Total weight of every cycle is even if and only if the total weight of every triangle is even. Integer weights are written on the edges of a complete graph. Prove that the total weight of every cycle is even if and only if the total weight of every triangle is even.
Is there any hints on how to get started here?
 A: Clearly if the weight of all cycles are even then the weights of all triangles are also even (since triangles are cycles themselves). 
For the converse, let us induct on cycle length as Quimey suggests. For $k=3$ this holds by assumption. Suppose this holds for all cycles of length $k$ and consider $C$ with length $k+1$, say $$C = (v_1,\ v_2,\ v_3,\ \cdots,\ v_{k+1}, v_1)$$
Let us insert an edge, say $(v_1,\ v_3)$. This splits $C$ into $T = (v_1,\ v_2,\ v_3,\ v_1)$ is a triangle and $$C^\prime = (v_1,\ v_3,\ v_4,\ \cdots,\ v_{k+1},\ v_1)$$ which is a cycle of length $k$. 
Denoting $e_{i,j}$ as the edge weight of $(v_i,\ v_j)$, we then have
$$e_{1,2} + e_{2,3} + e_{3,1} \equiv 0 \pmod{2}$$
by assumption. The edge weight of $C^\prime$ is also even
$$e_{1,3} + e_{3,4} + \cdots + e_{k,k+1} + e_{k+1,1} \equiv 0 \pmod{2}$$
by inductive hypothesis. Together then we have
$$(e_{1,2} + e_{2,3} + e_{3,1}) + (e_{1,3} + e_{3,4} + \cdots + e_{k,k+1} + e_{k+1,1}) \equiv 0\pmod{2}$$
The above counts $e_{3,1} = e_{1,3}$ twice, so we in fact have
$$e_{1,2} + e_{2,3} + e_{3,4} + \cdots + e_{k,k+1} + e_{k+1,1} \equiv 0\pmod{2}$$
which is exactly what we wanted to prove.
A: It is obvious that if every cycle has even total weight, then so does any triangle.  
For the other direction, suppose to the contrary that the total weight of each triangle is even, but there is a cycle with odd total weight. Then there is a shortest cycle $\mathcal{C}$ with odd total weight. 
Cycle $\mathcal{C}$ has length $\ge 4$. Let $p$, $q$, and $r$ be consecutive vertices on this cycle. 
The cycle obtained from $\mathcal{C}$ by deleting vertex $q$, and replacing the edges $pq$ and $qr$ by the edge $pq$, is a cycle $\mathcal{C}'$ shorter than $\mathcal{C}$.  By the minimality assumption, the total weight of $\mathcal{C}'$ is even.
For any edge $uv$, let $W(u,v)$ be the weight of $uv$. 
Let the weight of $\mathcal{C}$ be $W(\mathcal{C})$, and the weight of $\mathcal{C}'$ be $W(\mathcal{C}')$.
We have 
$$
\begin{align}W(\mathcal{C})&=W(\mathcal{C}')+W(p,q)+W(q, r)-W(p,r)\\
&= W(\mathcal{C}')+W(p,q)+W(q,r)+W(p,r) -2W(p,r).\end{align}\tag{$1$}$$
But $W(\mathcal{C}')$ is even. Also, $W(p,q)+W(q,r)+W(p,r)$ is even, since it is the sum of the edge weights of a triangle.  And  of course $2W(p,r)$ is even. 
Thus the right-hand side of $(1)$ is even, contradicting the fact that $W(\mathcal{C})$ is odd.  
Remark: It is straightforward to recast the above argument as a conventional induction argument. 
