Graph Problems(Euler,Hamilton,Color)

Let be $n_1$, $n_2$ such natural numbers that $n_1\geq 3$ and $n_2\geq 3$ and let be $G_{n1,n2}$ a graph that takes shape by taking $G_{n_1}$, the cycle of $n_1$ vertices, and $G_{n_2}$ the cycle of $n_2$ vertices, and linking all of $G_{n_1}$ vertices with all of $G_{n_2}$ vertices.

a) For which values of $n_1,n_2$ has the graph $G_{n1,n2}$ an Euler cycle ?

b) For which values of ${n_1,n_2}$ has the graph $G_{n1,n2}$ a Hamilton cycle ?

c) For which values of ${n_1,n_2}$ is the graph $G_{n1,n2}$ a 5-colorable but not 4-colorable?

Indication: Show first that in each legal(valid) coloration of $G_{n1,n2}$ the sets of colours that are used for $G_{n_1}$ vertices and $G_{n_2}$ vertices they have to to be foreign from each other.

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There is a well-known criterion for when graphs are Eulerian. It should be easy to apply here. For Hamiltonian and 4-coloring, try some examples and you'll figure it out. – aaron Mar 6 '11 at 0:01
A graph G contains an Eulerian circuit if and only if the degree of each vertex is even.I also think that the graph Gn1,n2 is complete and bipartite. – Nick Mar 6 '11 at 0:07
The graph $G_{n1,n2}$ is not complete unless $n1$ and $n2$ are less than 4, as it is missing the diagonals of the two cycles. – Ross Millikan Mar 6 '11 at 0:11
I think each of n1,n2 has to be even or else not all the degree of each vertex of Gn1,n2 will be even thus no euler..so in mathimatical terms how to prove the above if it is right of course – Nick Mar 6 '11 at 0:13
@Ross i believe you are 100% didn't see it...So what about for having an Euler Cycle that both Gn1 ,Gn2 have to be even Cycles?Is there a way to prove it if i am right? – Nick Mar 6 '11 at 0:21

Added after the first two comments: The description of Euler cycle is just a restatement of one of the comments to the question, and is correct. For the Hamiltonian cycle, it is true there always is one. To prove it, you should be able to describe one. For example, if $n1=n2$, you could number the vertices in each cycle from 1 to n in order around the cycle. The Hamiltonian cycle is from vertex $i$ in $G_{n1}$ to vertex $i$ in $G_{n2}$, then from vertex $i$ in $G_{n2}$ to vertex $i+1$ in $G_{n1}$ and finally from vertex $n$ in $G_{n2}$ to vertex $1$ in $G_{n1}$. Can you find a construction that works if $n1 \ne n2$?
Further addition, now that the due date has probably passed: For b) there is always a Hamiltonian cycle. Start at one vertex of $G_{n1}$ and follow the cycle to the last vertex before you close. Then go to a vertex of $G_{n2}$, traverse the $G_{n2}$ cycle until the last vertex and go back to the vertex you started at. This is a Hamiltonian cycle. For c), coloring a cycle takes $2$ colors if the cycle is even and $3$ if it is odd. As the colors in the two cycles must be distinct, the whole graph takes $4$ if $n1, n2$ are both even, $5$ if one is even and one is odd, $6$ if they are both odd.