COMBINATORIAL ANALYSIS (MATRIX PROBLEMS, ORDER THEORY)

NEW

Problem: Reconstructed statement: for every integer $n>7$ and every $d\in\{0,2\}$, the friendship graph $F_n$ admits a super $(a,d)$-edge-antimagic total labeling for some integer $a$. Here $F_n$ is the graph formed from $n$ triangles sharing one common vertex. A super $(a,d)$-edge-antimagic total labeling of a graph with $p$ vertices and $q$ edges is a bijection

$$\lambda:V(G)\cup E(G)\to \{1,\dots,p+q\}$$

such that $\lambda(V)=\{1,\dots,p\}$ and the edge weights

$$w(uv)=\lambda(u)+\lambda(uv)+\lambda(v)$$

are exactly $a,a+d,\dots,a+(q-1)d$.

Result: The reconstructed universal existence statement is false. Already $F_8$ admits no such labeling for either $d=0$ or $d=2$.

Let $F_8$ have center $c$. Then $p=17$, $q=24$, vertex labels are $1,\dots,17$, and edge labels are $18,\dots,41$. Put $s=\lambda(c)$. Since $\deg(c)=16$ and every other vertex has degree $2$,

$$\sum_{uv\in E(F_8)}(\lambda(u)+\lambda(v)) =16s+2(153-s)=14s+306.$$

Also

$$\sum_{e\in E(F_8)}\lambda(e)=18+\cdots+41=708.$$

Thus the total sum of edge weights is $14s+1014$.

If $d=0$, all 24 weights equal $a$, so

$$24a=14s+1014.$$

Hence $s\equiv 3\pmod {12}$, so $s=3$ or $s=15$.

For $s=3$, the edge vertex-sums must be $a-41,\dots,a-18=3,\dots,26$. The spoke sums are

$$4,5,7,8,\dots,20.$$

Thus the eight rim sums must be

$$\{3,6,21,22,23,24,25,26\}.$$

But rim sum $3$ forces the pair $\{1,2\}$, after which rim sum $6$ is impossible, since it would require $\{1,5\}$ or $\{2,4\}$.

For $s=15$, the edge vertex-sums must be $10,\dots,33$. The spoke sums are $16,\dots,32$ except $30$, so the rim sums must be

$$\{10,11,12,13,14,15,30,33\}.$$

Rim sum $33$ forces $\{16,17\}$, after which rim sum $30$ is impossible, since it would require $\{13,17\}$ or $\{14,16\}$. Hence no $d=0$ labeling exists.

If $d=2$, the weights are $a,a+2,\dots,a+46$, so

$$24a+552=14s+1014,$$

again forcing $s\equiv3\pmod{12}$, hence $s=3$ or $s=15$.

Write each edge label as $18+k$, $0\le k\le23$, and let $t$ be the vertex-sum of that edge. Then $t+k=w-18$.

For $s=3$, one gets $a=21$, so the values $t+k$ must be

$$3,5,7,\dots,49.$$

The value $3$ forces $t=3,k=0$, hence rim pair $\{1,2\}$. Then value $5$ forces the spoke sum $4$ with $k=1$, and value $7$ forces the spoke sum $5$ with $k=2$. Now value $9$ has no possible unused representation: sums $3,4,5$ are already used, sum $6$ is impossible, and sums $7,8,9$ would require already-used $k=2,1,0$. Contradiction.

For $s=15$, one gets $a=28$, so the values $t+k$ must be

$$10,12,\dots,56.$$

The value $56$ forces $t=33,k=23$, hence rim pair $\{16,17\}$. Then $54$ forces the spoke sum $32$ with $k=22$, and $52$ forces the spoke sum $31$ with $k=21$. Now value $50$ has no possible unused representation: sums $33,32,31$ are already used, sum $30$ is impossible after pairing $\{16,17\}$, and the remaining candidates require already-used $k=23,22,21$. Contradiction.

Therefore $F_8$ is a counterexample for both $d=0$ and $d=2$, so the claimed existence for all $n>7$ is false.

Citation: No literature citation used; the counterexample and proof above are self-contained.