I have such graph: $$ whose \ vertex\ set\ is\ V=\{1,2,..,82\} \ vertices\ i\ and\ j\ are\ adjacent\ iff\ (i-j)\ mod\ 4\ =0\\a)\ Calculate\ the\ chromatic\ number\ of\ G.\ b)Is\ G \ Eulerian?\ c)Is\ G\ Planar?$$ So I think the graph has 4 subgraphs where each vertex is connected to another. $$1st:\ V=\{1,5,9, 4k+1\}\ where\ k\in [0,20]\\2nd:\ V=\{2,6,10, 4k+2\}\ where\ k\in [0,20]\\ 3rd: \ V=\{3,7,11, 4k+3\}\ where\ k\in [0,19]\\4th:\ V=\{4,8,12, 2k\}\ where\ k\in [0,19]\\$$ However first, second, third, and forth graph are not connected to each other. Chromatic number will be number of first graph vertices $$\gamma (G) = 21$$ b) Well the graph is not connected so it cannot be Eulerian, moreover in 4th graph each vertex has odd degree.
c) The graph is not connected so it cannot be planar
Am I right?
The graph as you notice is a disjoint union of several large cliques.
a) you are correct
b) right again
c) right answer, wrong reason. There are disconnected planar graphs (just take two planar graphs and consider their disjoint union). However, the graph you have certainly contains a $K_5$ and thus is not planar.