# Solution

(related to Problem: Circling the Squares)

Though this problem might strike the novice as being rather difficult, it is, as a matter of fact, quite easy, and is made still easier by inserting four out of the ten numbers.

First, it will be found that squares that are diametrically opposite have a common difference. For example, the difference between the square of $14$ and the square of $2,$ in the diagram, is $192;$ and the difference between the square of $16$ and the square of $8$ is also $192.$ This must be so in every case. Then it should be remembered that the difference between squares of two consecutive numbers is always twice the smaller number plus $1,$ and that the difference between the squares of any two numbers can always be expressed as the difference of the numbers multiplied by their sum. Thus the square of $5$ ($25$) less the square of $4$ ($16$) equals $(2 \times 4) + 1,$ or $9;$ also, the square of $7$ ($49$) less the square of $3$ ($9$) equals $(7 + 3) \times (7 - 3),$ or $40.$

Now, the number $192,$ referred to above, may be divided into five different pairs of even factors: $2 \times 96,$ $4 \times 48,$ $6 \times 32,$ $8 \times 24,$ and $12 \times 16,$ and these divided by $2$ give us, $1 \times 48,$ $2 \times 24,$ $3 \times 16,$ $4 \times 12,$ and $6 \times 8.$ The difference and sum respectively of each of these pairs in turn produce $47, 49;$ $22, 26;$ $13, 19;$ $8, 16;$ and $2, 14.$ These are the required numbers, four of which are already placed. The six numbers that have to be added may be placed in just six different ways, one of which is as follows, reading round the circle clockwise: $16,$ $2,$ $49,$ $22,$ $19,$ $8,$ $14,$ $47,$ $26,$ $13.$

I will just draw the reader's attention to one other little point. In all circles of this kind, the difference between diametrically opposite numbers increases by a certain ratio, the first numbers (with the exception of a circle of $6$) being $4$ and $6,$ and the others formed by doubling the next preceding but one. Thus, in the above case, the first difference is $2,$ and then the numbers increase by $4,$ 6, $8,$ and $12.$ Of course, an infinite number of solutions may be found if we admit fractions. The number of squares in a circle of this kind must, however, be of the form $4n + 6;$ that is, it must be a number composed of $6$ plus a multiple of $4.$

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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