Here is the basic idea of edge magic graphs from Wolfram Math World (in case they is new to you):

"An edge-magic graph is a labeled graph with e graph edges labeled with distinct elements {1,2,...,e} so that the sum of the graph edge labels at each graph vertex is the same. "

And because a picture is worth a lot of words, here is an example from the same source:

In this example the four vertices along each edge have a sum of 24. Notice that this one uses 1-12 with 7 and 11 omitted. It seems that C. W. Trigg proved there was no solution with 1-10 as early as 1960 in response to a question posed in Pi Mu Epsilon.

I think I have made it make sense in my head that if 1-10 doesn't work, then no consecutive sequence of 10 numbers would. So now I'm wondering if it is possible to do it with some non-consecutive arithmetic sequence (1,3,5...) or 2,4,6 etc.. Maybe I've overlooked an easy reason why this would also be impossible, but I'll let my readers educate me, as they do so often.

It is possible to solve it using all prime numbers, and they don't get beyond two digits (although using consecutive primes is also possible, it gets pretty massive for paper and pencil work).

## 1 comment:

If it's an arithmetic progression of common difference d, then we could divide everything by d ... so I think your proof for consecutive integers applies just as well to any arithmetic progression, and they all fit together because every vertex touches the same number of edges.

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