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?

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.
