CHAPTER V THE PORISMS AND OTHER ASSUMPTIONS IN DIOPHANTUS I HAVE already mentioned (in Chapter I.) the three explicit references made by Diophantus to "The Porisms" and the possibility that, if these formed a separate work, it may have been from that work that Diophantus took a number of other propositions relating to properties of numbers which he enunciates or tacitly takes for granted in the Arithmetica. I begin with the three propositions for which he expressly refers to "The Porisms." Porism 1. In V. 3 he says, "We have it in the Porisms that, 'If each of two numbers and their product when severally added to the same given number produce squares, the squares with which they are so connected are squares of two consecutive numbers1."" That is to say, if x+a=m2, y+a=n2, and if xy+a is also a square, then m~n = 1. The theorem is not correctly enunciated, for it would appear that m~ n = 1 is not the only condition under which the three expressions may be simultaneously squares. For suppose x+a=m3, y+a = n2, xy+a=p3. By means of the first two equations we have may be a square certain conditions must be satisfied. One sufficient condition is 1 Literally "(the numbers) arise from two consecutive squares" (yeyóvaoir áæò dúo τετραγώνων τῶν κατὰ τὸ ἑξῆς). But we may also regard m2n2 — a (m2 + n2 − 1 ) + a2 = p2 as an indeterminate equation in m of which we know one solution, namely m=n± I. Other solutions are then found by substituting ≈ + (n + 1) for m, whence we obtain the equation or + (n3 − a)(n + 1)2 — a (n2 − 1 ) + a2 = p3‚ (n2 − a) x2 + 2 (n2 − a) (n ± 1) z + {n (n ± 1) − a}2 = p2, which is easy to solve in Diophantus' manner, since the absolute term is a square. But in the problem v. 3 three numbers are required, such that each of them, and the product of each pair, when severally added to a given number, produce squares. Thus if the third number be 2, three additional conditions have to be satisfied, namely z+a=u2, zx + a = v3, zy+a=w3. The two last conditions are satisfied, if m+ 1 = n, by putting and perhaps this means of satisfying the conditions may have affected the formulation of the Porism1. The problem v. 4 immediately following assumes the truth of the same Porism with a substituted for + a. Porism 2. In v. 5 Diophantus says, “Again we have it in the Porisms that, 'Given any two consecutive squares, we can find in addition a third number, namely the number greater by 2 than the double of the sum of the two squares, which makes the greatest of three numbers such that the product of any pair of them added to either the sum of that pair or the remaining number gives a square."" That is, the three numbers m2, (m+1)2, 4(m2 +m+ 1) 1 Euler has a paper describing and illustrating a general method of finding such "porisms" the effect of which is to secure that, when some conditions are satisfied, the rest are simultaneously satisfied ("De problematibus indeterminatis quae videntur plus quam determinata" in Novi Commentarii Acad. Petropol. 1756–57, Vol. vi. (1761), p. 85 sqq. Commentationes arithmeticae collectae, I. pp. 245-259). This particular porism of Diophantus appears as a particular case in § 13 of the paper. In fact, if have the property that the product of any two plus either the sum of those two or the remaining number gives a square. X, Y, Z denote the numbers respectively, XY+X+Y= ( m2 + m + 1)2, XY+Z= ( m2 + m + 2)2, YZ + X = (2m3 + 3m + 2)2, ZX + Y = (2m2 + m + 1)2. Porism 3 occurs in v. 16. Unfortunately the text is defective and Tannery has had to supply three words1; but there can be no doubt that the correct statement of the Porism here in question is "The difference of any two cubes is also the sum of two cubes," i.e. can be transformed into the sum of two cubes, or two cubes can be found the sum of which is equal to the difference between any two given cubes. Diophantus contents himself with the enunciation of the proposition and does not show how to prove it or how he effected the transformation in practice. The subject of the transformation of sums and differences of cubes was investigated by Vieta, Bachet and Fermat. Vieta (Zetetica, IV. 18-20) has three problems on the subject. (1) Given two cubes, to find in rational numbers two other cubes such that their sum is equal to the difference of the given cubes'. As a solution of a3 — b3 =x3 +3, he finds 1 ἔχομεν δὲ ἐν τοῖς Πορίσμασιν ὅτι " πάντων δύο κύβων ἡ ὑπεροχὴ κύβων < δύο σύνθεμά ἐστιν >. ? The solution given by Vieta is obtainable thus. The given cubes being a3, 63, where a>b, we assume x - b, a - kx as the sides of the required cubes. Vieta's second problem is similarly solved by taking a+x, kx-b as the sides of the required cubes, and the third problem by taking x-b, kx-a as the sides of the required cubes respectively. (2) Given two cubes, to find in rational numbers two others such that their difference is equal to the sum of the given cubes. Solving a + b = x3 —ƒa3, we find that (3) Given two cubes, to find in rational numbers two cubes such that their difference is equal to the difference of the given cubes. For the equation a3 - b3=x3-y, Vieta finds x= b (2a3-b3) , y = a (2b3-a3) as a solution1. In the solution of (1) x is clearly negative if 263 > a3; therefore, in order that the result may be “rational," a must be > 26. But for a "rational" result in (3) we must, on the contrary, have a3< 2ba3. Fermat was apparently the first to notice that, in consequence, the processes in (1) and (3) exactly supplement each other, so that by employing them successively we can effect the transformation required in (1) even when a is not > 26. The process (2) is always possible; therefore, by a suitable combination of the three processes, the transformation of a sum of two cubes into a difference of two cubes, or of a difference of two cubes into a sum or a difference of two other cubes is always + (465) = (472)3, but not the simpler solution 33+43+53=63. Euler ac 37 37 1 Vieta's formulae for these transformations give any number of very special solutions (in integers and fractions) of the indeterminate equation x3+μ3+23±3, including solutions in which one of the first three cubes is negative. These special solutions are based on the assumption that the values of two of the unknowns are given to begin with. Euler observed, however, that the method does not give all the possible values of the other two even in that case. Given the cubes 33 and 43, the method furnishes the solution 33 +43 + cordingly attacked the problem of solving the equation x3+μ3+33=23 more generally. He began with assuming only one, instead of two, of the cubes to be given, and, on that assumption, found a solution much more general than that of Vieta. Next he gave a more general solution still, on the assumption that none of the cubes are given to begin with. Lastly he proceeded to the problem To find all the sets of three integral cubes the sum of which is a cube and showed how to obtain a very large number of such sets including sets in which one of the cubes is negative (Novi Commentarii Acad. Petropol 1756-57, Vol. VI. (1761), p. 155 sq. = Commentationes arithmeticae, 1. pp. 193–207). The problem of solving 23+53=23+23 in integers in any number of ways had occupied Frénicle, who gave a number of solutions (Oeuvres de Fermat, 111. pp. 420, 535); but the method by which he discovered them does not appear. practicable'. Fermat showed also how, by a repeated use of the several processes as required, we can transform a sum of two cubes into a sum of two other cubes, the latter sum into the sum of two others and so on ad infinitum®. Besides the "Porisms" there are many other propositions assumed or implied by Diophantus which are not definitely called 1 Fermat (note on IV. 2) illustrates by the following case : Given two cubes 125 and 64, to transform their difference into the sum of two other cubes. Here a=5, b=4, and so 263>a3; therefore we must first apply the third process by which we obtain 3 cubes 3 , we can, by the first process, turn the difference between the into the sum of two cubes. "In fact," says Fermat, "if the three processes are used in turn and continued ad infinitum, we shall get a succession ad infinitum of two cubes satisfying the same condition; for from the two cubes last found, the sum of which is equal to the difference of the two given cubes, we can, by the second process, find two more cubes the difference of which is equal to the sum of the two cubes last found, that is, to the difference between the two original cubes; from the new difference between two cubes we can obtain a new sum of two cubes, and so on ad infinitum.” As a last illustration, to show how a difference between cubes can be transformed into the difference between two other cubes even where the condition for process (3) is not satisfied, Fermat takes the case of 8-1, i.e. the case where * Suppose it required to solve the fourth problem of transforming the sum of two cubes into the sum of two other cubes. Let it be required so to transform 23+ 13 or 9. First transform the sum into a difference of two cubes by process (2). This gives The latter two cubes satisfy the condition for process (3) and, applying that process, we get The cubes last found satisfy the condition for process (1), and accordingly the difference between the said last cubes, and therefore the sum of the original cubes, is at last transformed into the sum of two other cubes. |