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Fibonacci

Leonardo of Pisa · Leonardo Pisano, called Bigollo

A course in two strands: the life of the Pisan merchant’s son who learned Indian numerals in a North African customs house, and the mathematics of his books, from place-value arithmetic and commercial algebra to the rabbit sequence and a remarkable treatise on squares.

Born c. 1170, probably Pisa Died after 1240, probably Pisa Liber abaci 1202, revised 1228 Level undergraduate → advanced undergraduate
Squares of side 1, 1, 2, 3, 5, 8, 13 tile a 21 × 13 rectangle, because each new side is the sum of the two before. The arcs approximate a golden spiral. The picture is modern; Leonardo drew nothing like it.
Lesson 00

How to use this course

Most people meet Leonardo of Pisa through a sequence of numbers he did not invent and a nickname he never used. His real achievement was larger. His Liber abaci of 1202 taught Latin Europe to compute with the Hindu-Arabic numerals and with the algebra of the Islamic world, and his Liber quadratorum of 1225 is the most original work of number theory written in Europe between antiquity and Pierre de Fermat.

Prerequisites for Part II
  • School algebra, including quadratic equations.
  • Proof by induction and basic modular arithmetic (congruences, gcd).
  • For Lesson 07, comfort with linear recurrences; for Lesson 08, the rational root test and Newton’s method.

The biographical record is thin. Almost everything we know comes from three kinds of source: what Leonardo says about himself in the prologue to the Liber abaci and in his dedications, the surviving manuscripts of his books, and one decree of the Commune of Pisa from 1240. The lessons below say which source supports each claim and mark later tradition as such. All dates are CE. Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, arguments, and exercises with folded solutions. Leonardo wrote everything in words; the symbols are ours.

Reference

Timeline

c. 1170Born, probably in Pisa, son of Guglielmo (Guilielmus), of the Bonacci family. The year is an estimate; some historians prefer c. 1175.
c. 1180s–90sHis father serves as a public notary for Pisan merchants at the customs house of Bugia (Béjaïa, Algeria); Leonardo joins him and studies calculation with the Indian figures (prologue of the Liber abaci).
c. 1190sTravels on business to Egypt, Syria, Greece, Sicily and Provence, learning local methods (prologue).
c. 1200Settles back in Pisa.
1202Completes the Liber abaci. No copy of this first version survives.
1220Practica geometriae, dedicated to Dominicus Hispanus (some scholars date it 1223).
c. 1225Presented to Emperor Frederick II at Pisa; Johannes of Palermo poses problems. Flos, the Liber quadratorum and the letter to Master Theodorus follow.
1228Revised Liber abaci, dedicated to Michael Scot; this version survives.
1240The Commune of Pisa grants “Master Leonardo Bigollo” an annual salary for accounting work and teaching. Last documentary trace.
after 1240Dies, probably in Pisa; the year is unknown (c. 1250 is often given).
1506A Pisan notary, Perizolo, writes “Lionardo Fibonacci”, the earliest known use of the nickname.
1838Guglielmo Libri popularizes the name “Fibonacci” in his history of mathematics in Italy.
1857–62Baldassarre Boncompagni publishes the Scritti di Leonardo Pisano, the first printed edition.
1863Statue by Giovanni Paganucci erected in Pisa; now in the Camposanto Monumentale.
1870sÉdouard Lucas attaches Leonardo’s name to the rabbit sequence.
1963The Fibonacci Association begins publishing The Fibonacci Quarterly.
2002Laurence Sigler’s complete English translation of the Liber abaci.
Part I · Formation
Lesson 01

Pisa and Bugia

Leonardo was born around 1170, probably in Pisa. No birth record exists; the date is inferred backwards from the 1202 Liber abaci and from his father’s career. Pisa was then one of the great maritime republics of Italy, a rival of Genoa and Venice, with trading colonies across the Mediterranean. Its merchants bought wool, leather, wax and alum in North African ports and needed officials there to represent them.

His father, Guglielmo, belonged to a family known as the Bonacci. In the prologue of the Liber abaci Leonardo tells us that his father was a public scribe or notary at the customs house of Bugia, on the coast of what is now Algeria, acting for the Pisan merchants who traded there. Bugia (modern Béjaïa) was a prosperous port under the Almohad caliphate. Leonardo says his father sent for him while he was a boy, because he wanted him to learn something useful for his future.

The lesson in the customs house

At Bugia, Leonardo writes, he was taught for some days the art of the nine Indian figures, and he came to like it above everything else. In paraphrase, he found the new methods so superior that he considered the older Latin practices, and even the methods of Pythagoras, as errors by comparison. This is the earliest and most reliable fact we have about his education, and it comes from his own pen.

The name

He called himself Leonardus Pisanus, Leonardo of Pisa, and sometimes Leonardus Bigollo. The meaning of “Bigollo” is disputed; suggestions include a traveller and a person of little account. “Fibonacci” is a contraction of filius Bonacci, son of Bonacci. Leonardo never used it. A Pisan notary wrote “Lionardo Fibonacci” in 1506, and the historian Guglielmo Libri made the name standard in 1838.

DiscussionAlmost all we know of Leonardo’s youth comes from one paragraph that he wrote himself, as an advertisement for his book. What should a historian accept from such a source, and what should be treated with caution?

Lesson 02

Mediterranean apprenticeship

The prologue continues: after Bugia, Leonardo travelled on business to Egypt, Syria, Greece, Sicily and Provence, and in each place he studied the calculating methods used there and debated them with local practitioners. He gives no dates. The journeys probably fill the 1190s, and he says he returned to Pisa afterwards; c. 1200 is the usual estimate.

These were the trading routes of a Pisan merchant, and they were also the routes along which Arabic mathematics reached Europe. In Egypt and Syria he would have met calculators who wrote numbers with the Indian digits. Sicily, ruled by Norman and then Hohenstaufen kings, had Greek, Arabic and Latin communities side by side. Provence lay on the way to Spain, where translators at Toledo had been turning Arabic science into Latin since the mid-twelfth century.

What he read

His books show his sources more clearly than his biography does. The algebra of the Liber abaci follows the six standard equation types of Muhammad ibn Musa al-Khwarizmi, and many of its problems and methods follow the Egyptian algebraist Abu Kamil, whose work extended al-Khwarizmi’s. His geometry rests on Euclid’s Elements and on On Divisions, a Euclidean work known through Arabic. He also knew Latin translations of Arabic arithmetic. He refers to the double false-position method by its Arabic name, elchataym.

Whether Leonardo read Arabic is not known. His commercial problems use the coins and goods of North Africa and the Levant, which suggests direct contact with Arabic-speaking calculators rather than only Latin books.

DiscussionCompare Leonardo’s route into mathematics with that of a university-trained scholar of the same period, who would have studied the quadrivium in Latin. What did each route make easy, and what did it leave out?

Lesson 03

The Liber abaci, 1202

Back in Pisa, Leonardo wrote the Liber abaci, the “book of calculation”. Despite the title it is not about the abacus board. It teaches computation with written Indian numerals, the method that would replace the counting board in Italian commerce. He completed it in 1202. That first version is lost. What survives is a revised version of 1228, dedicated to the court scholar Michael Scot, in at least nineteen manuscripts, three of them complete.

The book is long, about six hundred pages in Sigler’s translation, and it has fifteen chapters:

  1. Reading and writing numbers with the Indian figures (chapter 1);
  2. multiplication, addition, subtraction and division of whole numbers (chapters 2–5);
  3. calculation with fractions and mixed numbers (chapters 6–7);
  4. pricing of goods, barter, partnerships and alloying of coins (chapters 8–11);
  5. a large collection of problems, including the rabbits (chapter 12);
  6. double false position, elchataym (chapter 13);
  7. square and cube roots and Euclid’s binomials and apotomes (chapter 14);
  8. geometric rules and algebra (chapter 15).

The audience was merchants and their sons, but the book is not a simple manual. Leonardo proves his rules, often geometrically in Euclid’s style, and he cannot resist hard recreational problems. This mixture of commerce and proof is the subject of Part II.

DiscussionThe 1202 text is lost and we read the 1228 revision. How might this affect claims about what Leonardo knew “in 1202”, for example the rabbit problem?

Part II · The mathematics
Lesson 04

The nine Indian figures

For the general reader. Roman numerals are fine for recording numbers and hopeless for calculating with them. Europeans calculated on counting boards and wrote down only the answers. The Indian system, with nine figures and a zero, lets you calculate on paper, and the written work can be checked afterwards. Leonardo’s first chapter opens by naming the nine figures and the sign 0, which he says the Arabs call zephirum; the word gave us both “cipher” and “zero”.

Goals
  • State precisely what positional notation is and why it makes algorithms uniform.
  • Prove the casting-out-nines check that abacus arithmetic relied on.
  • Understand trial division, which the Liber abaci describes for factoring.

Place value

Theorem (base-\(b\) representation)

Let \(b\ge2\). Every positive integer \(N\) has a unique representation \(N=\sum_{k=0}^{m} d_k b^k\) with digits \(0\le d_k\le b-1\) and \(d_m\ne0\).

Proof. Existence: divide, \(N=qb+d_0\) with \(0\le d_0<b\), and apply induction to \(q<N\). Uniqueness: reducing modulo \(b\) fixes \(d_0\); subtract it, divide by \(b\), and repeat. With \(b=10\) this is Leonardo’s system. The algorithms for addition and multiplication are the same at every position, which is what made them teachable to merchants’ sons.

Checking a calculation

Leonardo checks multiplications by remainders, the casting out of nines, and also by other numbers such as 7 and 11. The principle is a congruence.

Proposition

If \(N=\sum d_k10^k\), then \(N\equiv\sum d_k \pmod 9\) and \(N\equiv\sum (-1)^k d_k\pmod{11}\). Consequently, if \(AB=C\) then the digit sums satisfy \(s(A)s(B)\equiv s(C)\pmod 9\).

Proof. \(10\equiv1\pmod9\) and \(10\equiv-1\pmod{11}\), so \(10^k\equiv1\) and \(10^k\equiv(-1)^k\) respectively. Reduction modulo \(n\) respects products. The check is one-sided: a wrong product with the right residue passes, for instance a transposition of two digits.

Trial division

The Liber abaci contains the first known description in Europe of factoring by trial division. The method rests on a simple bound.

Proposition

If \(N>1\) has no prime divisor \(p\le\sqrt N\), then \(N\) is prime.

If \(N=ab\) with \(1<a\le b\), then \(a^2\le N\), so \(a\le\sqrt N\), and any prime factor of \(a\) is also at most \(\sqrt N\).

Exercise 4.1Use casting out nines to test the claim \(3457\times 2689 = 9295873\). Then test it modulo 11.

Show solution
Digit sums: \(3+4+5+7=19\equiv1\), \(2+6+8+9=25\equiv7\), product of residues \(7\). For \(9295873\): \(9+2+9+5+8+7+3=43\equiv7\). The test passes. Modulo 11, alternate sums from the units digit: \(3457\): \(7-5+4-3=3\); \(2689\): \(9-8+6-2=5\); product \(15\equiv4\). \(9295873\): \(3-7+8-5+9-2+9=15\equiv4\). Both checks pass, and in fact \(3457\times2689=9295873\) is correct.

Exercise 4.2Find a wrong product that passes the nines check but fails the elevens check, and explain why using both checks is stronger than using either alone.

Show solution
Swap two adjacent digits of the correct answer: \(9295837\) has the same digit sum, so it passes modulo 9, but its alternating sum is \(7-3+8-5+9-2+9=23\equiv1\not\equiv4\), so it fails modulo 11. Passing both checks means the error is divisible by \(\operatorname{lcm}(9,11)=99\), a much rarer accident than an error divisible by 9 alone.

Exercise 4.3Show by trial division that 10403 is not prime, and state how many trial divisors were needed at most.

Show solution
\(\sqrt{10403}\approx101.99\), so test primes up to 101. Divisibility fails for all primes up to 97, and \(10403=101\times103\). At most 26 trial divisions (the primes \(\le101\)) are needed.
Lesson 05

Fractions, composite and Egyptian

For the general reader. Medieval money and measures were not decimal. A Pisan pound had 20 shillings and a shilling 12 pence. Leonardo invented a fraction notation suited to such systems, a kind of mixed-radix number, and he also taught how to write any fraction as a sum of distinct unit fractions, the old Egyptian practice that was still alive in his sources.

Goals
  • Read Leonardo’s composite fractions as mixed-radix expansions.
  • Prove that the greedy algorithm always produces an Egyptian fraction.
  • Compute expansions and see where the greedy method is inefficient.

Composite fractions

Leonardo writes fractions to the left of whole numbers and reads them from right to left. A composite fraction is a row of numerators over a row of denominators under one bar. In modern terms,

\[ \frac{a_1\ a_2\ \cdots\ a_k}{d_1\ d_2\ \cdots\ d_k} \;=\; \frac{a_k}{d_k}+\frac{a_{k-1}}{d_{k-1}d_k}+\cdots+\frac{a_1}{d_1d_2\cdots d_k}, \qquad 0\le a_i<d_i. \]

So a sum in pounds, shillings and pence is a composite fraction with denominators \(12,20\): a quantity of \(s\) shillings and \(p\) pence is \(\frac{p\ s}{12\ 20}\) of a pound. Every rational number in \([0,1)\) whose denominator divides \(d_1\cdots d_k\) has exactly one such expansion, by the same division argument as Lesson 04.

Egyptian fractions

Theorem (the greedy algorithm; Leonardo, Liber abaci ch. 7)

Let \(0<p/q<1\) in lowest terms. Let \(n\) be the least integer with \(1/n\le p/q\), that is \(n=\lceil q/p\rceil\). Then \(p/q-1/n=p'/q'\) with \(0\le p'<p\). Repeating the step ends after at most \(p\) steps and writes \(p/q\) as a sum of distinct unit fractions.

Proof. \(\frac pq-\frac1n=\frac{np-q}{qn}\). Since \(n-1<q/p\), we have \((n-1)p<q\), so \(np-q<p\). The numerators strictly decrease, so the process stops. The unit fractions are distinct because, after subtracting \(1/n\), the remainder is less than \(\frac1{n-1}-\frac1n\le\frac1n\) when \(n\ge2\), so the next denominator is larger than \(n\). The method is now called the Fibonacci–Sylvester expansion, after James Joseph Sylvester, who studied it in 1880. Leonardo gives it as one of several rules and prefers other decompositions when they give smaller denominators.

Exercise 5.1Find the greedy expansion of \(4/13\).

Show solution
\(\lceil13/4\rceil=4\): \(4/13-1/4=3/52\). \(\lceil52/3\rceil=18\): \(3/52-1/18=2/936=1/468\). So \(4/13=1/4+1/18+1/468\). The numerators went \(4,3,1\), as the proof predicts.

Exercise 5.2Show that the greedy algorithm gives \(5/121=1/25+1/757+1/763309+1/873960180913+\cdots\) (five terms in all), while \(5/121=1/33+1/121+1/363\). What does this say about the algorithm?

Show solution
Check the short form: with common denominator 363, \(11+3+1=15\), and \(15/363=5/121\). For the greedy form, \(\lceil121/5\rceil=25\) and \(5/121-1/25=4/3025\); then \(\lceil3025/4\rceil=757\), and so on, with denominators growing roughly by squaring. The greedy algorithm always terminates but can be very far from optimal.

Exercise 5.3Write 7 shillings and 4 pence as a composite fraction of a pound, and convert it to an ordinary fraction.

Show solution
It is \(\frac{4\ 7}{12\ 20}\), meaning \(\frac7{20}+\frac4{12\cdot20}=\frac{84+4}{240}=\frac{88}{240}=\frac{11}{30}\) of a pound.
Lesson 06

Commercial problems and false position

For the general reader. Before symbolic algebra, the standard way to solve a linear problem was to guess. Guess an answer, see how far off it is, guess again, and combine the two errors. Leonardo called this method by its Arabic name, elchataym, and devoted chapter 13 of the Liber abaci to it. Chapter 12 collects hundreds of problems, many of them older than Leonardo and still told today.

Goals
  • Prove that double false position is exact for linear problems.
  • Solve an indeterminate problem in whole numbers, as in the problem of the birds.
  • See the remainder problems of chapter 12 as the Chinese remainder theorem.
Theorem (double false position)

Let \(f(x)=ax+b\) with \(a\ne0\), and suppose we want \(f(x)=c\). For two guesses \(g_1\ne g_2\) with errors \(e_i=f(g_i)-c\), the solution is

\[ x=\frac{g_1e_2-g_2e_1}{e_2-e_1}. \]

Proof. \(e_i=a(g_i-x)\) where \(x\) is the true solution. Then \(g_1e_2-g_2e_1=a(g_1g_2-g_1x-g_2g_1+g_2x)=a x(g_2-g_1)\) and \(e_2-e_1=a(g_2-g_1)\). Divide. For non-linear \(f\) the same formula is the secant method, which is not exact but converges.

Remainders

Chapter 12 asks for a number which leaves given remainders when divided by 3, 5 and 7. The rule Leonardo gives, also found in the Chinese Sunzi suanjing centuries earlier, multiplies the remainders by 70, 21 and 15, adds, and reduces modulo 105.

Theorem (Chinese remainder theorem)

If \(m_1,\dots,m_k\) are pairwise coprime and \(M=m_1\cdots m_k\), then for any \(r_1,\dots,r_k\) the system \(x\equiv r_i\pmod{m_i}\) has a unique solution modulo \(M\), namely \(x\equiv\sum_i r_i\,e_i\) where \(e_i\equiv1\pmod{m_i}\) and \(e_i\equiv0\pmod{m_j}\) for \(j\ne i\).

For moduli 3, 5, 7, the idempotents are \(e_1=70\), \(e_2=21\), \(e_3=15\): each is a multiple of the other two moduli and is \(\equiv1\) modulo its own.

Exercise 6.1The letter to Master Theodorus (c. 1225) poses a bird problem: 30 birds are bought for 30 coins, sparrows at \(\tfrac13\) coin each, turtledoves at \(\tfrac12\) and pigeons at 2. Find all solutions in whole numbers with at least one bird of each kind.

Show solution
\(s+t+p=30\) and \(\tfrac s3+\tfrac t2+2p=30\). Multiply the second by 6: \(2s+3t+12p=180\). Subtract twice the first: \(t+10p=120\), so \(t=120-10p\) and \(s=9p-90\). Positivity forces \(10<p<12\), so \(p=11\), \(t=10\), \(s=9\): nine sparrows, ten turtledoves, eleven pigeons. This is the solution Leonardo gives.

Exercise 6.2A merchant’s goods, after a toll of one-tenth and a further toll of 2 coins, are worth 25 coins. Find the original value by double false position with guesses 20 and 40, and check the formula.

Show solution
\(f(x)=\tfrac{9}{10}x-2\), target 25. \(f(20)=16\), error \(e_1=-9\); \(f(40)=34\), error \(e_2=9\). Then \(x=\frac{20\cdot9-40\cdot(-9)}{9-(-9)}=\frac{540}{18}=30\). Check: \(27-2=25\).

Exercise 6.3Find the least positive number leaving remainders 2, 3, 2 on division by 3, 5, 7.

Show solution
\(2\cdot70+3\cdot21+2\cdot15=140+63+30=233\equiv233-210=23\pmod{105}\). Check: \(23=7\cdot3+2=4\cdot5+3=3\cdot7+2\). This is the example of the Sunzi suanjing.
Lesson 07

The rabbits and the sequence

For the general reader. In chapter 12 of the Liber abaci, Leonardo asks how many pairs of rabbits one pair produces in a year, if each pair bears a new pair every month starting in its second month. The answer is 377, and the monthly totals form the sequence now named after him. It was one recreational problem among hundreds, and it was not new: Indian scholars of poetic metre had met the same numbers centuries before.

Goals
  • Derive the recurrence from the rabbit model and recover Leonardo’s 377.
  • Prove Binet’s formula, Cassini’s identity and the gcd property.
  • State Zeckendorf’s theorem and the convergence of ratios to the golden ratio.

The problem as Leonardo sets it

He starts with one pair, which gives birth in the first month, so that after one month there are 2 pairs. He lists the totals month by month: 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, and he explains that each number is the sum of the two before. He adds that the method works for any number of months. With the modern indexing \(F_0=0,\ F_1=1,\ F_{n+1}=F_n+F_{n-1}\), his total after \(n\) months is \(F_{n+2}\), and 377 is \(F_{14}\).

Why the recurrence holds. Let \(A_n\) be the number of adult pairs and \(Y_n\) the number of young pairs after \(n\) months. Each adult pair produces one young pair, and young pairs become adults a month later: \(A_{n+1}=A_n+Y_n\), \(Y_{n+1}=A_n\). The total \(T_n=A_n+Y_n\) then satisfies \(T_{n+1}=A_{n+1}+Y_{n+1}=T_n+A_n=T_n+T_{n-1}\).

Closed form

Theorem (Binet’s formula)

Let \(\varphi=\frac{1+\sqrt5}2\) and \(\psi=\frac{1-\sqrt5}2\), the roots of \(t^2=t+1\). Then for all \(n\ge0\),

\[ F_n=\frac{\varphi^n-\psi^n}{\sqrt5}, \qquad\text{and}\qquad \lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\varphi. \]

Proof. Both \(\varphi^n\) and \(\psi^n\) satisfy the recurrence, since \(\varphi^{n+1}=\varphi^n+\varphi^{n-1}\) follows from \(\varphi^2=\varphi+1\). So does any linear combination. The combination \((\varphi^n-\psi^n)/\sqrt5\) takes the values 0 and 1 at \(n=0,1\), and a solution is determined by two initial values. Since \(|\psi|<1\), \(F_n\) is the integer nearest to \(\varphi^n/\sqrt5\), and the ratio tends to \(\varphi\). The formula is named after Jacques Binet (1843) but was known earlier to Abraham de Moivre and Daniel Bernoulli. Johannes Kepler remarked on the limit of the ratios in 1611.

Identities

Theorem

(a) Matrix form. \(\begin{pmatrix}1&1\\1&0\end{pmatrix}^{n}=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}\) for \(n\ge1\).

(b) Cassini’s identity (1680). \(F_{n+1}F_{n-1}-F_n^2=(-1)^n\).

(c) Addition. \(F_{m+n}=F_{m+1}F_n+F_mF_{n-1}\).

(d) Divisibility. \(\gcd(F_m,F_n)=F_{\gcd(m,n)}\).

Proofs. (a) by induction, multiplying by the matrix once more. (b) take determinants in (a): the matrix has determinant \(-1\). (c) compare entries of \(Q^{m+n}=Q^mQ^n\), where \(Q\) is the matrix in (a). (d) From (b), consecutive Fibonacci numbers are coprime. From (c), \(F_n\mid F_{kn}\) by induction on \(k\), and \(\gcd(F_{m+n},F_n)=\gcd(F_{m+1}F_n+F_mF_{n-1},F_n)=\gcd(F_mF_{n-1},F_n)=\gcd(F_m,F_n)\), using \(\gcd(F_{n-1},F_n)=1\). So the pair of indices can be reduced exactly as in the Euclidean algorithm, ending at \(F_{\gcd(m,n)}\).

Theorem (Zeckendorf)

Every positive integer is uniquely a sum of distinct Fibonacci numbers \(F_k\) with \(k\ge2\), no two of them consecutive.

Before Leonardo

Indian prosodists counted the rhythms of a line of verse made of short syllables (one beat) and long syllables (two beats). The number of patterns lasting \(n\) beats is \(F_{n+1}\). The idea goes back to Pingala’s treatise on metre (dated variously between c. 450 and c. 200 BCE). Virahanka (c. 700) gave the rule explicitly, and Gopala (c. 1135) and Hemachandra (c. 1150) restated it. There is no evidence that Leonardo knew this tradition.

Exercise 7.1Show that the number of ways to write \(n\) as an ordered sum of 1s and 2s is \(F_{n+1}\). Check \(n=4\) by listing.

Show solution
Let \(c_n\) be the count. The first summand is 1 (then \(c_{n-1}\) ways) or 2 (then \(c_{n-2}\) ways), so \(c_n=c_{n-1}+c_{n-2}\), with \(c_1=1=F_2\), \(c_2=2=F_3\). For \(n=4\): 1111, 112, 121, 211, 22, five ways, and \(F_5=5\).

Exercise 7.2Verify Cassini’s identity for \(n=6\), and use it to explain the classical paradox in which an \(8\times8\) square is cut into four pieces and rearranged into a \(5\times13\) rectangle of area 65.

Show solution
\(F_7F_5-F_6^2=13\cdot5-64=1=(-1)^6\). The pieces are cut along lines of slopes \(3/8\) and \(2/5\). These are close but unequal, and the rearranged “rectangle” hides a thin parallelogram along its diagonal of area exactly \(F_7F_5-F_6^2=1\).

Exercise 7.3Compute \(\gcd(F_{12},F_{18})\) directly and with the theorem. Then find the Zeckendorf representation of 100.

Show solution
\(F_{12}=144\), \(F_{18}=2584\); \(\gcd(144,2584)=8\), and \(F_{\gcd(12,18)}=F_6=8\). For 100, take the largest Fibonacci number \(\le100\) repeatedly: \(100=89+8+3\), with indices 11, 6, 4, none consecutive.
Lesson 08

Algebra and the cubic of Flos

For the general reader. The last chapter of the Liber abaci teaches algebra in the Arabic manner, with words instead of symbols and geometric proofs for the rules. Around 1225 Johannes of Palermo, a scholar of Frederick II’s court, challenged Leonardo with a cubic equation. Leonardo proved the solution could not be found by the usual means, then gave a numerical answer correct to about ten decimal places, without saying how he found it.

Goals
  • Solve quadratics in al-Khwarizmi’s six canonical forms and justify the rule.
  • Prove that the cubic \(x^3+2x^2+10x=20\) has no rational root and no root of the form \(\sqrt r\) with \(r\) rational.
  • Reproduce Leonardo’s sexagesimal approximation numerically.

Quadratics in words

With only positive coefficients allowed, al-Khwarizmi had distinguished six types: squares equal roots, squares equal numbers, roots equal numbers, and three mixed types such as “squares and roots equal numbers”, \(x^2+bx=c\). Leonardo calls the unknown res or radix and its square census; he also uses causa, apparently the first Latin author to do so. For \(x^2+bx=c\) the rule is to halve the roots, square, add the number, take the root, and subtract half the roots:

\[ x=\sqrt{\left(\tfrac b2\right)^2+c}-\tfrac b2 , \]

justified by completing a square of side \(x+\tfrac b2\) in a figure.

The cubic

Johannes’s problem, in modern form, was to solve \(x^3+2x^2+10x=20\). The same equation appears in the algebra of Omar Khayyam, who solved such cubics by intersecting conic sections. Leonardo argued in several steps that the root is not an integer, not a fraction, not the square root of a fraction, and not one of the other irrational lines classified in Euclid’s Book X. He then stated, in sexagesimal fractions,

\[ x\approx 1;22,7,42,33,4,40 \;=\; 1+\frac{22}{60}+\frac{7}{60^2}+\frac{42}{60^3}+\frac{33}{60^4}+\frac{4}{60^5}+\frac{40}{60^6}\approx 1.3688081078532, \]

while the true root is \(1.3688081078214\ldots\). The error is about \(3\times10^{-11}\). He did not explain his method. Historians have suggested a method of the kind used in Arabic sources for roots, but there is no direct evidence.

Proposition

The real root of \(x^3+2x^2+10x-20=0\) is unique, lies in \((1,2)\), and is irrational. It is not of the form \(\sqrt r\) with \(r\in\mathbb Q\).

Exercise 8.1Prove the proposition.

Show solution
The derivative \(3x^2+4x+10\) has negative discriminant, so \(f\) is strictly increasing and has one real root. \(f(1)=-7<0<16=f(2)\), so it lies in \((1,2)\). A rational root of a monic integer polynomial is an integer dividing 20, and there is no integer in \((1,2)\). If \(x=\sqrt r\), then \(x^3+10x=x(r+10)\) and \(20-2x^2=20-2r\), so \(x=\frac{20-2r}{r+10}\) would be rational, which we have just excluded.

Exercise 8.2Starting from \(x_0=1.5\), apply Newton’s method to \(f(x)=x^3+2x^2+10x-20\). How many steps are needed to reach the accuracy of Leonardo’s answer?

Show solution
\(x_{n+1}=x_n-\frac{f(x_n)}{3x_n^2+4x_n+10}\). One gets \(x_1\approx1.373626\), \(x_2\approx1.3688148\), \(x_3\approx1.36880810783\), \(x_4\approx1.3688081078214\). Three steps give about ten correct digits. Leonardo had no such method in general form, which makes his result impressive.

Exercise 8.3Solve \(x^2+10x=39\) by the rule above, and explain the geometric figure behind it.

Show solution
\(x=\sqrt{25+39}-5=8-5=3\). Figure: a square of side \(x\) with two rectangles \(x\times5\) attached to two sides has area \(x^2+10x=39\). Completing the corner with a \(5\times5\) square gives a square of area 64 and side \(x+5=8\). This is al-Khwarizmi’s own example.

Exercise 8.4The third problem of Flos: three men own a pile of money in shares \(\tfrac12,\tfrac13,\tfrac16\). Each takes some money; the first returns half of what he took, the second a third, the third a sixth. The returned money is split equally among the three, and each then has exactly his share. Show that a pile of 47, with the men taking 33, 13 and 1, works.

Show solution
Returned: \(16\tfrac12+4\tfrac13+\tfrac16=21\), so each receives 7. The first holds \(16\tfrac12+7=23\tfrac12=\tfrac{47}2\); the second \(8\tfrac23+7=15\tfrac23=\tfrac{47}3\); the third \(\tfrac56+7=7\tfrac56=\tfrac{47}6\). The problem is indeterminate: any multiple of this solution also works.
Lesson 09

The Book of Squares

For the general reader. Johannes of Palermo also asked Leonardo for a square number which stays a square when 5 is added or subtracted. Leonardo answered with \(\left(\tfrac{41}{12}\right)^2\), and around the problem he built the Liber quadratorum (1225), dedicated to Frederick II. It survives in a single fifteenth-century manuscript in Milan. It is a book of number theory in the line of Diophantus, though Leonardo shows no knowledge of him, and its results anticipate Fermat and Leonhard Euler.

Goals
  • Generate squares and Pythagorean triples from sums of odd numbers.
  • Prove the two-squares identity and the congruum formula.
  • Solve the problem of 5 and connect it to congruent numbers.

Odd numbers and triples

The book starts from the fact that \(1+3+\cdots+(2n-1)=n^2\), so that \(n^2+(2n+1)=(n+1)^2\). If \(2n+1=m^2\) is an odd square, this gives a Pythagorean triple. For example, \(9=2\cdot4+1\) gives \(16+9=25\).

Theorem (two-squares identity; Liber quadratorum)

For all numbers \(a,b,c,d\), \[(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2.\] So a product of sums of two squares is a sum of two squares, usually in two ways.

The identity was known to Diophantus implicitly and to Brahmagupta in a more general form; it is often called the Brahmagupta–Fibonacci identity. Leonardo proves it for integers and uses it to produce numbers that are sums of two squares in several ways.

Congrua

A number \(c\) is a congruum if there is a square \(x^2\) with \(x^2-c\) and \(x^2+c\) both squares: three squares in arithmetic progression with common difference \(c\). Leonardo found all of them.

Theorem (Leonardo, 1225)

For integers \(m>n>0\),

\[ (m^2+n^2)^2\pm 4mn(m^2-n^2)=(m^2-n^2\pm2mn)^2, \]

so \(4mn(m^2-n^2)\) is a congruum with middle square \((m^2+n^2)^2\). Every integer congruum arises, up to a square factor, in this way. Every congruum of this form is divisible by 24.

Proof of the identity. Expand: \((m^2-n^2\pm2mn)^2=(m^2-n^2)^2\pm4mn(m^2-n^2)+4m^2n^2\), and \((m^2-n^2)^2+4m^2n^2=(m^2+n^2)^2\). Geometrically, \(m^2-n^2,\ 2mn,\ m^2+n^2\) is a Pythagorean triple and the congruum is four times the area of that right triangle. Leonardo’s own statement uses the product \(ab(a+b)(b-a)\) with \(a+b\) even, which is the same family after substituting \(a=m-n\), \(b=m+n\).

The problem of 5

We want \(x^2\pm5\) to be rational squares. Multiply by \(k^2\): it suffices to find a congruum equal to \(5k^2\). Take \(m=5,\ n=4\): \(4\cdot5\cdot4\cdot9=720=5\cdot12^2\), with middle square \(41^2\). Indeed \(41^2-720=961=31^2\) and \(41^2+720=2401=49^2\). Dividing by \(144\),

\[ \left(\tfrac{41}{12}\right)^2-5=\left(\tfrac{31}{12}\right)^2,\qquad \left(\tfrac{41}{12}\right)^2+5=\left(\tfrac{49}{12}\right)^2. \]

Squares are not congrua

Leonardo stated that no congruum is a square. Sources differ on whether his argument counts as a proof; the first complete proof is Fermat’s, by infinite descent, in the form that the area of a right triangle with rational sides is never a square. Equivalently, \(x^4-y^4=z^2\) has no solution in positive integers. In modern language a number is congruent if it is the area of a rational right triangle, so the result says 1 is not a congruent number. The congruent number problem, for which Leonardo’s 5 was the first worked case, is now tied to elliptic curves and the Birch and Swinnerton-Dyer conjecture through Tunnell’s theorem.

Exercise 9.1Prove that \(4mn(m^2-n^2)\) is divisible by 24.

Show solution
Write it as \(4\,mn(m-n)(m+n)\). Divisibility by 3: if neither \(m\) nor \(n\) is divisible by 3, then \(m\equiv\pm n\pmod3\), so 3 divides \(m-n\) or \(m+n\). Divisibility by 2: if \(m,n\) are both odd, \(m-n\) is even; otherwise one of them is even. So \(mn(m-n)(m+n)\) is divisible by 6, and the congruum by 24.

Exercise 9.2Find a rational \(x\) with \(x^2-6\) and \(x^2+6\) both squares.

Show solution
We need a congruum \(6k^2\). Take \(m=2,n=1\): \(4\cdot2\cdot1\cdot3=24=6\cdot2^2\), middle square \(5^2\): \(25-24=1\), \(25+24=49\). Divide by 4: \(x=\tfrac52\), with \(\tfrac{25}4-6=\tfrac14\) and \(\tfrac{25}4+6=\tfrac{49}4\). (6 is the area of the 3–4–5 triangle.)

Exercise 9.3Use the two-squares identity to write \(65=5\cdot13\) as a sum of two squares in two ways.

Show solution
\(5=1^2+2^2\), \(13=2^2+3^2\). Then \((1\cdot2+2\cdot3)^2+(1\cdot3-2\cdot2)^2=64+1\) and \((1\cdot2-2\cdot3)^2+(1\cdot3+2\cdot2)^2=16+49\). So \(65=8^2+1^2=4^2+7^2\).
Lesson 10

Roots and practical geometry

For the general reader. The Practica geometriae (1220) is a surveyor’s and builder’s geometry in eight chapters, dedicated to Dominicus Hispanus. It begins with definitions and constructions from Euclid, computes with Pisan units of measure, extracts square and cube roots, measures fields and solids, finds heights with similar triangles, and ends with pentagons and decagons. The letter to Master Theodorus adds a pentagon inscribed in a triangle.

Goals
  • Derive and bound the standard linear approximation for square roots.
  • Apply the same idea to cube roots.
  • Use similar triangles for indirect measurement and compute the side of a regular decagon.

Approximating roots

The medieval rule, going back to Heron and used throughout Arabic arithmetic, is to write \(N=a^2+r\) with \(a^2\) the largest square below \(N\) and take \(\sqrt N\approx a+\frac r{2a}\). Repeating the step is Heron’s iteration.

Proposition

If \(N=a^2+r\) with \(a>0\) and \(0\le r\), then \(\sqrt N\le a+\frac r{2a}\), with error at most \(\frac{r^2}{8a^3}\). The iteration \(x_{k+1}=\frac12\left(x_k+\frac N{x_k}\right)\) satisfies \(x_{k+1}-\sqrt N=\frac{(x_k-\sqrt N)^2}{2x_k}\), so the number of correct digits roughly doubles each step.

Proof. \(\left(a+\frac r{2a}\right)^2=N+\frac{r^2}{4a^2}\ge N\). Then \(a+\frac r{2a}-\sqrt N=\frac{r^2/(4a^2)}{a+\frac r{2a}+\sqrt N}\le\frac{r^2}{8a^3}\). The iteration identity is direct algebra: \(\frac12(x+N/x)-\sqrt N=\frac{(x-\sqrt N)^2}{2x}\).

For cube roots the same linearization gives \(\sqrt[3]{a^3+r}\approx a+\frac r{3a^2}\). The Practica gives rules for cube roots of this general type; the exact rule Leonardo used is discussed in Hughes’s edition.

Pentagons and decagons

Theorem (Euclid XIII.9–10)

The side of a regular decagon inscribed in a circle of radius \(R\) is \(s_{10}=\frac{\sqrt5-1}2R=R/\varphi\). The side of the inscribed regular pentagon satisfies \(s_5^2=R^2+s_{10}^2\).

In the isosceles triangle formed by the centre and one side of the decagon, the apex angle is \(36^\circ\) and the base angles \(72^\circ\). Bisecting a base angle produces a smaller similar triangle, which gives \(s_{10}^2=R(R-s_{10})\). Solve the quadratic.

Exercise 10.1Approximate \(\sqrt{10}\) by the rule, then do one Heron step, and compare with \(\sqrt{10}=3.16227766\ldots\).

Show solution
\(10=3^2+1\), so \(\sqrt{10}\approx3+\tfrac16=3.1\overline6\), error \(\approx0.0044\le\tfrac1{216}\approx0.0046\). Heron: \(\tfrac12\left(\tfrac{19}6+\tfrac{60}{19}\right)=\tfrac{721}{228}\approx3.162281\), error \(\approx3\times10^{-6}\).

Exercise 10.2A pole 2 units tall casts a shadow 3 units long while a tower casts a shadow of 45 units. How tall is the tower? Then prove \(s_{10}=\frac{\sqrt5-1}2R\) from \(s_{10}^2=R(R-s_{10})\), and compute \(s_{10}\) for \(R=10\).

Show solution
Similar triangles: \(h/45=2/3\), so \(h=30\). For the decagon, \(s^2+Rs-R^2=0\) gives \(s=\frac{-R+\sqrt{5R^2}}2=\frac{\sqrt5-1}2R\). For \(R=10\), \(s=5(\sqrt5-1)\approx6.180\).
Part III · Life and legacy
Lesson 11

The emperor’s court and the city’s salary

Frederick II of Hohenstaufen, king of Sicily, was crowned emperor in Rome in November 1220. His court gathered scholars of many languages: the astrologer and translator Michael Scot, the court philosopher Master Theodorus, and Johannes of Palermo. Frederick founded the University of Naples in 1224. When the court was at Pisa, around 1225, Leonardo was presented to the emperor. According to the dedication of the Liber quadratorum, it was Master Dominicus who brought him to Frederick’s attention, and Johannes of Palermo put problems to him in the emperor’s presence.

Leonardo answered in writing. Flos (“flower”) contains his solutions to three of the problems and other work; the Liber quadratorum grew out of the problem of 5; and a letter to Master Theodorus treats the birds problem, a pentagon inscribed in an isosceles triangle with sides 10, 10 and 12, and a system of linked equations for five men’s money. In that last problem one variant has a negative solution, which Leonardo interprets as a debt, an early and careful use of negative quantities in Latin mathematics.

In 1228 he revised the Liber abaci and sent it to Michael Scot. Two other works are mentioned by Leonardo and are lost: a book of commercial arithmetic, Di minor guisa, and a commentary on Book X of Euclid’s Elements.

The decree of 1240

The last document about him is a decree of the Commune of Pisa from 1240. It grants the “serious and learned” Master Leonardo Bigollo an annual salary, in recognition of his services to the city and its citizens in matters of accounting and teaching. The decree shows that Leonardo had become a public figure: the city valued him for exactly the commercial skills his first book taught. Nothing is known of his death. It is usually placed after 1240 and before about 1250, probably in Pisa.

DiscussionThe court problems were posed in public, as a kind of contest. Compare this with the cubic-equation contests in sixteenth-century Italy between Niccolò Tartaglia and Antonio Fior. What did such contests reward, and what did they discourage?

Lesson 12

Abbaco, Lucas and the afterlife of a name

Leonardo’s numerals spread slowly. From the late thirteenth century, Italian cities supported abbaco schools, where masters taught merchants’ sons arithmetic with Indian numerals, commercial rules and some algebra, in the vernacular. Hundreds of abbaco treatises survive. Many draw on the Liber abaci, directly or at second hand, though historians such as Jens Høyrup have argued that the abbaco tradition also drew on other sources from the Mediterranean. Luca Pacioli’s Summa of 1494, the first printed encyclopedia of this mathematics, used Leonardo’s material and named him as a source. Full adoption of written numerals in European accounting took until the sixteenth century.

His number theory had less effect. The Liber quadratorum was barely read in the Middle Ages, and its results were found again by Francesco Maurolico, Fermat and Euler. Its single manuscript was identified and published by Baldassarre Boncompagni in the nineteenth century, together with the other works, in the Scritti di Leonardo Pisano (1857–62).

The sequence and the name

The fame of “Fibonacci” comes from the nineteenth century. The French number theorist Édouard Lucas, working in the 1870s on recurrences and primality, attached Leonardo’s name to the rabbit sequence and studied the companion sequence \(2,1,3,4,7,11,\ldots\), now called the Lucas numbers. Lucas’s methods led to tests for the primality of Mersenne numbers; in 1876 he proved \(2^{127}-1\) prime. In 1963 the Fibonacci Association began publishing The Fibonacci Quarterly, founded by Verner Hoggatt and Alfred Brousseau, devoted to the sequence and its relatives.

Pisa set up a statue of Leonardo by Giovanni Paganucci in 1863; it now stands in the Camposanto Monumentale. Modern English translations by Laurence Sigler (the Book of Squares, 1987; the Liber abaci, 2002) and Barnabas Hughes (the Practica geometriae, 2008) have made all his surviving books readable.

Assessment

  • Arithmetic: the most influential European account of Indian numerals and their algorithms, written for practical use and with proofs.
  • Algebra: transmission of al-Khwarizmi’s and Abu Kamil’s algebra, and a careful study of a cubic three centuries before the Italian solutions.
  • Number theory: the congruum theorem, the two-squares identity and the first worked congruent-number problem.
  • Name: a sequence older than Leonardo now carries a name he never used.

Final projectChoose one problem from Part II. Write a two-page account for a strong undergraduate that states it as Leonardo did (in words), solves it as he would have, then solves it in modern notation and explains what the notation adds.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Leonardo Pisano Fibonacci, MacTutor History of Mathematics.
  2. Fibonacci, Wikipedia.
  3. K. Devlin, The Man of Numbers: Fibonacci’s Arithmetic Revolution, Walker, 2011.
  4. A. F. Horadam, Fibonacci’s mathematical letter to Master Theodorus, The Fibonacci Quarterly 29 (1991).
  5. J. Høyrup, Jacopo da Firenze’s Tractatus Algorismi and Early Italian Abbacus Culture, Birkhäuser, 2007.

Texts and mathematics

  1. B. Boncompagni (ed.), Scritti di Leonardo Pisano, 2 vols., Rome, 1857–1862.
  2. L. E. Sigler, Fibonacci’s Liber Abaci, Springer, 2002.
  3. L. E. Sigler, Leonardo Pisano Fibonacci: The Book of Squares, Academic Press, 1987.
  4. B. Hughes, Fibonacci’s De Practica Geometrie, Springer, 2008.
  5. Liber Abaci, The Book of Squares and Congruum, Wikipedia.
  6. T. Phillips, The rabbit problem in the Liber abaci (Latin text with translation), Stony Brook.
  7. Fibonacci and square numbers, MAA Convergence.
  8. Flos: the three problems, ProofWiki.
  9. Fibonacci sequence, Wikipedia (history and Indian precursors).
  10. The Fibonacci Quarterly, the Fibonacci Association.