Proving a fact about 4-regular graphs 
Prove that 4-regular graphs have no bridges.

How can I proceed? This has no solution on the textbook, and it is hard to think of any invariant or theorem involving 4-reg graphs in particular.
 A: Let's reduce this problem a bit.  It suffices to consider $4$-regular connected graphs (take the connected components) and then prove that these graphs are $2$-edge connected (a graph has no bridge if and only if it has no cut edges).
As noted by RGB in the comments, the key observation here is that even graphs (of which $4$-regular graphs are a special case) have an Eulerian circuit.  Deleting any edge leaves us with an Eulerian trail, and so every two vertices are connected by a trail (since we can just walk along the trail between any two points).  Thus, the graph is bridgeless: the deletion of any edge leaves the graph connected.
A: $E$ is the number of edges.
$d(v)$ is the degree of the vertex $v$;(it is $4$ in our case for each $v$) then
$$E = \frac12\sum_{i=1}^n d(v_i)$$
If $4$-regular graph had bridge, then removing it would create a component with some $k$ vertices,in which $d(v) = 3$ for one vertex and $d(v) = 4$ for all the others, but this can't happen as
$E = (3 + (k-1)\cdot4) /2$ is not a natural number.
