Proof that if $G$ doesn't contain cycle with length $1 \mod l$ then $\chi(G) \le l$ Let $G$ be a graph and $l \ge 2$. Proof that if $G$ doesn't contain cycle with length $1 \mod l$ then $\chi(G) \le l$, where $\chi(G)$ is the minimum number of colours needed to properly colour $G$ (no adjacent vertices in $G$ assigned the same colours).
I want to tell something about my current approach but after 1 hour of drawing cycles I didn't see anything interesting. Moreover I think that cycles it is really weird that we consider length modulo. How it can works that for cycle with length $l, l+2$ it would work, $l+1$ doesn't...
 A: Use induction. Let $v$ be a vertex in $G$. Color properly $G \setminus \{v\}$ with $\ell$ colors $0,1,2,\ldots \ell-1$. We now consider two cases.
Case 1: All of $\ell$ colors are used to $N_G(v)$. Then pick an arbitrary color $k$, and a neighbour $w$ of $v$. Change $w$'s color to $k-1$, and in turn all of $w$'s neighbours colored $k-1$, to $k-2$, and so on and so forth [arithmetic done mod $\ell$], terminating when there are no more vertices left to change colors. Then if this process terminates in a finite number of steps the resulting coloring is still a proper coloring. We claim that no vertex $u$ in $G \setminus \{v\}$ gets recolored twice. [Indeed, the only way a vertex $u$ (colored $k'$ in the original coloring say) could be recolored twice would be if in the original coloring there is a path $P$ of $j\ell+1$ vertices; $j$ a positive integer, such that $P$ starts at $u$ and also (a) writing $P = uu_1\ldots u_{j\ell}$, vertex $u_r$ was colored $k'-r$ in the original coloring, and (b) $u_{j\ell}$ is adjacent to $u$. But then $u$ and $u_{j\ell}$ would have been given the same coloring $k'$ in the original coloring, contradicting the assumption that the original coloring was proper.[Not to mention, contradicting the assumption that there are no cycles of length $j\ell+1$ in $G$]] So the process indeed terminates in a finite number of steps and thus the resulting new coloring is proper.
Furthermore, we claim that no vertex in $N_G(v)$ that wasn't colored $k$ before in the original coloring, get colored $k$. [Indeed, if there were such a vertex $w' \in N_G(v)$ then $w'$ would have been assigned $k+1$ in the original coloring. Thus, using a similar line of reasoning as in the above paragraph, this would imply a path $P$ of $j\ell$ vertices in $G \setminus \{v\}$ with one endpoint as $w$ and the other endpoint $w'$, which would there imply that is a cycle of length 1 mod $\ell$ in $G$; namely $P$ plus the two edges $w'v$ and $vw$]. 
Thus, we can repeat the process in the above two paragraphs until we achieve a proper coloring of $G \setminus \{v\}$ where there are no vertices in $N_G(v)$ colored $k$, and then we assign the color $k$ to $v$.
Case 2: there is a color $k$ in $0,1,\ldots, \ell-1$ such that no vertex in $N_G(v)$ is colored $k$. Then assign $v$ the color $k$.
This answer was enhanced by bof's observation that in the recoloring no vertex would be recolored twice even if there were cycles of length $j\ell+1$.
