Consider an even order, balanced(both partitions have same vertices) bipartite regular graph of order greater than or equal to $12$ and degree atleast six and divisible by $6$. Then is the graph of Type 1(totally colorable by $\Delta+1$ colors where $\Delta$ is the maximum degree)?


By petersen theorem, the graph has 2-factor( in fact k-factors for $k\le2n$, where $2n$ be the total number of vertices). Again, it is a union of disjoint $1$ factors(perfect matchings).Could these be used to provide a total coloring?


Apart from this, since the degree is also divisible by $3$(i.e. divisible by $6$), then, we may be able to find cycles of order divisible by $3$($6$); and since cycles whose order is divisible by $3$ can be totally colored(each pair of adjacent or incident elements of graph receive different colors) with $3$ colors, therefore we can also totally color each disjoint cycle in the graph using just three colors. Can we use this fact to totally color the whole graph with ($\Delta+1$) colors?



Total Coloring of even regular bipartite graphs
No it is not always possible to total color a bipartite graph with $\Delta+1$ colors, even with the given restrictions on $\Delta$ and the number of vertices. This is a counterexample. Let $G$ be a complete bipartite graph with $n$ vertices on each side, where $n$ can be any integer you want as long as it is sufficiently large. [So the degree of each vertex is $n$.]


Suppose there were a proper total coloring $\chi$ using only $n+1$ colors, Let $X$ and $Y$ be the parts of $G$ [to be clear $X$ and $Y$ are sets of vertices], and let $X\_{n+1}$ be the entire subset of $X$ that is colored with the $(n+1)$-st colour by $\chi$, and assume WLOG that $X\_{n+1}$ is nonempty. Next, let $E\_{n+1}$ be the subset of edges colored with the $(n+1)$-st color by this proper total coloring $\chi$. Then $E\_{n+1}$ is a matching of cardinality $|X|-|X\_{n+1}|=|Y|-|X\_{n+1}| < |Y|$. So let $U$ be a subset of $Y$ not incident to an edge in $E\_{n+1}$. Then as $G$ is a complete graph, every vertex in $U$ is adjacent to a vertex in $X\_{n+1}$, so no vertex in $U$ can 
be colored with the $(n+1)$st color, and furthermore, as every vertex in $U$ is also not incident to an edge in $E\_{n+1}$, it follows that for each $u \in U$ there are only $n$ colors to color $u$ and its $n$ edges incident to $U$ with all different colors. This is impossible.


