Is it possible to number the vertices of a cube from $1$ to $8$ in such a way that the value of the sum on every edge is different

Question

Is it possible to number the vertices of a cube from $1$ to $8$ in such a way that the value of the sum on every edge is different?
I tried using PHP but as far as I can get, there are $13$ distinct sums. So does this say that such a case will exist, or is there more to it?

Answer 1

As you note, the value of the sum of an edge can take $13$ distinct values; any value from $3$ to $15$ is possible. As there are $12$ edges, if every edge has a distinct sum then there is precisely one value between $3$ and $15$ that is not the sum of an edge, say $k$. Then the total sum of all edges is $$3+4+5+6+7+8+9+10+11+12+13+14+15-k=117-k.$$ On the other hand, the total sum of all edges is three times the total sum of all vertices, which is $$3\times(1+2+3+4+5+6+7+8)=108,$$ which shows that $k=9$. This also means that all other sums are the sum of an edge.

To get the sums $3$ and $4$ we see that $1$ shares an edge with $2$ and $3$. Then $2$ and $3$ do not share an edge, so to get the sum $5$ we see that $1$ must also share an edge with $4$. Then getting the sum $6$ is impossible; we conclude that no such configuration exists.