Place different integers on the vertices of an octagon so that the sum of the integers in any two vertices joined by one of its edges is a power of 2. Do so in such a way that the largest integer used is as small as possible.

Do likewise on the vertices of a decagon.


Here is one possible arrangement:

6, -4, 5, -1, 2, 0, 4, -2

Proof of optimality:

Here is the graph for integers $\leq 5$:
enter image description here
Note that integers $\leq -4$ do not occur, as each one can connect to at most one integer $\leq 5$.
The graph is so simple that an exhaustive analysis can be done to show that there is no simple loop of length $8$.

Slightly better proof:

We look at the edge connecting 0 and 4. If this edge is used in the loop, then at least one of 1, 2 cannot present in the loop, and at least one of -2, -3 cannot present in the loop. This results in at most $7$ numbers in the loop.

Now assume that the edge connecting 0 and 4 is not used in the loop. We may thus delete that edge from the graph.

Since the loop is of length $8$, one number does not belong to the loop. However, in the current graph, removing any number will result in one of its neighbors having at most $1$ edge. Hence that neighbor cannot occur in the loop either. This leads to at most $7$ numbers in the loop.

In view of the answer by loopy walt, this also proves optimality for the case of decagon.


My best shot:

      3      -3

   -2           4

      6       0


And for 10:

        6  -4
    -2         5

   3            -3

    -1         4
        2   0



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