COMBINATORIAL ANALYSIS (MATRIX PROBLEMS, ORDER THEORY)

NEW

Problem: Reconstructed statement: with the standard graph-theoretic convention that $P_n$ is the path on $n$ vertices, the ladder

has $2n$ vertices and $3n-2$ edges. The conjecture asserts that for every even $n\ge2$ and every $d\in\{0,2\}$, $L_n$ has a super $(a,d)$-edge-antimagic total labeling: a bijection

with $f(V)=\{1,\dots,|V|\}$, such that edge-weights

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

Result: The literal conjecture is false.

Take $n=2$. Then

Here $|V|=4$, $|E|=4$. In any super total labeling, the vertex labels are $1,2,3,4$, and the edge labels are $5,6,7,8$.

Summing the four edge-weights over all edges of $C_4$, each vertex label is counted twice, and each edge label once. Hence

If $d=0$, all four weights equal $a$, so their sum is $4a$, impossible since $46$ is not divisible by $4$.

If $d=2$, the four weights are $a,a+2,a+4,a+6$, whose sum is

Thus $4a+12=46$, so $a=17/2$, impossible because all weights are integers.

Therefore $L_2$ is not super $(a,d)$-edge-antimagic total for either $d=0$ or $d=2$. Since $2$ is even and satisfies $n\ge2$, the conjecture as stated is disproved.

The likely formulation defect is the missing exclusion of the boundary case $n=2$; the natural repair would be to require even $n\ge4$, or to use an indexing convention where $L_2$ is not $C_4$.

Citation: No external citation is needed; the disproof is the elementary $L_2\cong C_4$ counterexample above.