Structural Properties and Labeling of Graphs

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Problem: I interpret $C_m\times P_n$ as the Cartesian product $C_m\square P_n$, the generalized prism. A graph $G$ with $p$ vertices and $q$ edges is super $(a,d)$-edge-antimagic total if there is a bijection

with $f(V)=\{1,\dots,p\}$ and edge weights

equal, as a multiset, to $a,a+d,\dots,a+(q-1)d$.
The reconstructed conjecture is: for every even $m\ge4$, every $n\ge3$, and every $d\in\{0,2\}$, $C_m\square P_n$ is super $(a,d)$-edge-antimagic total for some $a$.

Result: The conjecture is false. In fact, $C_4\square P_n$ has no super $(a,0)$-edge-antimagic total labeling for any $n\ge3$.

Let $G=C_4\square P_n$. Then

Also $G$ is bipartite and $\Delta(G)\le4$.

Assume, for contradiction, that $G$ has a super $(a,0)$-edge-antimagic total labeling $f$. Since $d=0$, every edge has the same weight $a$. The edge labels are exactly $p+1,\dots,p+q$, so the vertex-sums

over all edges are $q$ distinct consecutive integers. Since vertex labels are $1,\dots,p$, every edge-sum lies between $3$ and $2p-1$. Because $q=2p-4$, the only possible consecutive intervals are

Now use bipartiteness. If the edge-sums contain $\{3,4,\dots,p+1\}$, then the vertex labeled $1$ must be adjacent to every vertex labeled $2,\dots,p$: indeed, sums $3$ and $4$ force edges $1\!-\!2$ and $1\!-\!3$, so labels $2,3$ lie in the opposite bipartition class from label $1$; inductively, for sum $k+1$, every possible pair not involving $1$ has both endpoints already in that same bipartition class, so the edge must be $1\!-\!k$. Thus $\deg(1)=p-1$, impossible because $\Delta(G)\le4$ and $p-1=4n-1>4$.

Similarly, if the edge-sums contain $\{p+1,\dots,2p-1\}$, applying the same argument to the complementary labels $p+1-f(v)$ forces the vertex labeled $p$ to have degree $p-1$, again impossible.

Both possible edge-sum intervals lead to contradiction. Hence $C_4\square P_n$ is not super $(a,0)$-edge-antimagic total. Taking $n=3$, the graph $C_4\square P_3$ satisfies the conjecture’s hypotheses $m=4$ even, $n=3$, $d=0$, but fails the conclusion.

Verification audit: the counterexample is within the stated parameter range; no extra hypotheses were introduced; the argument uses only the standard SEAT definition and the bipartiteness/degree properties of the Cartesian generalized prism.

Citation: No known literature citation is used for this disproof; it is a direct counterexample from the standard definitions.