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Is it possible to create a fourth-order magic square consisting of consecutive composite numbers that don't form an arithmetic sequence? If possible, give an example . If not, provide a proof.

Clarification:

In case someone is not sure what consecutive composite numbers are, here is an example: 4, 6, 8, 9 and 10 are five consecutive composite numbers because they are all the composite numbers from 4 to 10 and they are listed in order.

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  • $\begingroup$ Doesn't every set of consecutive numbers form an arithmetic sequence? Or do you mean that for example 4 and 6 are consecutive because we're only looking at composite numbers? $\endgroup$
    – Jaap Scherphuis
    May 13, 2020 at 8:48
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    $\begingroup$ @JaapScherphuis See edit. $\endgroup$
    – Pedja
    May 13, 2020 at 8:59

1 Answer 1

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It is possible.

First start with the following standard magic square.

8  11 14 1
13 2  7  12
3  16 9  6
10 5  4  15

You can then transform it by

decrementing the numbers 1 to 8

which gives the following magic square which is not in arithmetic progression:

7  11 14 0
13 1  6  12
2  16 9  5
10 4  3  15

Now we just need to add the right number to make them all composite:

It has numbers 0-7 and 9-16, missing number 8. We need to find a prime p with a prime gap of at least 9 on both sides, and then add p-8. This makes the missing number the prime p, and all the other numbers composite. The first such prime is 211.

This gives the final answer:

210   214   217   203
216   204   209   215
205   219   212   208
213   207   206   218

This has magic constant

844

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