Use Tutte's synthesis to prove that the Harary graph $H_{3,n}$ is 3-connected $\forall n>4$. 
Use Tutte's synthesis to prove that the Harary graph $H_{3,n}$ is 3-connected $\forall n>4$.

I thought I could prove this by induction; I was able to prove the base case $(H_{3,4})$, but I couldn't go further could someone give me some hints on how to proceed?
 A: The important thing here is that you need to exploit the differences in definition between $H_{3,2k}$ and $H_{3,2k+1}$. Induction seems a good idea, indeed.
The induction base is trivial, since $H_{3,4}$ and $H_{3,5}$ actually are wheels.
Case 1: $n=2k+1$ ($k\geq 3$).
$H_{3,2k}$ is 3-connected by the induction hypothesis.
Note that $H_{3,2k}$ is $C_{2k}$ with long diagonals, i.e. an edge between antipodal vertices $i$ and $i+k$.
Now add an edge between $k$ and $2k-1$ (these are nonadjacent since $k\geq 3$).
Then split vertex $2k-1$ in $2k-1$ and $2k$.
The new vertex $2k$ takes the edges to $0$ and the edge to $k$ from $2k-1$.
The result is $H_{3,2k+1}$.
Case 2: $n=2k$ ($k\geq 3$).
$H_{3,2k-1}$ is 3-connected by the induction hypothesis.
Now we first shift all labels $k,\ldots,2k-1$ one up, then split vertex $k-1$ into $k-1$ and $k$.
The new vertex $k$ takes the edges to $0$ and the edge to $k+1$ from $k-1$.
The result is $H_{3,2k}$.
This may look a bit complicated but if you just draw a picture while reading, taking $k=3$ in both cases, it will be rather obvious.
