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 x 6,377,352 squared + 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 8/7 ft., and of the smaller 3/7 ft., if we are to have the answer in the smallest possible figures.  Here the lineal measurement is 11/7 ft.—­that is, 1+4/7 ft.  What are the cubic contents of the two cubes?  First 8/7 x 3/7 x 8/7 = 512/343, and secondly 3/7 x 3/7 x 3/7 = 27/343.  Add these together and the result is 539/343, which reduces to 11/7 or 1+4/7 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 ab + c squared = denominator, and a squared — c squared, b squared — c squared, and a squared — b squared will be the three numerators.  Thus, using the generators 3, 2, 1, we get 8/7, 3/7, 5/7 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 6n + 1, 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:  99999999/99990001 and 19999/99990001.