prove no two cycles in a graph will share an edge if all the cycles in the graph are of odd length I am trying to use induction but cant make any headway on this problem.
Was considering using the property that biparted graphs have no cycles of odd length but Im not sure if that helps
Suppose there are two (odd) cycles in your graph, say $C_1 = [u_1, u_2, \dots, u_n, u_1]$ and $C_2 = [v_1, v_2, \dots, v_m, v_1]$, that have k common edges. These edges lie between $k+1$ vertices in either cycle.

Now, we know that $n$ and $m$ are odd, so $n=2a-1$ and $m=2b-1$, for some $a,b \in \mathbb{N}$. Therefore the number of vertices in the big cycle (the one obtained by forgetting the common edges) is equal to $n-(k-1))+m-(k-1) = 2a-1+2b-1-2k+2=2(a+b-k)$, which is an even number, a contradiction.
Suppose that we have two cycles $C_1$ and $C_2$ that share an edge. Without loss of generality, these cycles intersect in a path. Why? If not, then we have two points $v, w$ in both $C_1$ and $C_2$, and two paths $v, v_1, \ldots, v_n, w$ and $v, w_1, \ldots, w_m, w$ in $C_1$ and $C_2$ respectively. Consider the least $i$ such that $v_i \neq w_i$, and the least $j \ge i$ such that $v_j$ belongs on the other path (i.e. is equal to $w_k$ for some $k \ge i$). Then
$$v_{i-1}, v_i, \ldots,v_j, w_{k-1},\ldots,w_{i-1}$$
is a cycle, where $v_{i-1} = w_{i - 1}$ is $v$ when $i = 1$, that intersects with $C_1$ (and $C_2$) in a path.

So, we have three paths $P_1, P_2, P_3$ such that $C_1 = P_1 \cup P_2$ and $C_2 = P_2 \cup P_3$. Consider the cycle $C_3$ made up of $P_1$ and $P_3$. Then,
\begin{align*}
|C_1| &amp;= |P_1| + |P_2| \\
|C_2| &amp;= |P_2| + |P_3| \\
|C_3| &amp;= |P_1| + |P_3|.
\end{align*}
Each $|C_i|$ is odd, but when we sum these expressions,
$$|C_1| + |C_2| + |C_3| = 2(|P_1| + |P_2| + |P_3|),$$
which is even. This is a contradiction. Hence, the cycles cannot share an edge.