} \\\hline 10 & 11\\ f-a\ (12) & h-a\ (6)\\ h-d\ (4) & d-f\ (8)\\ \text{Totally $12+4=16$ elements} & \text{Totally $6+8=14$ elements} \\\hline \end{array} $$ Here, an information like $d-a\ (12)$ suggests that we can build $12$ different differences using $\pm(d-a)$, and replacing $a,d$ by elements in their classes in $\bar R$. Now we can go further in detail. We already have $a=00\in R$. Using the group $G$ we can arrange for $b=02\in R$, and using the part $\begin{bmatrix}* &0\\* & 1\end{bmatrix}$ of $G$ we can also arrange for $b=20$. In the block $00\in \bar R$ we then have in $\Bbb D$ the marked red elements $$\begin{matrix} \color{gray}{00}& \color{red}{02}& \color{gray}{04}& \color{red}{06}\\ \color{red}{20}& 22& 24& \color{red}{26}\\ \color{gray}{40}& 42& \color{gray}{44}& 46\\ \color{red}{60}& \color{red}{62}& 64& 66 \end{matrix}$$ Now the four elements $\pm(d-e)$, $\pm (f-g)$ are also in this list, avoiding the red and gray entries. So $e=d+t$, $g=f+v$, where $t,v$ have order two and avoid the red entries and "themselves". Set also $d=01+s$, $f=10+u$, $h=11+w$, where $s,u,w$ have order two. We are already in position to let the computer do a quick search. Let us stop here, and postpone the discussion.

• Assume the remained situation, that in the block $00$ there are $2$ elements (and that this is maximal), so we have to cover the partition $8=2+2+2+2$, the elements are placed in blocks as follows: $$ \begin{array}{|l|l|} \hline 00\ :\ a,b & 01\ :\ c,d\\\hline 10 \ :\ e,f& 11\ :\ g,h\\\hline \end{array} $$ and the differences (the set $\Bbb D$) are then distributed as follows:$$ \begin{array}{|l|l|} \hline 00 & 01\\ a-b\ (2) & c-a\ (8)\\ c-d\ (2) & g-e\ (8)\\ e-f\ (2) &\\ g-h\ (2) &\\ \text{Totally $2+2+2+2=8$ elements} & \text{Totally $8+8=16$ elements} \\\hline 10 & 11\\ e-a\ (8) & g-a\ (8)\\ g-c\ (8) & c-e\ (8)\\ \text{Totally $8+8=16$ elements} & \text{Totally $8+8=16$ elements} \\\hline \end{array} $$ As above, we can assume $a=00$, $b=02$, $c=01+s$, $d=c+t=01+s+t$, $e=10+u$, $f=e+v=10+u+v$, $g=11+w$, $h=g+x=11+w+x$, where $s,t,u,v,w,x$ are from the $00$ block, and $t,v,x$ have order two, so that the eight elements $\pm(02),\pm t,\pm v,\pm x$ are different. Using the $\begin{bmatrix}* &0\\* & 1\end{bmatrix}$ part of $G$, we can arrange for $t=20$. After a possible exchange of $e,f$, and/or $g,h$, we can change sign in $v,x$. We are again in position to start a quick computer search, if we want to do it.

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A computer search is a matter that separates mathematicians. People like me that consider that life is too short, and writing code speeds up the final answer, will always type that lines of code. But people that never write code will consider this, even in this century, as unacceptable. I will try to cover both worlds.

---

Here are two pieces of sage code that cover in a quick+lazy way the two isolated cases $3+2+2+1$ and $2+2+2+2$. The first case:

A  = IntegerModRing(8)
A1 = [ A(0), A(2), A(4), A(6) ]    # elements of additive order < 8 in A

R  = [ vector([a,b]) for a in A  for b in A  ]
R1 = [ vector([a,b]) for a in A1 for b in A1 ]

def r(a,b): return vector([A(a), A(b)])

a, b ,c = r(0,0), r(0,2), r(2,0)

T = [ r(s,t) for (s,t) in ( (2,2), (2,4), (4,2), (4,6), (6,2), (6,4) ) ]

for t in T:
    for v in T:
        if v in (t, -t): continue
        for d1, f1 in cartesian_product( [R1, R1] ):
            d = r(0,1) + d1
            e = d + t
            f = r(1,0) + f1
            g = f + v
            S1 = [a,b,c,d,e,f,g]
            if len(set([tuple(x-y) for x in S1 for y in S1 if x != y ])) != 42:
                continue
            print('Possible choice of seven elements: %s' % S1)
            for h1 in R1:
                h = r(1,1) + h1            
                S = S1 + [h, ]
                if len(set(tuple(x-y) for x in S1 for y in S1 if x != y)) != 56:
                    continue
                print('SOLUTION: %s' % S)

We obtain several possibilities for $a,b,c,d,e,f,g$, but there is no $h$ completing the set.

---

The case $2+2+2+2$:

A  = IntegerModRing(8)
A1 = [ A(0), A(2), A(4), A(6) ]    # elements of additive order < 8 in A

R  = [ vector([a,b]) for a in A  for b in A  ]
R1 = [ vector([a,b]) for a in A1 for b in A1 ]

def r(j,k): return vector([A(j), A(k)])

vx_choices = [ r(j,k)
               for (j,k) in ( (2,2), (2,4), (2,6), (4,2) ) ]

a, b, t = r(0,0), r(0,2), r(2,0)
for v, x in cartesian_product( [vx_choices, vx_choices] ):
    elements_in_ID = [ b, -b, t, -t, v, -v, x, -x ]
    if len( set( [ tuple(elem) for elem in elements_in_ID ] ) ) != 8:
        continue
    print(v, x)
    for c1, e1 in cartesian_product( [R1, R1] ):
        c = r(0,1) + c1
        d = c + t
        e = r(1,0) + e1
        f = e + v
        S1 = [a,b,c,d,e,f]
        if len( set( [ tuple(X-Y) for X in S1 for Y in S1 if X != Y ] ) ) != 30:
            continue
        for g1 in R1:
            g = r(1,1) + g1
            h = g + x
            S = [a,b,c,d,e,f,g,h]
            if len( set( [ tuple(X-Y) for X in S for Y in S if X != Y ] ) ) != 56:
                continue
            print('SOLUTION: %s' % S)

No solution is found.

---

To have also a human taste for the final search, let us consider the case $3+2+2+1$. We use $a',b',c',d',e',f',g',h'$ in $(2\Bbb Z/8)^2$, so that $a=00+a'=a'$, $b=00+b'=b'$, $c=00+c'=c'$, $d=01+d'$, $e=01+e'$, $f=10+f'$, $g=10+g'$, $h=11+h'$. If we write now the different elements of $\Bbb D$, in the different blocks where they appear as above, but using $a',b',\dots, h'$ explicitly, then we have some "twists" in the writing, coming from the fact that - for example - if $d=01+d'$, then $-d=07-d'=01-d'+06$.

Here are the elements: $$ \begin{array}{|l|l|} \hline 00 +\dots & 01 + \dots\\ a'-b'\ (6) & d'-a'\ [6]\ a'-d'+06\ [6]\\ d'-e'\ (2) & h'-f'\ [2]\ f'-h'+06\ [2]\\ f'-g'\ (2) & \\\hline 10+\dots & 11+\dots\\ f'-a'\ [6]\ a'-f'+60\ [6] & h'-a'\ [3]\ a'-h'+66\ [3]\\ h'-d'\ [2]\ d'-h'+60\ [2] & d'-f'+60\ [4]\ f'-d'+06\ [4]\\\hline \end{array} $$ Here, an information like $a'-h'+66\ [3]$ means that we build $[3]$ elements of the same shape, replacing $a',h'$ by corresponding elements. (But $h'-a'$ is an other case.)

Note that the sum of the elements in the $11$ block is $00$, since it is $\sum\pm (a-h)+\sum\pm(d-f)$. But on the other side, if we use the explicit information in $2\Bbb Z/8$ that we have, we get $$\sum(h'-a')+\color{red}{\sum(a'-h'+66)}+\sum(d'-f'+06)+\sum(f'-d'+60)\ ,$$ and the red sum has three elements, all $a',b',\dots,h'$ are cancelling, and we get the result from the twisting $$ 3\cdot (6,6)+ 4\cdot (0,6)+4\cdot (6,0)\ne (0,0)\ .$$

This gives a contradiction, so there is no solution in the $3+2+2+1$ case. (The case $2+2+2+2$ needs an argument of a different type, the same is not working. I will search for it, come back and edit if i find a quick human path to the solution, else the computer search remains, there was structural push as much as possible.)

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I have to submit without a double check, hope that the above arguments do not have a hidden trap... the daily job...