The Puzzle Of The Doctor Of Physic
(
CANTERBURY PUZZLES)
This Doctor, learned though he was, for "In all this world to him there was none like To speak of physic and of surgery," and "He knew the cause of every malady," yet was he not indifferent to the more material side of life. "Gold in physic is a cordial; Therefore he lovéd gold in special." The problem that the Doctor propounded to the assembled pilgrims was this. He produced two spherical phials, as shown in our illustration, and pointed out that one phial was exactly a foot in circumference, and the other two feet in circumference.
"I do wish," said the Doctor, addressing the company, "to have the exact measures of two other phials, of a like shape but different in size, that may together contain just as much liquid as is contained by these two." To find exact dimensions in the smallest possible numbers is one of the toughest nuts I have attempted. Of course the thickness of the glass, and the neck and base, are to be ignored.
Answer:
Here we have indeed a knotty problem. Our text-books 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 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. , "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 13 + 23 = 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 Théorie des Nombres, p. xxx.
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.
Informational
