Answer
:
Here we have indeed
a knotty problem.
Our textbooks tell us
that all
spheres are similar, and that similar solids are as the cubes of
corresponding lengths.
Therefore, as the circumferences of the two
phials
were one foot and two feet respectively and the cubes of one and two
added together make nine, what we have to find is two other numbers
whose
cubes added together make nine.
These numbers
clearly must be
fractional.
Now, this little
question has really engaged the
attention of learned
men
for two hundred and fifty years; but although Peter de Fermat showed in
the seventeenth century how an answer may be found in two fractions
with
a denominator of no fewer than twentyone 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.
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.
Some years ago I published a solution for the case of
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.
