An extension of the principle involved in this puzzle is given in No. 42, "The Riddle of the Pilgrims." See also "The Eight Villas" and "A Dormitory Puzzle" in _A. in M._ 18.--_The Shipman's Puzzle._ There are just two hundred and sixty-four different ways in which the ship _Magdalen_ might have made her ten annual voyages without ever going over the same course twice in a year. Every year she must necessarily end her tenth voyage at the island from which she first set out. [Illustration] 19.--_The Puzzle of the Prioress._ The Abbot of Chertsey was quite correct. The curiously-shaped cross may be cut into four pieces that will fit together and form a perfect square. How this is done is shown in the illustration. See also p. 31 in _A. in M._ 20.--_The Puzzle of the Doctor of Physic._ 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. 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.