Let $\kappa$ be an infinite cardinal. Consider the following example to $2^\kappa\nrightarrow (3)^2\_\kappa$.


$V$ is a set of vertices, each of which is an element of $2^\kappa$. Color the edge between two vertices $f,g\in 2^\kappa$ by the least ordinal on which $f,g$ disagree. It follows that there is no homogeneous set of size $3$, i.e. there is no triangle with all three edges the same color. 


More is true: There is no closed walk of length 3,5,7,... which is monochromatic, i.e. all edges are the same color.


**Definition**:A closed walk consists of a sequence of vertices starting and ending at the same vertex.


**Question**: Can you find an example to $2^\kappa\nrightarrow (3)^2\_\kappa$, therefore no monochromatic triangles are allowed, but which has closed walks of length $n$, for any odd number $n>3$?



Example to $2^\kappa\nrightarrow (3)^2_\kappa$, plus closed walks of odd length?
The answer is yes. In fact, we can get an example by slightly modifying your example.


Let $V = 2^\kappa \cup \{a,b,c,d,e\}$. Color the edges $ab$, $bc$, $cd$, $de$, and $ea$ with some ordinal $< \kappa$, say $0$. Color all other edges from $a$ with a different color, say $A$. Color all other edges from $b$ with a different color, say $B$. Similarly for $c$, $d$, and $e$. The remaining edges are all between elements of $2^\kappa$, and we color them just as you did in your example, with the least ordinal on which they disagree. This gives a coloring of $V$ with $\kappa$ colors, namely $\kappa \cup \{A,B,C,D,E\}$. We have a $5$-cycle in color $0$. We obviously have no triangles in colors $A,B,C,D,E$, and we have no triangles in our other colors $> 0$ because there are no triangles in your example. We have no triangles in color $0$ because in that color we only added a $5$-cycle that is not connected to anything else.


This shows that we can extend your example to one including a $5$-cycle in color $0$ by using five new colors and five new vertices. But we didn't use any specific properties of your example -- any example would have done just fine -- and the number $5$ wasn't special either. So after adding a $5$-cycle in color $0$, we can then add a $7$-cycle some other color, a $9$-cycle in some other color, etc. In fact, we can add monochromatic $n$-cycles for any set of odd $n$ we like, using any colors we like. You can even iterate the construction transfinitely ($\kappa$ steps) to get a graph of size $2^\kappa$ colored in $\kappa$ colors with no monochromatic triangles, but $\kappa$ $n$-cycles in every color for every odd $n > 3$.


