- 1 This paper was presented to the conference on “Greek Geometrical Analysis : Problems and Prospects (...)
- 2 The beginning of this passage is also found in the pseudo-Heronean Definitions [Heron 1912, 120.21 (...)
- 3 Sidoli points out that the Proclean scheme is suitable for the simpler problems of the Elements, b (...)
- 4 Cf. [Netz 1999, 293]. However, some scholars still think that the Euclidean format applies to Diop (...)
- 5 Cf. Acerbi’s statement “In the Diophantine problems construction is almost absent” [Diofanto 2011, (...)
- 6 References to the Greek books of Diophantus are given according to Tannery’s edition [Diophantus 1 (...)
- 7 See “Index graecitatis apud Diophantum” [Diophantus 1893-1895, II, 281], s.v. προσδιορισμός, and T (...)
1According to a well-known passage in Proclus’s Commentary on the First Book of Euclid’s Elements [Proclus 1873, 203.1–207.25], the formal division of a mathematical proposition comprises six parts : πρότασις (proposition), ἔκθεσις (setting-out, exposition), διορισμός (definition of goal, problem/proof-specification), κατασκευή (construction), ἀπόδειξις (proof, demonstration), and συμπέρασμα (conclusion).2 No matter whether this scheme is suitable for describing the canonical exposition of any geometrical (theorem/problem) or arithmetical (like those of Books VII–IX of the Elements) proposition,3 it is certainly not appropriate to describe a problem worked-out by algebra, like those in Diophantus’s Arithmetica.4 For example, Diophantus’s solutions do not feature any stage corresponding to κατασκευή, with its accompanying diagram.5 Moreover, the crucial stage of ἀπόδειξις, in the sense of a textual unit containing chains of deductive steps, like the proofs we read in the Elements, is downgraded in Diophantus to a mere test-proof, that is, a verification, which is very often skipped in the Greek books of the Arithmetica. Likewise, the part that is designated by διορισμός almost never appears in the Greek books, and only in the books which are preserved in Arabic translation it appears with remarkable consistency. The word itself does not appear in the Greek books, in which we find to be employed four times the cognate noun προσδιορισμός [Diophantus 1893-1895, I, 36.3 ; 340.9],6 and the verb διορίζεσθαι [424.14 ; 428.21], but with the totally different meaning of the condition of solubility (determination).7 By virtue of the above, the scheme described by Proclus is evidently not suitable to describe the structure of the Diophantine proposition.
2Recognizing the unsuitability of the Proclean scheme, Sesiano proposed another one, according to which the presentation of a Diophantine worked-out problem comprises the following parts : πρότασις (described as “statement of the problem, in terms of required magnitudes and given ones, if any”) ; διορισμός (described as “limiting condition”) ; ἔκθεσις (described as “numerical setting of the given magnitudes”) ; ἀνάλυσις (not described at all) ; σύνθεσις (not described at all) ; and συμπέρασμα (described as “final statement”) [Sesiano 1982, 49]. Sesiano, therefore, regards analysis and synthesis as components of the Diophantine resolutory procedure.
- 8 Of course this feature is common to any algebraic solution as well. More specifically a solution i (...)
- 9 Cf. however n. 23 below.
3It is true that a characteristic of the Diophantine procedure is to operate with the quaesitum—provided that the latter has been named—and this makes the Diophantine solutions apparently analytical [cf. Acerbi in Diofanto 2011, 49].8 Interpretations of the Diophantine solutions through the machinery of the ancient analysis are witnessed in the scholarship. Besides Sesiano, one might mention for example the study of [Netz 2012], while the influential [Klein 1968] still remains the locus classicus for the idea of contextualising Diophantus’s approach within the framework of the ancient analysis.9 Concerns about this viewpoint have been raised by Acerbi in the introduction of his [Diofanto 2011]. Acerbi rightly remarks that there are modalities in the Diophantine approach which do not square with the analytical paradigm [Diofanto 2011, 49], and wonders if Diophantus himself would have considered his own solutions as analyses. He points out that “he never speaks of ‘synthesis’, but only of ‘demonstration’, which however is always ‘manifest’ ” [Diofanto 2011]. Besides, even if one were to “qualify the approach in Arithmetica not as ‘analytic’ but as operating by ‘reduction’ ”, as Acerbi suggests, even then the reductions cannot be regarded as relating to an analytical undertaking, since the reasoning is devoid of any connotation of search for preconditions [Diofanto 2011, 50].
- 10 However, Acerbi’s reservations are expressed within an interpretative framework which considers Di (...)
4The concerns expressed by Acerbi are legitimate.10 Be that as it may, the issue deserves further investigation, especially since the previous discussions fail to take into account the premodern algebra as interpretative context for Diophantus’s problem solving. The present paper has therefore a twofold aim : first, to provide additional arguments against the idea that Diophantus proceeds in his solutions by analysis-and-synthesis ; second, to discuss the very few propositions of the Arithmetica which are referred to as “analysis” or “synthesis”.
- 11 The term “metrical analysis” was introduced by [Sidoli 2004] to designate analytical practices con (...)
- 12 For recent studies on the meaning and use of the concept of “given” in Greek analytical practices, (...)
5Starting with the first, it is important to note that linguistic analysis does not support subsuming the Diophantine solutions into the field of ancient analysis. An essential feature of any ancient Greek analytical text, either geometric or metrical,11 is the use of the technical language of “givens” [δοθέν and/or δεδομένον] and the unfolding of mathematical arguments shaped as chains of givens, through which the predicate “given” is transferred from one object to another.12 In Acerbi’s words, the language of the “givens” is “one of the major features of the proof format of analysis and synthesis” [Diofanto 2011, 119]. Yet, this is not the case in Diophantus’s Arithmetica. Although the term “given” appears quite often in it, it is not employed in the core part of the propositions, where the working out of the task set forth in the enunciation takes place, nor are the solutions carried out through any chain of givens.
- 13 Another use of the word “given”, as part of the formulaic expression “triangle given in form”, occ (...)
- 14 An example of a problem in which the word “given” is employed in the generic enunciation and the d (...)
6Let us be more precise. The technical term “given” which is used in analytical texts should be distinguished from the general usage of the term, as, for example, when we say “to write a given number as a sum of two square numbers” or “given a number, to find in how many ways it can be polygonal”, and so on. Such usages cannot be classified as related to analytical practices [cf. Sidoli 2018a, 354]. But this is precisely the case of Diophantus’s primary use of the word “given” in Arithmetica.13 When this word appears in a proposition it appears in the enunciation (generic and/or instantiated) and in the determination, and not in the core part of the solution, that is, the setting up and the solution of the equation.14 One might say, by repeating the words of Sidoli [Sidoli 2018a, 358], that, in practicing problem-solving by analysis, given objects are given by the problem-solver, whereas given objects (namely, numbers) in the Arithmetica are always given by the problem-setter. Therefore, Diophantus does not use the word “given” with the technical meaning it has in ancient analytical texts, with one exception, Problem 10 of the Greek Book V, to be discussed later.
- 15 See “Index graecitatis apud Diophantum” [Diophantus 1893-1895, II, 262], s.v. ἀναλύειν εἰς μόριον, (...)
- 16 [Diophantus 1893-1895, II, 283], s.v. σύνθεσις, translated by Tannery as “additio, summa”.
- 17 [Diophantus 1893-1895, II, 283], s.v. συντιθέναι, translated by Tannery as “addere”.
7Moreover, Diophantus’s use of the words ἀνάλυσις and σύνθεσις is different from that in the ancient analytical texts. The word ἀνάλυσις is never used in the Greek books of the Arithmetica with the technical meaning it has in ancient analysis, while the verb ἀναλύειν is used only with the meaning “to reduce in the same part” [ἀναλύειν εἰς μόριον].15 As for the word σύνθεσις, it is used in the Greek books with the meaning of “addition, sum”,16 and the corresponding verb συντιθέναι with the meaning “to add”.17 In the Arabic text the words “analysis” and “synthesis” are used a couple of times in Problems 37, 42a, and 43 of Book IV, which will be discussed in the second part of this paper.
- 18 See [Acerbi 2011, 138–141], [Berggren & Van Brummelen 2000, 5–16], [Fournarakis & Christianidis 20 (...)
- 19 For the meaning of these technical terms in the context of the premodern algebraic practice see [O (...)
8Besides linguistic arguments, there is another point on which the problem-solving of Diophantus stands out compared with the analytical problem-solving, as the latter is practiced in the preserved analytical texts. Taking for granted the traditional quadripartite division of a proposition worked-out by analysis-and-synthesis into “transformation”, “resolution”, “construction”, and “demonstration”,18 in Diophantus’s solutions we do not observe what is very common in almost every instance of an analyzed proposition, that is, the intermingling of steps belonging to different parts. In propositions treated by analysis, steps belonging to “transformation” are often intermingled with steps belonging to “resolution”, as well as steps belonging to “construction” are intermingled with steps belonging to “demonstration” [Sidoli 2018a, 359]. Such intermingling is not possible in Diophantus’s solutions, in fact, in any solution by premodern algebra. For example, the treatment of the equation—which includes its simplification (application of the “restoration” and “confrontation” steps)19 and solution—can only start when the stage of its setting-up has been accomplished. In other words, the production of the equation, and the solution of the equation are two tasks that cannot be intermingled with each other. Likewise, the “demonstration” (test-proof) can only occur at the very end of the solution. So, in a solution by premodern algebra the steps are clearly separated from one another.
9More differences between premodern algebra and the sub-category of ancient analysis called “metrical analysis” have been identified by Sidoli :
There are a number of fundamental differences between the methods of metrical analysis and those of premodern algebra—found, for example, in Diophantus’ Arithmetics. For one thing, Diophantus’ procedures are, at least in principle, purely numerical, and do not rely on any underlying geometric conception.
Next, metrical analysis has no special nomenclature for designating objects to be sought, as is introduced in premodern algebra. In ancient analysis, we deal only with geometric objects and with those same objects when they are given in various ways—hence, there is no notational, or conceptual, device that allows us to set up a relation containing both given elements and explicitly sought elements. A problem in premodern algebra, however, begins with an instantiation of the stated problem using a certain equation, in which terms to be sought are set into relation with actual numbers that are stated to be given.
Finally, although metrical analysis provides an algorithm for computing a definite value, it does so in a different manner than premodern algebra. Diophantus’ procedure for setting up the equation with which he will work, allows him to set known and sought values on both sides of his equation, so that he can apply operations mathematically equivalent to the arithmetical operations without regard to the epistemological, or ontological, status of any of the objects in his equation. Of course, in metrical analysis we also find equations, and proportions, with non-given objects on both sides of the equation, which are then subjected to arithmetical operations and ratio manipulations, but this is done only as an intermediate step before immediately asserting that some geometrical object is given in terms of some computable number. In metrical analysis, it is this series of claims about what is given that functions as the primary problem-solving procedure. [Sidoli 2018a, 393–394]
- 20 This rule is not general ; for example, it does not apply to locus problems.
10To the above arguments one should add that a feature occurring quite often in problem-solving by analysis, whatever geometric or metrical, is that analysis begins with the initial assumption that the problem is solved in its entirety.20 Transferred in the setting of problem-solving by premodern algebra, this would mean that all sought-after numbers called for in the enunciation must be assumed to be known, that is, they have to be named right from the outset of the solution. Against this background, one would expect that any algebraic solution begins with assigning names to all numbers, the finding of which the enunciation of the problem calls for. This happens, of course, if there is only one unknown, and in some cases if there are just a few. But in more complicated problems, where several numbers are required, the assignments are done gradually, as the resolutory procedure progresses. This is the case, for example, of all “derivative” assignments, that is, assignments which are based on other assignments. For example, the very first problem of the Arithmetica, whose instantiated enunciation we write today as
- 21 For the technical terms of premodern algebra, the notions of “assignment”, “named” and “unnamed” t (...)
11is solved by assigning the smaller sought-after number (designated above by y) to be “1 Number” (which we will write “1N”), thus making the greater sought-after number (designated by x) to be “1 Number, 40 units” (which we will write “1N 40u”). The first assignment is “direct”, the second is “derivative”. The former is made from the outset, the latter presupposes not only the first but also that the addition of “1N” and “40u”, stipulated by the second condition of the problem, has been performed, and the sum (“1N 40u”) has been found.21
12The above arguments are enough to dismiss the idea that the analysis-and-synthesis format provides a suitable interpretative framework for Diophantus’s method of solution. But there is also another reason for rejecting such an idea : Diophantus describes clearly his method of solution in the introduction of Book I. And he doesn’t have a single word to say about “analysis” and/or “synthesis”. On the contrary, he speaks about a “way” [ὁδός] for solving arithmetical problems, the stages of which are outlined as follows :
- Generic enunciation.
- Determination (if necessary).
- Instantiation (sometimes accompanied by the instantiated enunciation).
- Set up of the equation (comprising (a) naming the unknowns and (b) working through the operations stipulated in the problem on the names).
- Treatment of the equation (comprising (a) simplification and (b) solution of the simplified form).
- Answer to the problem.
- Test-proof.
- 22 See the references given in n. 8 above.
- 23 The idea of contextualizing Diophantus’s modus operandi within the conceptual framework of the anc (...)
13That the above list portrays the structure of the Diophantine worked-out problem is shown in a number of recent studies.22 Now, the core part of this structure, stages 4 and 5, is the main theme discussed by Diophantus in the introduction, and there is nothing in his text suggesting that his undertaking was conceived to be part of the field of analysis-and-synthesis.23
- 24 VG is the Greek Book V, while IVA is the Arabic Book IV.
14As said before there are four instances in the preserved books of the Arithmetica which might be regarded as linked to practices belonging to the field of analysis. They are found in Problems VG.10, and IVA.37, 42a, and 43.24 In what follows, we examine these problems, paying special attention to the relevant aspects.
15Problem VG.10 asks to cut unit into two parts, and add to each part a different given number, and make a square. Its instantiated enunciation corresponds to the problem we write in modern algebra as
- 25 The relation 9u=□+□' (i.e., the sum of the given numbers increased by one unit must be the sum of (...)
- 26 We use ℓ as the sign standing for “lacking”, by which we translate the Greek λείψει. The correspon (...)
- 27 Henceforth the expression “forthcoming equations” will be used to designate such half-constructed (...)
- 28 Tannery’s text has (the abbreviations being resolved) μονάδων γ λείψει ἀριθμοῦ τινος (3 units lack (...)
16The resolution runs as follows : The conditions of the problem entail that 9u=□+□’, with 2u<□<3u.25 Diophantus assigns the square designated by □ to be 1 Power (we write this assignment as □ ≔1P), with 2u < 1P < 3u. Then, he subtracts the 1P from 9u, to obtain 9u lacking 1P (which we write, 9u ℓ26 1P). Hence, 9u ℓ 1P=□'. This half-constructed equation27 is easily solved. We simply have to assign the side of the unnamed square to be an expression of the type “3u lacking some Ns”,28 or, in quasi-modern terms, “3u ℓ mN” , the multitude (i.e., the coefficient m) being any positive rational number. But 1P must lie between 2u and 3u. Diophantus takes two squares, one, the smaller, greater than 2u, the other, smaller than 3u, and sets out to establish the 1P in the interval between the two squares. 
(square of 
), and 
(square of 
), are such squares. So, if we make the multitude (m) to be such that the solution of the equation which emerges from the forthcoming equation satisfies the double inequality 
, we shall solve the problem.
17Now, the equation which arises from the forthcoming equation, after an assignment for the unnamed square of the type mentioned above, is, in quasi-modern terms, 9u ℓ IP = m2P 9u ℓ 6mN (corresponding to 9 − t2=m2t2+9 − 6mt), thus 1N is 
. Therefore, we are looking for a value of m which satisfies the double inequality 
. This is a subsidiary problem, which Diophantus solves by positing m ≔ 1N. Working through the operations we obtain the double inequality 
, from which two inequalities emerge :
18Diophantus solves the two inequalities by applying the rules for solving the corresponding trinomial equations. For the first he finds successively : Square ½72u → 1296u ; multiply 17u by 17u → 289u ; subtract this from 1296u → 1007u ; take the square root of this, it is 
; take the half of 72u → 36u ; add this to 
, the result is 
; divide this by 17u, the result is 
. Thus, the 1N is not greater than 
. Similarly, for the second inequality he finds that the 1N is not smaller than 
. He adopts for 1N the numerical value 3½u, so this is the value of the unknown number of the subsidiary problem designated by m.
19We are now in place to continue the solution of the original problem, by making a suitable assignment for the side of the □'. The complete solution is shown in the table below :
| Assignments |
Operations |
Equation |
|
□ ≔ 1P(2u < 1P < 3u)
|
|
|
|
Subtract 1P from 9u → 9u ℓ 1P
|
|
|
|
9u ℓ 1P =□’
|
|
 ≔ 3u ℓ 3½ N
|
|
|
|
Square 3u ℓ 3½ N → 12¼ P 9u ℓ 21N
|
|
|
|
9u ℓ 1P = 12¼ P 9u ℓ 21N
|
20The equation simplifies to 13¼ P = 21N, hence 
. The square designated by □ is 
, so, by subtracting 2u, we find the sought-after number designated by x to be 
, while the number designated by y is found to be 
.
21The solution above involves three problems :
- 29 For the Greek text see [336.17–338.9]. Tannery followed Bachet by adding the diagram suggested by (...)
22At the first stage, the problem P1 is reduced to P2. At this stage, Diophantus formulates an argument framed in the language of “givens”. The argument has the format of analysis, and this happens only once throughout the preserved books of the Arithmetica. The text is the following :29
Let a unit, AB, be set out, and let it be cut at Γ, and let a dyad, AΔ, be added to AΓ, and a hexad, BE, to ΓB. Therefore, each of ΓΔ, ΓE is a square. And since AB is 1 unit, while AΔ and BE together is an octad, therefore ΔE as a whole becomes 9 units, and these need to be divided into two squares, the (squares) ΓΔ and ΓE. But since one of the squares is greater than AΔ, that is, of a dyad, but smaller than ΔB, that is, of a triad, I am reduced to dividing a proposed square, as in the present case the 9, into two squares, ΔΓ and ΓE, so that one, ΓΔ, is in the interval between the dyad and the triad. Indeed, ΓΔ having been found, and the dyad AΔ being given, then the remainder AΓ is given. But AB is 1 unit ; therefore BΓ, a remainder, is given. Therefore, Γ, at which the unit is cut, is, also, given.
- 30 Proposition 4 of Euclid’s Data is repeatedly implied in the deductive steps.
- 31 Displaying all the above characteristics Problem VG.10 is closely aligned with Diophantus’s projec (...)
23The chain of givens at the end of this argument confirms that once the problem P2 is solved, the original problem P1 is also solved.30 So, its aim is not to solve the original problem, but to reduce it. Rather than a complete ἀνάλυσις, therefore, we have here an ἀπαγωγή, in which the analytical procedure stops halfway, reaching another problem to solve [cf. Iwata 2016]. Moreover, the analytical argument does not function as a primary problem-solving procedure. This part of the resolution has nothing to do with algebra. The algebraic part of the solution starts after the reduction, it concerns the reduced problem, and features all key characteristics of a premodern algebraic solution : naming the unknowns, working through the operations stipulated by the enunciation, setting up, simplifying, and solving an equation framed in the language of names, and computing the numerical values of the sought-after numbers.31
24The problem asks to find a cubic number such that if we multiply the square that comes from its side by two given numbers, and we add what is gathered from each of them to the cubic number, amounts to a square number. Its instantiated version corresponds to the problem we write today as
- 32 “Māl” is a technical term of Arabic algebra, corresponding to the term Power (δύναμις, the square (...)
25Diophantus’s solution runs as follows. He assigns the sought-after cube (x3) to be 1 Cube (we write this assignment, x3 ≔ 1C), therefore the square that comes from its side (x2) is 1 Māl (we write, x2 ≔ 1M).32 Working through the operations he obtains the pair of forthcoming equations :
- 33 “Thing” is a technical term of Arabic algebra, corresponding to the term Number (ἀριθμός, the alge (...)
- 34 For the Arabic text see [Sesiano 1982, lines 1115–1127, the emphasis is added].
26Next, Diophantus must find a square expression (corresponding to □) to equate to the 1C 5M. He wants the side of the square to be “Things”,33 that is, a multitude of Things. Likewise he must find a square expression (corresponding to □') to equate to the 1C 10M. In order to find conditions on the respective “multitudes” he argues as follows :34
- 35 The capitalized “Equals” translates the verb ‘adala which is used in the statement of an algebraic (...)
If we make a side of the square that is a Cube and five Māls be Things, then its square is Māls. Then, if we subtract the five Māls in common from both sides, it leaves a Cube Equals35 Māls. Thus it is clear that the number assigned in this problem to be a Thing is equal to the number of the remaining Māls. And also, if we make a side of the square that is a Cube and ten Māls be Things, then its square is Māls. And if we subtract the ten Māls in common from both sides, it leaves a Cube Equals Māls. Therefore, the number assigned to be a Thing in this analysis is equal to the number of remaining Māls. So the remaining Māls in the first Equation should be equal to the remaining Māls in the second Equation. But the remaining Māls in the first Equation is a square number less five units, and the remaining Māls in the second Equation is a square less ten units, and so we should find two square numbers such that if we subtract ten units from the greater and five units from the smaller, (they) are equal.
- 36 mT stands for the word “Things”, in plural, that is “a multitude of Things”.
27Diophantus deploys in this passage a mental reasoning which runs as follows : if we assign the side of the □ to be of the form mT,36 the square is m2M, hence the equation that arises from the first forthcoming equation is of the type 1C 5M = m2M (corresponding to t3+5t2 = m2t2), thus 1T is of the form m2 − 5 (*). Likewise, if we assign the side of the □' to be nT, the square is n2M, hence the equation that arises from the second forthcoming equation is of the type 1C 10M = n2M (corresponding to t3 + 10t2=n2t2), thus 1T is of the form n2 − 10(**). Now, from (*) and (**) we obtain that m2 − 5 equals n2 − 10, or, by adding 10 in both sides, m2 + 5 equals n2. Thus, two square numbers must be found whose difference is 5 (the smaller of which must be greater than 5). This is a subsidiary problem coinciding with II.10, so Diophantus gives directly the solution 
and 
for the squares n2 and m2 respectively. We assign these numbers to be in terms of a Māl, instead of a unit, so the assignments for the two unnamed squares are □ ≔ 
and □’ ≔ 
Thus the equations which arise from the two forthcoming equations are :
28After simplification, both equations become 
, therefore 
, so the sought-after cubic number is 
.
29In the above solution two problems are involved
30the latter (P2) being the 10th Problem of Greek Book II. The reduction, which aims at finding suitable assignments for the unnamed squares □ and □', is expressed as a mental process, and in this context we find the word “analysis” to be used—unlike the above discussed Problem VG.10, where the similar reduction from the second to the third problem involved was not described as “analysis”.
- 37 The enunciation calls for the difference to be a square, but it does not say which of the two is t (...)
31Two numbers are required, a cube and a square, such that both the sum and the difference of a cube of the cube and a square of the square are square numbers. In our notation the problem is written :37
32Diophantus proposes three solutions for this problem. The first, for which he gives only the instructions, is by the method of double-equality. Thus, the stage of the final calculation of the sought-after numbers is not actually pursued, Diophantus being satisfied with merely indicating it by the phrase “after knowing the Thing, we can synthesize everything in the problem”. The other two solutions are in fact identical, the only difference being that in the latter the pair of forthcoming equations is reduced to another pair with smaller “coefficients” and “powers”. Both solutions are described without actually working through the calculations. The phrase marking the transition to the stage of the final calculation of the required numbers is, in the second solution, “Then we return to perform the synthesis of the problem”, and in the third, “Once we know it [i.e., the numerical value of the “Thing”], we return to synthesize the problem according to the way we set it up in the analysis”. Below we present the third solution, by adding all implicit calculations.
- 38 CCC, abbreviation of “Cube Cube Cube”, corresponding to the 9th power of the algebraic unknown.
- 39 MM, abbreviation of “Māl Māl”, corresponding to the 4th power of the algebraic unknown.
- 40 CCM, abbreviation of “Cube Cube Māl”, corresponding to the 8th power of the algebraic unknown.
33We assign the side of the sought-after cube (x3) to be 2 Things (we write this assignment, x ≔ 2T), so the cube is 8C and the cube of the cube is 512CCC.38 Likewise, we assign the side of the sought-after square (y2) to be 4 Māls (we write, y ≔ 4M), so the square is 16MM39 and the square of the square is 256CCM.40 Working through the operations we obtain the pair of forthcoming equations
34In order to find suitable assignments for the two unnamed squares Diophantus proceeds in the way we saw in the previous problem. The text runs as follows :
- 41 For the Arabic see [Sesiano 1982, lines 1391–1409, added emphasis].
If we wish, we can say : five hundred twelve Cube Cube Cubes and two hundred fifty-six Cube Cube Māls Equal a square, and five hundred twelve Cube Cube Cubes less two hundred fifty-six Cube Cube Māls Equal a square. And for any square divided by a square, the result of the division is a square. Thus we divide the five hundred twelve Cube Cube Cubes and the two hundred fifty-six Cube Cube Māls by a square, which can be a Cube Cube Māl or four Cube Cube Māls or nine Cube Cube Māls or sixteen Cube Cube Māls or by whatever we wish among the square numbers after we make any one of them Cube Cube Māls. As for Cube Cube Māls, the result of its division by Cube Cube Māls is a number, and for Cube Cube Cubes, the result is Things. Let us suppose that we divide them by sixteen Cube Cube Māls. Then the result of the division is thirty-two Things and sixteen units. And by the same amount we divide this square, let us divide the other square, which is five hundred twelve Cube Cube Cubes less two hundred fifty-six Cube Cube Māls, to get thirty-two Things less sixteen units. Thus thirty-two Things and sixteen units are a square, and thirty-two Things less sixteen units are a square. Let us look for a number that, if we add a given (mafrūd) number to it, which is sixteen, it gives a square, and if we subtract a given (mafrūd) number from it, which is sixteen, it gives a square. Once we have found that number, we divide it by thirty-two, so what results from the division is the Thing. Once we know it, we return to synthesize the problem according to the way we set it up in the analysis.41
35After reduction of “powers” and “coefficients” of the forthcoming equations, by dividing all terms by 16CCM, a new pair of forthcoming equations is obtained :
36Following the same procedure as in Problem 37, Diophantus assigns the □'' to be a square multitude of units, say m2u, so the 1T is found to be 
, and, similarly, he assign the □''' to be a square multitude of units, say n2u, thus the 1T is 
. Thus, we are reduced again to a subsidiary problem of the type of II.10, and, the rest of the resolution is the same as before.
37The problem asks to find two numbers, a cube and a square, such that a cube of the cube, if we add a given multiple of a square of the square to it, amounts to a square number, and if we subtract a given multiple of a square of the square from it, leaves a square number. In modern notation its instantiated version is :
38We adopt the assignments x3 ≔ 1C, y ≔ 2M, and work through the operations to obtain the pair of forthcoming equations :
39In order to find suitable assignments for the unnamed squares Diophantus argues as follows :
- 42 For the Arabic text see [Sesiano 1982, lines 1474–1494 ; emphasis added].
So if we assign (fara.dnā) a side of the square Equated to the Cube Cube Cube and the twenty Cube Cube Māls to be Māl Māls, then its square is Māl Māls by Māl Māls, of which one of them is called a Cube Cube Māl. If we Equate it with <the Cube> Cube Cube and the twenty Cube Cube Māls, then we subtract the twenty <Cube Cube Māls> in common, it leaves a Cube Cube Cube Equals Cube Cube Māls, its number being equal to a square less twenty, and that is the assigned (mafrūd) number of a Thing in this solution. Likewise, if we assign (fara.dnā) a side of the square Equated to the Cube Cube Cube less twelve Cube Cube Māls to be Māl Māls, then its square is Cube Cube Māls. And if we add to it the twelve Cube Cube Māls lacking from the Cube Cube Cube and we make them added in common to both sides, it amounts to a Cube Cube Cube Equals Cube Cube Māls, its number being equal to a square and twelve, and that is (also) the assigned number of a Thing in the problem. So a [great] square less twenty Equals a small square and twelve. And we add the twenty in common to the two sides together, giving a small square and thirty-two (which) Equal a great square. The small square is four units, and when thirty-two is added to it, it amounts to thirty-six, which is the great square. So we make the square Equated to the Cube Cube Cube and twenty Cube Cube Māls be thirty-six Cube Cube Māls, and the square Equated to the other square be four Cube Cube Māls. Both Equations turn out to be, after the restoration and confrontation and division, a Thing Equals sixteen units. We will now perform the synthesis of the problem in the way we did the analysis.42
40Following the usual procedure Diophantus assigns the side of the □ to be, in quasi-modern terms, mMM, the square is m2CCM, so the equation 1CCC 20CCM=m2CCM (corresponding to t9 + 20t8 = m2t8) arises, which is simplified to 1CCC=(m2−20)CCM (i.e., t9= (m2−20)t8), and, after division by a CCM the 1T is found to be of the form m2−20. Likewise, if we assign the side of the □' to be nMM the square is n2CCM, and so the equation 1CCC ℓ 12CCM = n2CCM (corresponding to t9 − 12t8 = n2t8) arises, which is simplified to 1CCC=(n2+12)CCM, and, after division by a CCM the 1T is found to be of the form n2+12. So, m2−20 = n2+12. Thus, we are led to a subsidiary problem, namely to find two square numbers whose difference is 32. This problem was solved in II.10, so Diophantus gives directly the solution 36u for m2 and 4u for n2. We assign these numbers to be in terms of a Cube Cube Māl, so the assignments for the two unnamed squares are □36 ≔ CCM and □' ≔ 4CCM. Thus the equations which arise from the two forthcoming equations are :
41After simplification each equation becomes 1CCC=16CCM, and, by division, 1T=16u. Therefore, the sought-after cube is 4096u, and the sought-after square is 262,144u.
42As we saw the words “analysis” and “synthesis”/ “synthesize”, with a meaning which might be regarded as relating to ancient analysis, appear within Arithmetica only in Problems 37, 42a, and 43 of the Arabic Book IV. These problems belong to a larger group consisting of Problems IVA.34–43, whose enunciations are (in our notation) :
- 43 The unnamed squares indicated in the table by the symbols □ and □' do not always coincide with the (...)
43The solutions of these problems involve mentally conducted reductions to other problems, namely to Problems II.9 and II.10. The table below shows these reductions, giving : (1) the pair of forthcoming equations which result in the course of the resolutory procedure in each case ;43 (2) the assignments adopted ; and (3) the problem from Book II from which the assignments are found.
44Two kinds of assignment are used in the solutions of the above problems, which might be characterized as “effective” and “virtual” (or “assignments in the indeterminate”). “Effective” is, for example, the initial assignment in Problem IVA.37, which is introduced by the sentence “We assign the cube to be one Cube”. “Virtual” is the assignment, in the same problem, which is introduced by the conditional phrase “If we make a side of the square [...] be Things,” that is, “[...] be [a multitude of] Things”, the multitude being unspecified. Thus, we have in this problem six assignments, four of which are “effective” and two “virtual”. The “effective” assignments are, 1) x3 ≔ 1C ; 2) x2 ≔ 1M ; 3) □ ≔ 7⅓T ; 4) □ ≔ 7⅔T, while the “virtual” assignments, which we give in quasi-modern notation, are, 5) □ ≔ mT ; 6) □' ≔ nT. The last two assignments may be considered as “assignments in the indeterminate”, in the same sense that the solutions to the lemmas to Problems IVG.34–36 are “solutions in the indeterminate” [ἐν τῷ ἀορίστῳ], whose role is to provide assignments to be used in the problems that follow.
45Now, according to Klein, “the calculation ending ‘in the indeterminate’, which Diophantus uses only as an auxiliary procedure, must be understood as the true analogue to geometric (‘problematical’) analysis” [Klein 1968, 163]. Under such a perspective, the “virtual” assignments 5) and 6) in Problem IVA.37 may be considered as belonging to the “transformation” part of an analytical reasoning, which ends with a known problem, namely the problem n2 – m2=5 (II.10). The “resolution” part of this analysis would be framed this way : Once a solution of the problem n2 – m2=5 is given, the multitude (m) of Things in the assignment of the side of □, and the multitude (n) of Things in the assignment of the side of □' are also given. Of course such “resolution” does not exist in the text. Always under this perspective, the “construction” part of the procedure would be the production of the “effective” assignments 3) and 4) from the “virtual” assignments 5) and 6).
- 44 What in the aforesaid problems from the Arabic Book IV is described by the expressions “synthesis (...)
46If we adopt this explanation for the word “analysis” in the text of the three problems, then we must admit that the analytical reasoning refers not to the whole solution but only to a task pertaining to a specific stage of it, that of finding assignments for the two unnamed squares. In this sense, analysis would be a heuristic tool to carry out only this task. However, textual evidence does not support this interpretation. For in Problems 42a and 43, just after the finding of the numerical value of the algebraic unknown (the “Thing”), and before the next stage, which is the calculation of the numerical values of the sought-after numbers, there is a sentence which marks the transition from one stage to the next. In Problem 42a, for which the text offers three solutions, the sentence in the first solution is “After knowing the Thing, we can synthesize everything in the problem” ; in the second solution is “Then we return to perform the synthesis of the problem” ; and in the third solution is “Once we know it [i.e., the numerical value of the “Thing”], we return to synthesize the problem according to the way we set it up it in the analysis”. In Problem 43 the sentence is, “We will now perform the synthesis of the problem in the way we did the analysis”. These phrases seem to imply that the word “analysis” refers to the whole series of assignments, and not only to the assignments for the two unnamed squares. Indeed, only in this manner is the allusion to “the synthesis of the problem” consistent with the calculation of all numbers for which the finding the enunciation of the problem calls.44
47Let us recapitulate : Problems 37, 42a, and 43 belong to a group of ten problems of the Arabic Book IV of the Arithmetica, in which a mental process is employed to reduce the task involved in a specific step of the resolutory procedure into a known problem from Book II. The words “analysis” and “synthesis”/ “synthesize” occur only in these three problems of the group. With the exception of the first occurrence of the word “analysis” in Problem 37, all other occurrences appear in sentences which mark a transition to the final stage of the solution, which is the calculation of the sought-after numbers. The table below shows the wording of the transition in the ten problems of the group :
| Pb |
Without sentence marking the transition |
With sentence marking the transition |
| 341 |
Let us then divide all of that by a Māl, to get one Thing Equals five units. Therefore a side of the cube is five units, and the cube is one hundred twenty-five... |
|
| 342 |
Therefore, the one Thing Equals sixteen units and a fourth of a unit. Since we assigned a side of the cube to be one Thing, a side of the cube is sixteen and a fourth... |
|
| 35 |
[...] it gives three and four fifths and a fifth of a fifth of one Equal one Thing. Since we assigned the cube to be from a side of one Thing, a side of the cube is ninety-six parts of twenty-five parts... |
|
| 36 |
[...] to get one Thing Equals twenty-one. Since we assigned a side of the cube to be one Thing, a side of the cube is twenty-one... |
|
| 37 |
[...] it gives one Thing Equals forty-eight units and seven ninths of a unit. And since we assigned a side of the cube to be one Thing, its side is four hundred thirty-nine ninths... |
|
| 38 |
[...] and the one Thing Equals fourteen units. And since we assigned a side of the cube to be one Thing, its side is fourteen... |
|
| 39 |
Therefore the Thing is three fourths of one. The Cube is twenty-seven eighths of an eighth, and a square of a side of the Cube is thirty-six eighths of an eighth. |
|
| 40 |
Therefore the one Thing is what results from dividing one thousand six hundred by three hundred eighty-four, which is four units and a sixth of a unit. Since we assigned a side of the square to be two Things, a side of the square is eight units and a third of a unit... |
|
| 41 |
So the one Thing is three units and a fifth of a unit. Since we assigned the square to be from a side of two Things, <its side> is six units and two fifths of a unit... |
|
| 42a1 |
|
And once the Thing is found, we can return to the hypostases we established. After knowing the Thing, we can synthesize everything in the problem. |
| 42a2 |
|
[...] and from that we can find the Thing whose value we are looking for. Then we return to perform the synthesis of the problem. |
| 42a3 |
|
so what results from the division is the Thing. Once we know it, we return to synthesize the problem according to the way we set it up in the analysis. |
| 42b |
Therefore, the one Thing is twelve parts of twenty-five. And since we assigned a side of the cube to be two Things, a side of the cube is twenty-four parts of twenty-five parts of a unit... |
|
| 43 |
|
We will now perform the synthesis of the problem in the way we did the analysis. |
48The only occurrence of the word “analysis” in a non-transitional sentence is found in Problem 37. Referring to the second equation of the pair of forthcoming equations :
49
50the text has the phrase “the assigned number in this analysis is a Thing equal to the number of remaining Māls”. However, a few lines above, referring to the first forthcoming equation, the phrase has “problem” instead of “analysis” : “the number assigned in this problem to be a Thing is equal to the number of the remaining Māls”. It seems, therefore, that the use of word “analysis” is due to a scribal mistake. This is corroborated by the fact that all other problems of the group 37–43 which contain similar phrases, have “problem”, as shown in the table below :
| Pb |
Forthcoming equations |
|
| 342 |
1C 4M=□ |
“the number that was assigned in the problem to be one Thing...” |
|
1C ℓ 4M=□' |
“the number that was assigned in the problem to be a Thing...” |
| 35 |
4M 1C=□ |
“the number that was assigned in the problem as <a Thing>...” |
|
4M ℓ 1C=□' |
“the number assigned in the problem to be a Thing...” |
| 37 |
1C 5M=□ |
“the number assigned in this problem to be a Thing...” |
|
1C 10M=□' |
“the assigned number in this analysis is a Thing...” |
| 38 |
1C ℓ 5M=□ |
“the number taken to be Things in the problem” |
|
1C ℓ 10M=□' |
“the number taken to be Things in the problem” |
51As for the transitional sentences in Problems 42a–43 which contain the words “analysis” and “synthesis”/ “synthesize,” they are in all probability later additions. The extreme case that the problems as a whole are not genuine cannot be excluded, especially if one takes into account that in the Fahrī of al-Karajī, which borrows heavily from Diophantus, the last three problems of the Arabic Book IV are omitted [Sesiano 1982, 60].
52Let us provide a brief summary of what we discussed in this paper. In Section I we highlighted the unsuitability of two schemes which have been proposed for describing the structure of the propositions in Arithmetica, the Proclean and the analytical. A weakness that both schemes share in common is that they do not take into account that Diophantus’s solutions are solutions by premodern algebra. In Section II we investigated in particular some discrepancies between the analytical style of reasoning and Diophantus’s presentation. In Section III we discussed the only occurrence of an analytical argument in the preserved Arithmetica, namely the argument by “givens” in Problem 10 of the Greek Book V, and we showed that its function is to reduce the problem into another, and not to solve it. Lastly, in Section IV we investigated the arguments referred to as “analyses” and “syntheses” in three problems from the Arabic Book IV, and we concluded by questioning the genuineness of the relevant passages.
53The basic thesis that recurs like a leitmotif throughout this paper is that Arithmetica is a work of premodern algebra, and so the solutions it contains not only exhibit the technical vocabulary of Diophantus’s algebra but also are structured according to the norms of the premodern algebraic problem solving.
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