# Powerful Octagon

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 $$\leq 5$$:

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:


5
3      -3

-2           4

6       0
2



And for 10:


6  -4
-2         5

3            -3

-1         4
2   0