a common difference of 13 are 8092919380, 10692119380 and 12772919380 whole square numbers the common difference will always be divisible by 24, so it is obvious that our squares must be fractional. Readers should now try to solve the case where the common difference is 23. It is rather a hard nut.
129.—THE BATTLE OF HASTINGS.
Any number (not itself a square number) may be multiplied by a square that will give a product I less than another square. The given number must not itself be a square, because a square multiplied by a square produces a square, and no square plus 1 can be a square. My remarks throughout must be understood to apply to whole numbers, because fractional soldiers are not of much use in war.
Now, of all the numbers from 2 to 99 inclusive, 61 happens to be the most awkward one to work, and the lowest possible answer to our puzzle is that Harold's army consisted of 3,119,882,982,860,264,400 men. That is, there would be 51,145,622,669,840,400 men (the square of 226,153,980) in each of the sixty-one squares. Add one man (Harold), and they could then form one large square with 1,766,319,049 men on every side. The general problem, of which this is a particular case, is known as the "Pellian Equation"—apparently because Pell neither first propounded the question nor first solved it! It was issued as a challenge by Fermat to the English mathematicians of his day. It is readily solved by the use of continued fractions.
Next to 61, the most difficult number under 100 is 97, where 97×6,377,3522+1 = a square.
The reason why I assumed that there must be something wrong with the figures in the chronicle is that we can confidently say that Harold's army did not contain over three trillion men! If this army (not to mention the Normans) had had the whole surface of the earth (sea included) on which to encamp, each man would have had slightly more than a quarter of a square inch of space in which to move about! Put another way: Allowing one square foot of standing-room per man, each small square would have required all the space allowed by a globe three times the diameter of the earth.
130.—THE SCULPTOR'S PROBLEM.
A little thought will make it clear that the answer must be fractional, and that in one case the numerator will be greater and in the other case less than the denominator. As a matter of fact, the height of the larger cube must be 87 ft., and of the smaller 13/7 ft., if we are to have the answer in the smallest possible figures. Here the lineal measurement is 117 ft.—that is, 147 ft. What are the cubic contents of the two cubes? First 87×87×87=512343 and secondly 37×37×37=27343. Add these together and the result is 539343, which reduces to 117, or 147 ft. We thus see that the answers in cubic feet and lineal feet are precisely the same.
The germ of the idea is to be found in the works of Diophantus of Alexandria, who wrote about the beginning of the fourth century. These fractional numbers appear in triads, and are obtained from three generators, a, b, c, where a is the largest and c the smallest.
Then denominator, and ,, and will be the three numerators. Thus, using the generators 3, 2, 1, we get 37, 37, 57 and we can pair the first and second, as in the above solution, or the first and third for a second solution. The denominator must always be a prime number of the form , or composed of such primes. Thus you can have 13, 19, etc., as denominators, but not 25, 55, 187, etc.
When the principle is understood there is no difficulty in writing down the dimensions of as many sets of cubes as the most exacting collector may require. If the reader would like one, for example, with plenty of nines, perhaps the following would satisfy him: 9999999999990001 and 1999999990001.
131.—THE SPANISH MISER.
There must have been 386 doubloons in one box, 8,450 in another, and 16,514 in the third, because 386 is the smallest number that can occur. If I had asked for the smallest aggregate number of coins, the answer would have been 482, 3,362, and 6,242. It will be found in either case that if the contents of any two of the three boxes be combined, they form a square number of coins. It is a curious coincidence (nothing more, for it will net always happen) that in the first solution the digits of the three numbers add to 17 in every case, and in the second solution to 14. It should be noted that the middle one of the three numbers will always be half a square.
132.—THE NINE TREASURE BOXES.
Here is the answer that fulfils the conditions:—
| A=4 | B=3,364 | C=6,724 |
| D=2,116 | E=5,476 | F=8,836 |
| G=9,409 | H= 12,769 | I=16,129 |
Each of these is a square number, the roots, taken in alphabetical order, being 2, 58, 82, 46, 74, 94, 97, 113, and 127, while the required difference between A and B, B and C, D and E, etc., is in every case 3,360.
133.—THE FIVE BRIGANDS.
The sum of 200 doubloons might have been held by the five brigands in any one of 6,627 different ways. Alfonso may have held any number from 1 to 11. If he held 1 doubloon, there are 1,005 different ways of distributing the remainder; if he held 2, there are 985 ways; if 3, there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832 ways; if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8 doubloons, 388 ways; if 9 doubloons, 200 ways; if 10 doubloons, 60 ways; and if Alfonso held 11 doubloons, the remainder could be distri-
