Every other vertex shares $5$ neighbors with fixed vertex
Assume all vertices have even degree. Then, (a) the set of neighbors, $N$ say, of $v$ has even size, and (b) the vertices outside of $N \cup \{v\}$, call this $M$, has size $29-|N|$, which is odd.
We count the total number of stubs (edge-endpoints) in $N$: if every vertex in $N$ has even degree, the total number of stubs should be even. We count:
- Each $N$-to-$N$ edge contributes $2$ stubs (an even number in total).
- Each $v$-to-$N$ edge contributes $1$ stub (an even number in total).
- Each $M$-to-$N$ edge contributes $1$ stub (an odd number, $5|M|$, in total).
This gives a contradiction.
This is illustrated below:
