How to use this course
Diophantus is a name attached to a book. Almost nothing certain is known about the man, and even his century is disputed. The book, the Arithmetica, is the most important work on arithmetic and algebra to survive from Greek antiquity. Its problems ask for rational numbers satisfying equations, and its methods were read by Arabic algebraists, by François Viète and by Pierre de Fermat, whose notes in its margin started modern number theory. Today any polynomial equation to be solved in integers or rationals is called a Diophantine equation.
- High-school algebra: linear and quadratic equations, factoring.
- Elementary number theory: divisibility, congruences modulo small numbers.
- For Lessons 06 and 09, coordinate geometry of lines, circles and simple curves. Some familiarity with elliptic curves helps but is not required.
Part I separates what the sources say from what later tradition added. Each technical lesson in Part II opens with a paragraph for the general reader, states Diophantus’s method in modern notation, shows how he applied it to his own numbers, and ends with exercises whose solutions are folded. Diophantus wrote no general theorems. When a box below states a theorem, it is a modern statement of what his method achieves, and the lesson says so. Problems are cited as book.number in the Greek numbering used by Tannery and Heath unless marked as Arabic.
Timeline
No date in Diophantus’s life is documented. The first rows give the limits within which he must have written; the rest trace the book.
| c. 150 BC | Hypsicles defines polygonal numbers. Diophantus quotes this definition, so he wrote later. |
| c. AD 250 | The most common estimate for his working life, based on a possible link with Anatolius, a third-century bishop of Laodicea. MacTutor gives c. 200 to c. 284. |
| 4th c. | Theon of Alexandria quotes Diophantus, the earliest known reference. This sets the latest possible date, around 350. |
| c. 400 | Hypatia, Theon’s daughter, writes a commentary on the Arithmetica, now lost (reported by the Byzantine lexicon Suda). |
| c. 500 | The age riddle about Diophantus enters the collection later included in Book 14 of the Greek Anthology, attributed to the grammarian Metrodorus. |
| 9th c. | Qusta ibn Luqa translates part of the Arithmetica into Arabic. |
| 11th c. | Michael Psellus mentions Diophantus and Anatolius in a letter. |
| 1198 | Date of the copy of the Arabic Books IV–VII later found in Mashhad. |
| c. 1290s | Maximus Planudes in Constantinople edits the Greek text and comments on the first two books. |
| 1463 | Regiomontanus reports a Greek manuscript of Diophantus and notes that no Latin translation exists. |
| 1570 | Rafael Bombelli translates much of the text into Latin (unpublished) and uses many problems in his Algebra. |
| 1575 | Wilhelm Xylander’s Latin translation, the first printed edition. |
| 1591 | Viète recasts many Diophantine problems in symbolic form in his Zetetica. |
| 1621 | Claude-Gaspard Bachet publishes the Greek text with a Latin translation in Paris. |
| c. 1637 | Fermat writes his note beside Problem II.8. |
| 1670 | Samuel de Fermat reprints Bachet’s edition with his father’s 48 observations. |
| 1893–95 | Paul Tannery’s critical edition of the Greek text. |
| 1910 | Thomas Heath’s Diophantus of Alexandria, second edition, the standard English account for a century. |
| 1968–71 | The Arabic Books IV–VII are identified in the Astan Quds library in Mashhad, Iran. |
| 1982, 1984 | Editions of the Arabic books by Jacques Sesiano (English) and Roshdi Rashed (French). |
Alexandria and the dating problem
Diophantus is called “of Alexandria” in the manuscripts, and nothing contradicts it. Alexandria had been the centre of Greek mathematics since Euclid, and under Roman rule it remained a city of schools and libraries. Beyond the name and city, everything about his life must be inferred.
The outer limits
- Lower limit. In On Polygonal Numbers Diophantus quotes a definition of Hypsicles, who worked around 150 BC. So he wrote after that.
- Upper limit. Theon of Alexandria, writing in the fourth century AD, quotes Diophantus. So he wrote before about 350.
These are the only firm facts. They leave a window of five centuries.
The Anatolius link
In the eleventh century the Byzantine scholar Michael Psellus wrote a letter that mentions Diophantus together with an Anatolius, who had collected the essentials of the same method of calculation. Paul Tannery, the nineteenth-century editor of Diophantus, thought Psellus was drawing on a lost commentary, perhaps Hypatia’s. Earlier scholars identified this Anatolius with the bishop of Laodicea who lived in the third century, and on that basis placed Diophantus around AD 250. This date is still the one most often given.
The inference is fragile. Wilbur Knorr retranslated the passage in 1993 and argued that it may concern a different Diophantus, and that the author of the Arithmetica could be earlier, perhaps even before Heron of Alexandria in the first century. The Arithmetica itself is addressed to a “most honoured Dionysius”. Some historians have wondered whether he was the Dionysius who became bishop of Alexandria in the mid-third century, but the name was common and the identification is a guess.
The best summary is that Diophantus probably flourished in the third century AD, and that this rests on a chain of plausible identifications, not on any document.
Exercise 1.1Classify each claim as documented, inferred or traditional, and name the source: (a) he wrote after Hypsicles; (b) he lived around 250; (c) he wrote before 350; (d) he lived in Alexandria.
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The epigram and other traditions
The one “biographical” text about Diophantus is a riddle. Book 14 of the Greek Anthology, a Byzantine collection of short poems, contains a group of arithmetical epigrams attributed to the grammarian Metrodorus, of the fifth or sixth century. One of them is presented as the epitaph on Diophantus’s tomb. It says, in substance:
- his boyhood lasted one sixth of his life;
- his beard grew after a further twelfth;
- after a further seventh he married;
- five years later a son was born;
- the son lived half as long as his father;
- the father died four years after the son.
The puzzle leads to a linear equation with the answer 84. Whether any of it is true is unknown. The poem is several centuries later than Diophantus, belongs to a genre of school puzzles, and is attached to him because his name stood for arithmetic. MacTutor and most historians treat its details as possibly fictitious. A course can use it as an exercise, and should not use it as evidence.
Hypatia
The tenth-century lexicon Suda says that Hypatia, the Alexandrian philosopher and mathematician killed in 415, wrote a commentary on Diophantus. It is lost. Tannery and Heath suggested that her commentary covered only the first six books, and that this is why only six survived in Greek. The hypothesis is attractive, but nothing proves it.
Exercise 2.1Write the equation of the epigram and solve it. At what ages, by the riddle, did Diophantus marry and become a father?
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Exercise 2.2Why must the answer be a multiple of 84 before any arithmetic is done, and what does that suggest about how the riddle was composed?
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The Arithmetica as a book
Diophantus says in his introduction that the work has thirteen books. Six survive in Greek. In the late 1960s four more, in an Arabic translation, were identified in the library of the shrine of Imam Reza in Mashhad, Iran, in a copy dated 1198. The Arabic text names its books as the fourth to the seventh, and the translation is attributed to Qusta ibn Luqa, a ninth-century scholar of Baalbek and Baghdad. Their content is more elementary than the Greek books traditionally numbered IV to VI. The current view, following Rashed and Sesiano, is that the Greek books are I–III and VIII–X, and the Arabic ones fill IV–VII. Three books remain lost. Together the surviving books contain about 290 problems.
Form
The introduction, addressed to Dionysius, explains the names for powers of the unknown, the rules for signs, and the reduction of an equation to a single term on each side. Then come the problems. Each states a task in general terms (“to divide a given square into two squares”), immediately fixes numbers (“let the given square be 16”), chooses an unknown cleverly, reduces to an equation of first or second degree, and checks the answer. There is no theorem, no general proof, and no symbol for a general given number. The generality lives in the method, which the reader learns by repetition across many examples.
What counts as a solution
Diophantus wants positive rational numbers. He has no negative numbers and no zero. An equation whose solution would be negative or irrational is called “absurd”; his example is \(4=4x+20\). He accepts fractions freely. This preference for rational solutions, where earlier Greek geometry had dealt in lengths and ratios of any kind, is what makes the book a work of arithmetic and the ancestor of number theory.
Other works
- On Polygonal Numbers: a fragment, a preface and five propositions, in a geometric style closer to Euclid (Lesson 10).
- Porisms: a lost collection of lemmas, three of which are cited in the Arithmetica (Lesson 08).
- Attributions: a lost work On Parts is known from a scholium to Iamblichus. Knorr argued that the Definitions transmitted under Heron’s name are by Diophantus. Both remain uncertain.
Exercise 3.1Diophantus treats \(ax^2+c=bx\), \(ax^2=bx+c\) and \(ax^2+bx=c\) (with positive \(a,b,c\)) as three different kinds of equation. Explain why, using his rules on admissible numbers.
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A notation for the unknown
For the general reader. Before Diophantus, Greek problems were written entirely in words. He introduced abbreviations for an unknown number, for its powers and for subtraction. Historians call this “syncopated” algebra, halfway between prose and the symbols we use now. It let him write an equation on a single line.
- Read an expression in Diophantus’s notation.
- Understand the rule of signs he stated.
- See what the notation could not express, and why that mattered.
The vocabulary
The forms below follow Heath’s reading of the manuscripts. The unknown is the arithmos, “number”. Its powers have names, and the higher ones are built from the lower by addition of exponents.
| \(x\) | arithmos, written with a sign like ς |
| \(x^2\) | dynamis (power), \(\Delta^{\Upsilon}\) |
| \(x^3\) | kybos (cube), \(\mathrm{K}^{\Upsilon}\) |
| \(x^4\) | dynamodynamis, \(\Delta^{\Upsilon}\Delta\) |
| \(x^5\) | dynamokybos, \(\Delta\mathrm{K}^{\Upsilon}\) |
| \(x^6\) | kybokybos, \(\mathrm{K}^{\Upsilon}\mathrm{K}\) |
| \(1\) | monas (unit), \(\mathring{\mathrm{M}}\), marking the constant term |
| \(1/x\) | arithmoston, the reciprocal, marked with a superscript χ; likewise for other powers |
| \(-\) | leipsis (lack), a sign like an inverted ψ, ⋔ |
| \(=\) | an abbreviation of isos (equal) |
Coefficients are Greek alphabetic numerals written after the power: \(\alpha=1,\ \beta=2,\ \gamma=3,\ \delta=4,\ \varepsilon=5,\ \varsigma=6,\ \zeta=7,\ \eta=8,\ \theta=9,\ \iota=10\), with further letters for tens and hundreds. Terms to be added are simply written side by side. All the subtracted terms are gathered after a single minus sign. Heath’s standard example is
\[ \mathrm{K}^{\Upsilon}\alpha\ \varsigma\,\eta\ \pitchfork\ \Delta^{\Upsilon}\varepsilon\ \mathring{\mathrm{M}}\alpha\ \ \text{ἴσ}\ \varsigma\,\alpha \quad\longleftrightarrow\quad x^3+8x-(5x^2+1)=x. \]The sign for the unknown resembles the numeral \(\varsigma=6\). Position separates them: a numeral always follows the name of a power. In the example the unknown carries the coefficient \(\eta=8\).
Rules
The introduction states the rule that a “lack” multiplied by a “lack” gives a “presence” (hyparxis), and a lack times a presence gives a lack: the rule of signs for subtracted terms. It also gives the two operations Diophantus uses to simplify every equation: add the lacking terms to both sides, and subtract like from like, until one term remains on each side. These are the operations that al-Khwarizmi would later call al-jabr and al-muqabala.
There is one symbol for the unknown. A second unknown has to be described in words or expressed in terms of the first. There is no symbol for a given but arbitrary number, so the general problem cannot be written as a formula. Every problem is therefore solved for particular numbers, and the method has to be inferred. Viète removed this limitation in 1591 by using letters for given quantities as well (see his course).
Exercise 4.1Translate \(\Delta^{\Upsilon}\beta\ \mathring{\mathrm{M}}\gamma\ \pitchfork\ \varsigma\,\delta\ \ \text{ἴσ}\ \mathring{\mathrm{M}}\theta\) into modern notation and solve it.
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Exercise 4.2Write \(x^4-2x^3+5=3x^2\) in Diophantine form, and explain why he would first rewrite it with no subtraction on the left.
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Determinate problems
For the general reader. Book I is mostly about problems with a single answer: find two numbers with a given sum and product, or a given sum and difference. Babylonian scribes had solved such problems two thousand years earlier. What is new in Diophantus is the way he chooses the unknown so that the problem collapses to a very simple equation.
- Learn the “half-sum plus or minus” substitution.
- See the condition Diophantus attaches for a rational answer.
- Recognize the discriminant in his conditions.
Problem I.27
Find two numbers whose sum is 20 and whose product is 96. Diophantus does not call the numbers \(x\) and \(20-x\). He lets them be \(10+x\) and \(10-x\), symmetric about half the sum. The product is \(100-x^2=96\), so \(x^2=4\), \(x=2\), and the numbers are 12 and 8. No quadratic formula is needed, because the linear term has been designed away.
He states the condition in advance: the square of half the sum must exceed the product by a square. In modern terms this is the requirement that the discriminant be a perfect square.
Given rationals \(s,p\), there are rationals \(u,v\) with \(u+v=s\) and \(uv=p\) if and only if \(\big(\tfrac s2\big)^2-p\) is the square of a rational \(d\). Then \(\{u,v\}=\{\tfrac s2+d,\ \tfrac s2-d\}\).
Proof. Put \(u=\frac s2+x\), \(v=\frac s2-x\). Then \(uv=\frac{s^2}{4}-x^2\), so \(uv=p\) exactly when \(x^2=\frac{s^2}{4}-p\). Positivity of both numbers requires \(0\le d<\frac s2\), which holds when \(p>0\).
The quadratic in general
When a problem leads to a full quadratic, Diophantus solves it by the equivalent of completing the square, and he shows that he knows the rule for the positive root in each of the three types of Lesson 03. In some places he computes the root without showing working, as if the procedure were standard. He accepts only one root, even where both are positive.
Exercise 5.1Find two numbers whose sum is 20 and the sum of whose squares is 208, by Diophantus’s substitution.
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Exercise 5.2Find two numbers whose difference is 4 and whose product is 96, choosing the unknown symmetrically.
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Exercise 5.3Why is \(4=4x+20\) “absurd” to Diophantus? Give a condition on positive \(a,b,c\) for \(ax+b=c\) to be admissible.
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Two squares and the chord method
For the general reader. Problem II.8 asks to split 16 into two squares of fractions. It looks like a puzzle, but the trick Diophantus uses finds every rational point on a circle, and it works for any conic that has one rational point. It is the oldest method in what is now called arithmetic geometry.
- Follow Diophantus’s solution of II.8.
- Interpret it as intersecting a conic with lines through a known point.
- Derive the parametrization of Pythagorean triples from it.
Problem II.8
To divide a given square into two squares. Let the square be 16. Let one of the squares be \(x^2\). Then the other is \(16-x^2\), which must be a square. Diophantus takes the side of that square to be “some number of \(x\) less the side of 16”, and chooses \(2x-4\). Then
\[ 16-x^2=(2x-4)^2=4x^2-16x+16 \ \Longrightarrow\ 5x^2=16x \ \Longrightarrow\ x=\tfrac{16}{5}. \]The squares are \(\frac{256}{25}\) and \(\frac{144}{25}\), and their sum is \(\frac{400}{25}=16\). The constant terms cancel by design, so the equation reduces to a linear one in \(x\).
The geometric reading
Write the condition as \(x^2+y^2=16\). The point \((0,-4)\) is on the circle. Diophantus’s choice \(y=2x-4\) is the line of slope 2 through it. A line through a rational point of a conic meets the conic in one more point, and that point is rational because the second root of a quadratic with rational coefficients and one rational root is rational. Varying the slope gives all rational points.
Let \(C\) be a conic given by a quadratic equation with rational coefficients, and let \(P\) be a rational point of \(C\). Then the rational points of \(C\) other than \(P\) correspond one-to-one with the rational slopes (including the vertical direction) of lines through \(P\), except for the tangent at \(P\). For the circle \(x^2+y^2=r^2\) and \(P=(0,-r)\), they are
\[ (x,y)=\Big(\frac{2rm}{1+m^2},\ \frac{r(m^2-1)}{1+m^2}\Big),\qquad m\in\mathbb Q. \]Proof sketch. Substitute \(y=mx-r\) into \(x^2+y^2=r^2\): \((1+m^2)x^2-2rmx=0\). The root \(x=0\) is \(P\); the other is \(x=\frac{2rm}{1+m^2}\), rational. Conversely the line joining \(P\) to any rational point has rational slope.
Diophantus never stated this theorem, but he used the same move, with an arbitrary coefficient in front of \(x\), across dozens of problems. Fermat’s famous note was written beside this problem (Lesson 12).
Exercise 6.1Redo II.8 with the side \(3x-4\). What squares do you get?
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Exercise 6.2Take \(r=1\) and \(m=p/q\) in lowest terms. Derive the formula for primitive Pythagorean triples.
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Exercise 6.3Diophantus also writes 13, given as \(4+9\), as a sum of two other squares. Use the rational point \((2,3)\) on \(x^2+y^2=13\) and the substitution \(x\mapsto x+2\), other side \(2x-3\).
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The double equation
For the general reader. Many problems ask for one number that makes two different expressions into squares at the same time. Diophantus had a standard technique for this, the “double equality”. It turns two conditions into one factorization.
- Learn the difference-of-squares method for two linear expressions.
- Reproduce one of his solutions.
- See its geometric meaning as a curve of genus 0 or 1.
The method
Suppose \(ax+b=u^2\) and \(cx+d=v^2\). Then \(u^2-v^2=(a-c)x+(b-d)\). Factor this difference as a product \(PQ\) of two expressions chosen so that the resulting equation is easy, and set \(u=\frac{P+Q}{2}\), \(v=\frac{P-Q}{2}\), since then \(u^2-v^2=PQ\). Substituting into either condition gives an equation for \(x\).
Find \(x\) so that \(10x+9\) and \(5x+4\) are both squares. The difference is \(5x+5=5\cdot(x+1)\). Take \(P=x+1\), \(Q=5\), so \(u=\frac{x+6}{2}\). Then
\[ \Big(\frac{x+6}{2}\Big)^2=10x+9\ \Longrightarrow\ x^2+12x+36=40x+36\ \Longrightarrow\ x=28. \]Check: \(10\cdot28+9=289=17^2\) and \(5\cdot28+4=144=12^2\).
The constant terms cancel because \(u=\frac{x+6}{2}\) has constant term 3, and \(3^2=9\) is the constant of \(10x+9\). Diophantus’s skill lies in choosing factors that make this happen.
Geometric meaning
The two conditions define the curve \(\{(x,u,v): u^2=ax+b,\ v^2=cx+d\}\). Eliminating \(x\) gives a single quadric \(c\,u^2-a\,v^2=cb-ad\) in \((u,v)\), a conic, so once one rational point is known the chord method of Lesson 06 produces infinitely many. When the expressions are quadratic in \(x\), the intersection of two quadrics in three-dimensional space is in general a curve of genus 1, an elliptic curve, and the problem becomes much harder. Diophantus solves such cases only for special numbers.
Exercise 7.1Find a positive integer \(x\) with \(x+4\) and \(x+13\) both squares.
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Exercise 7.2In the worked example, factor the difference instead as \(5x+5=(x+1)\cdot5\) with \(P=5\), \(Q=x+1\), but substitute into \(5x+4=v^2\). Do you get the same answer?
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Sums of squares and cubes
For the general reader. Which numbers are sums of two squares? Of three? Of four? Diophantus never asked these questions in general form. But the conditions he attaches to his problems show that he knew several of the answers, and Fermat, reading him, turned them into the theorems of modern number theory.
- Use the two-square identity as Diophantus did.
- Prove the easy halves of the two-, three- and four-square conditions.
- Meet the lost Porisms through a cube identity.
Products of sums of two squares
In Book III Diophantus uses the fact that 65, being the product of \(13=2^2+3^2\) and \(5=1^2+2^2\), is a sum of two squares in two ways, \(65=1^2+8^2=4^2+7^2\). This is an application of the identity
\[ (a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2=(ac+bd)^2+(ad-bc)^2, \]which in modern terms is the multiplicativity of the norm \(|a+bi|^2\) on the Gaussian integers.
Conditions that betray knowledge
Some of his problems carry conditions on the given numbers, often in a damaged text. Historians have read them as showing that he knew that no number of the form \(4n+3\) is a sum of two squares, and that none of the form \(24n+7\) is a sum of three squares. He also writes numbers as sums of four squares freely, as if he believed that every number can be so written. Bachet stated this explicitly in 1621, Fermat claimed a proof, and Joseph-Louis Lagrange proved it in 1770 using work of Leonhard Euler.
(1) A positive integer is a sum of two squares if and only if every prime \(\equiv3\pmod4\) divides it to an even power (Fermat, Euler). (2) A positive integer is a sum of three squares if and only if it is not of the form \(4^k(8m+7)\) (Legendre, Gauss). (3) Every positive integer is a sum of four squares (Lagrange, 1770).
A porism on cubes
One lemma that Diophantus cites from his Porisms says that for any rationals \(a>b>0\) there are positive rationals \(c,d\) with \(a^3-b^3=c^3+d^3\): a difference of two cubes is a sum of two cubes. One explicit solution, valid when \(a^3>2b^3\), is
\[ c=\frac{a(a^3-2b^3)}{a^3+b^3},\qquad d=\frac{b(2a^3-b^3)}{a^3+b^3}. \]Viète and Fermat both worked on these cube problems; they are now understood as moving along the cubic curve \(X^3+Y^3=a^3-b^3\) by the chord-and-tangent operations of Lesson 09.
Exercise 8.1Use the identity to write \(5\cdot13=65\) as a sum of two squares in both ways, and find two representations of \(13\cdot17=221\).
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Exercise 8.2Prove that no integer \(\equiv3\pmod4\) is a sum of two squares, and that no integer \(\equiv7\pmod8\) is a sum of three squares. Deduce the claim about \(24n+7\).
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Exercise 8.3Apply the cube formula to \(a=2\), \(b=1\) and check the result.
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Cubic problems and the tangent
For the general reader. When the expression to be made a square is a cubic, the geometry changes. The solutions lie on what is now called an elliptic curve. Diophantus found rational points on such curves by choosing his square so cleverly that the equation dropped to a single unknown. Read in coordinates, his choice is the tangent line to the curve, a method that became the foundation of the arithmetic of elliptic curves.
- Solve a cubic problem from Book VI as Diophantus did.
- Recognize his choice as a tangent line.
- State the chord-and-tangent construction and Mordell’s theorem.
An example
MacTutor cites a problem in which Diophantus needs \(x^3-3x^2+3x+1\) to be a square. He sets it equal to the square of \(\frac32x+1\):
\[ x^3-3x^2+3x+1=\tfrac94x^2+3x+1 \ \Longrightarrow\ x^3=\tfrac{21}{4}x^2 \ \Longrightarrow\ x=\tfrac{21}{4}. \]The constant and linear terms cancel, leaving two terms. The value is then a square: \(\big(\frac32\cdot\frac{21}{4}+1\big)^2=\big(\frac{71}{8}\big)^2\).
Why it works
Consider the curve \(E:\ y^2=f(x)\) with \(f(x)=x^3-3x^2+3x+1\). It contains the obvious point \(P=(0,1)\). A line \(y=1+mx\) through \(P\) meets \(E\) where \(f(x)=(1+mx)^2\), a cubic equation with root \(x=0\). If the slope \(m\) equals the slope of \(E\) at \(P\), then \(x=0\) is a double root, and the third root is rational. The slope at \(P\) is \(f'(0)/(2\cdot1)=\frac32\), exactly Diophantus’s coefficient.
Let \(E: y^2=x^3+ax^2+bx+c\) with rational coefficients and no repeated root of the cubic. If \(P,Q\) are rational points of \(E\), the line through them (the tangent if \(P=Q\)) meets \(E\) in a third point, which is rational. These operations make the rational points of \(E\), together with a point at infinity, into an abelian group \(E(\mathbb Q)\). Mordell’s theorem (1922) says that \(E(\mathbb Q)\) is finitely generated.
The third root is rational because the sum of the three roots of the cubic in \(x\) is minus the coefficient of \(x^2\), a rational number. Diophantus used the tangent and chord constructions in individual problems. Fermat used them deliberately, Newton described them geometrically, and Henri Poincaré in 1901 recognized the group structure. Mordell proved finite generation in 1922.
Exercise 9.1Verify that \(x=\frac{21}{4}\) makes \(x^3-3x^2+3x+1\) a square, by direct computation.
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Exercise 9.2On the same curve, use the chord through \(P=(0,1)\) and \(Q=(\frac{21}{4},\frac{71}{8})\) to find the sum of the \(x\)-coordinates of the three intersection points, and hence a third rational point.
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On Polygonal Numbers
For the general reader. Triangular numbers 1, 3, 6, 10, … count dots in triangles; square numbers count dots in squares; pentagonal numbers in pentagons; and so on. Diophantus’s short treatise proves, in the style of Euclid’s geometry, a test for whether a given number is polygonal of a given kind.
- Derive the general formula for polygonal numbers.
- Prove Diophantus’s criterion.
- Use it to recover the side of a polygonal number.
The \(n\)-th \(a\)-gonal number (\(a\ge3\)) is the sum of the first \(n\) terms of the arithmetic progression \(1,\ 1+(a-2),\ 1+2(a-2),\dots\). Hypsicles defined them this way. Summing,
\[ P_a(n)=\frac{n\big[(a-2)(n-1)+2\big]}{2}. \]If \(P\) is the \(n\)-th \(a\)-gonal number, then
\[ 8(a-2)P+(a-4)^2=\big[(a-2)(2n-1)+2\big]^2. \]Conversely, a positive integer \(P\) is \(a\)-gonal exactly when the left side is a square \(s^2\) with \(s>0\) and \(\frac{s-2}{a-2}\) an odd positive integer; then the side is \(n=\frac{1}{2}\big(\frac{s-2}{a-2}+1\big)\).
Proof. Expand: \(8(a-2)P=4(a-2)n[(a-2)(n-1)+2]=4(a-2)^2n^2-4(a-2)^2n+8(a-2)n\). Write \(k=a-2\). The right side is \((2kn-k+2)^2=4k^2n^2-4k^2n+k^2+8kn-4k+4\). The difference is \(k^2-4k+4=(k-2)^2=(a-4)^2\). Diophantus proved the same thing with line segments, and the fragment breaks off while he is treating the converse.
Exercise 10.1Show that the criterion for triangular numbers (\(a=3\)) is the familiar test “\(8P+1\) is a square”, and check \(P=36\).
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Exercise 10.2Is 70 a pentagonal number? If so, find its side.
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From Alexandria to Bachet
The Arithmetica survived through two routes, one Greek and one Arabic, which met again only in the twentieth century.
The Arabic route
In the ninth century Qusta ibn Luqa, a Christian scholar from Baalbek working in Baghdad, translated at least seven books into Arabic. The translation came after al-Khwarizmi’s algebra and was read in its light: the Arabic text uses the vocabulary of al-jabr. Algebraists of the following centuries, al-Karaji among them, took over many of Diophantus’s indeterminate problems. The four books later found in Mashhad come from this tradition.
The Greek route
In Constantinople the Greek text was copied and studied. Around the end of the thirteenth century Maximus Planudes edited it and wrote a commentary on the first two books. In the fourteenth and fifteenth centuries John Chortasmenos added scholia, including a complaint about the difficulty of the problems. Greek manuscripts reached Italy in the fifteenth century. In 1463 Regiomontanus wrote that he had found a Diophantus and that no one had yet translated it into Latin.
Printed Diophantus
- Rafael Bombelli, with Antonio Maria Pazzi, translated much of the text around 1570. The translation was never printed, but Bombelli placed many Diophantine problems in his Algebra (1572), giving them their first wide circulation in the West.
- Wilhelm Xylander (Holtzmann) published the first Latin translation at Basel in 1575.
- François Viète reworked problems from Diophantus in symbolic form in his Zeteticorum libri quinque (1591), proving that the new algebra could solve them in general.
- Claude-Gaspard Bachet de Méziriac published the Greek text with a Latin translation and commentary in Paris in 1621. This is the edition Fermat read.
DiscussionLagrange called Diophantus the inventor of algebra; Kurt Vogel and others reject the title, pointing to Babylonian and Arabic predecessors and successors. Using Lessons 04–07, argue for one position. What exactly did Diophantus add?
Fermat’s margin and Diophantine equations
Fermat read Bachet’s Diophantus closely and wrote observations in its margins. Beside Problem II.8, the splitting of a square into two squares, he wrote that it is impossible to split a cube into two cubes, or any higher power into two like powers, and that he had found a marvellous proof which the margin was too narrow to hold. His son Samuel printed the notes with a new edition of Bachet in 1670; the annotated copy itself is lost. The proof, by Andrew Wiles with Richard Taylor, came in 1994 and was published in 1995.
Fermat’s other notes on Diophantus contain his two-square theorem, his claim of the four-square theorem, his proof by infinite descent that a right triangle with rational sides cannot have square area, and problems on cubes. Euler, Lagrange, Legendre and Gauss spent much of the next century proving them.
The modern meaning
By the nineteenth century “Diophantine equation” meant any polynomial equation to be solved in integers or rationals. Some landmarks:
- 1900. David Hilbert’s tenth problem asks for a general procedure deciding whether a Diophantine equation has an integer solution. In 1970 Yuri Matiyasevich, completing work of Martin Davis, Hilary Putnam and Julia Robinson, proved that none exists.
- 1922. Louis Mordell proves that the rational points of an elliptic curve form a finitely generated group (Lesson 09).
- 1983. Gerd Faltings proves that a curve of genus at least 2 has only finitely many rational points. Diophantus’s chord method works on conics (genus 0), his tangent method on cubics (genus 1); in higher genus such methods must fail.
- 1995. Wiles’s proof of Fermat’s Last Theorem.
The Arabic books, edited by Jacques Sesiano (1982) and Roshdi Rashed (1984), changed the picture of the Arithmetica as a whole. A complete English translation of all ten surviving books, with commentary, by Jean Christianidis and Jeffrey Oaks appeared in 2023.
Final projectChoose one Diophantine method from Part II. State it in modern form, solve three of Diophantus’s problems with it in his own way, and then explain the geometry behind it: which curve the problem defines, and why the method works on that curve.
Connections
People linked to Diophantus of Alexandria across this collection. A name without a link has no page here yet. See Diophantus of Alexandria in the network of all 337 lives →
Influenced by
- Euclid
- Hypsicles
Influenced
- François Viète
- Hypatia of Alexandria
- Joseph-Louis Lagrange
- Leonhard Euler
- Pierre de Fermat
- Rafael Bombelli
- Claude-Gaspard Bachet de Méziriac
- Maximus Planudes
- Qusta ibn Luqa
- Theon of Alexandria
Colleagues and collaborators
- Anatolius of Laodicea
Also mentioned in this life
- Muhammad ibn Musa al-Khwarizmi
- André Weil
- René Descartes
- Carl Friedrich Gauss
- David Hilbert
- Adrien-Marie Legendre
- Isaac Newton
- Henri Poincaré
- Regiomontanus
- Andrew Wiles
Fields
- arithmetic 4 lives
- algebra 29 lives
- number theory 45 lives
On CoursesHub
Sources and further reading
Life and transmission
- J. J. O’Connor and E. F. Robertson, Diophantus of Alexandria, MacTutor History of Mathematics.
- Diophantus and Arithmetica, Wikipedia (with references to the editions).
- T. L. Heath, Diophantus of Alexandria: A Study in the History of Greek Algebra, 2nd ed., Cambridge, 1910.
- W. R. Knorr, Arithmêtikê stoicheiôsis: on Diophantus and Hero of Alexandria, Historia Mathematica 20 (1993).
- N. Schappacher, Diophantus of Alexandria: a text and its history, 2005.
- J. Christianidis and J. Oaks, The Arithmetica of Diophantus: A Complete Translation and Commentary, Routledge, 2023.
Texts and mathematics
- P. Tannery (ed.), Diophanti Alexandrini Opera omnia cum Graecis commentariis, 2 vols., Teubner, 1893–1895.
- J. Sesiano, Books IV to VII of Diophantus’ Arithmetica in the Arabic Translation Attributed to Qusṭā ibn Lūqā, Springer, 1982.
- R. Rashed (ed.), Diophante: Les Arithmétiques, vols. 3–4, Les Belles Lettres, 1984.
- C.-G. Bachet (ed.), Diophanti Alexandrini Arithmeticorum libri sex, Paris, 1621; reissued with Fermat’s notes, Toulouse, 1670.
- I. G. Bashmakova, Diophantus and Diophantine Equations, Mathematical Association of America, 1997.
- A. Weil, Number Theory: An Approach Through History from Hammurapi to Legendre, Birkhäuser, 1984.
- J. H. Silverman and J. Tate, Rational Points on Elliptic Curves, 2nd ed., Springer, 2015.
- Y. Matiyasevich, Hilbert’s Tenth Problem, MIT Press, 1993.