seventeenth century how an answer may be found in two fractions with
a denominator of no fewer than twenty-one figures, not only are all the
published answers, by his method, that I have seen inaccurate, but nobody
has ever published the much smaller result that I now print. The cubes of
(415280564497 / 348671682660) and (676702467503 / 348671682660) added
together make exactly nine, and therefore these fractions of a foot are
the measurements of the circumferences of the two phials that the Doctor
required to contain the same quantity of liquid as those produced. An
eminent actuary and another correspondent have taken the trouble to cube
out these numbers, and they both find my result quite correct.

If the phials were one foot and three feet in circumference respectively,
then an answer would be that the cubes of (63284705 / 21446828) and
(28340511 / 21446828) added together make exactly 28. See also No. 61,
"The Silver Cubes."

Given a known case for the expression of a number as the sum or
difference of two cubes, we can, by formula, derive from it an infinite
number of other cases alternately positive and negative. Thus Fermat,
starting from the known case 1^{3} + 2^{3} = 9 (which we will call a
fundamental case), first obtained a negative solution in bigger figures,
and from this his positive solution in bigger figures still. But there is
an infinite number of fundamentals, and I found by trial a negative
fundamental solution in smaller figures than his derived negative
solution, from which I obtained the result shown above. That is the
simple explanation.

We can say of any number up to 100 whether it is possible or not to
express it as the sum of two cubes, except 66. Students should read the
Introduction to Lucas's _Theorie des Nombres_, p. xxx.

Some years ago I published a solution for the case of

6 = (17/21)^3 + (37/21)^3,

of which Legendre gave at some length a "proof" of impossibility; but I
have since found that Lucas anticipated me in a communication to
Sylvester.

[Illustration]

21.--_The Ploughman's Puzzle._

The illustration shows how the sixteen trees might have been planted so
as to form as many as fifteen straight rows with four trees in every row.
This is in excess of what was for a long time believed to be the maximum
number of rows possible; and though with our present knowledge I cannot
rigorously demonstrate that fifteen rows cannot be beaten, I have a
strong "pious opinion