a. Suppose that two graphs are isomorphic and one of them is bipartite. Then must the other graph necessarily also be bipartite? Explain.
If two graphs are isomorphic and one of them contains a cycle of a particular length, then the same must be true of the other graph (haven't proved this yet). Bipartite graphs can only contain even cycles. So if graphs $A, B$ are isomorphic and $A$ is bipartite, then $B$ must also only contain even cycles and so $B$ is also bipartite.
If the above doesn't work, consider that two isomorphic graphs must have the same degree sequence (yet to prove it) and so if one of the graphs is bipartite, the other one must be as well.
b. Show that a graph that has $17$ vertices and $73$ edges cannot be bipartite.
If we have two parts s.t. $|X| = 8, |Y| = 9$, then there are at most $72$ edges. Do we consider all the cases?
c. Suppose that in a bipartite graph, all of the vertices except possibly one of them have the same degree $d$, and the remaining vertex has an unknown degree $x$. Prove that $x$ must be a multiple of $d$.
Suppose $d \nmid x.$ Then either $d > x$ or $d$ and $x$ have different parity. Then we can do case work to show that two parts (whatever they are) will never have equal sums of degrees.
Please point out if there are mistakes on here and if there's a much better/faster/more obvious way to do these. Thanks.
$2$. Let the two parts be $X$ and $Y$,$ |X|+|Y|=17$, $|X|,|Y|$ are integers, just prove that the maximum value of $|X||Y|$ cannot be more than $72$. (https://math.stackexchange.com/questions/1098587/maximum-no-of-edges-in-a-bipartite-graph )
$3$. The graph is divided into two parts $X$ and $Y$, so the vertex with unknown degree must lie in any one of the parts, say it lies in part $X$, Also, sum of the degrees in part $X$ and part $Y$ are same. And sum of the degree of part $Y$ equals $d|Y|$ hence the unknown degree must be divisible by $d$.
– Arpan1729 Jul 14 '17 at 01:32