The solution is not enough: Fibonacci’s inheritance

Original article by Salvatore Mosaico: “La soluzione non basta”, published in Alice & Bob, no. 17, on 18 May 2010. The proof credited to ChatGPT is a later addition and was not part of the original publication.

Abstract

I have always been impressed by the imagination of ancient mathematicians. They devised beautiful problems, yet their solutions are often not proved: they simply state the result with which the conditions of the problem can be checked. Using modern techniques, we can instead seek a solution with equations, systems, or other tools.

I happened, however, to read modern solutions to two Fibonacci problems from the Liber Abaci. Although correct, they left me unconvinced: they find the numbers but do not solve the problem in its essence, because they do not explain why the mechanism works. In this article I offer an explanation and a generalisation.

Introduction

While reading Mathematical Games of the Middle Ages, edited by Nando Geronimi, I was puzzled by the proposed solution to the following problem.

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/7 of what remains.”

To the second son he said: “You shall take 2 bezants and 1/7 of what remains.”

To the third son he said: “You shall take 3 bezants and 1/7 of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/7 of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Note. The problem appears in part seven of chapter XII of the Liber Abaci, under the title “The inheritance of a man’s property”, on folio 70 recto of the manuscript.

The proposed solution

The proposed solution calculates what the first son receives and what the second son receives, then requires those two amounts to be equal. Let N be the number of sons and S the amount received by each; N · S is the entire inheritance.

First son:  1 + (N · S − 1) / 7
Second son: 2 + ((N − 1) · S − 2) / 7

Because both receive S, the following system results:

1 + (N · S − 1) / 7 = S
2 + ((N − 1) · S − 2) / 7 = S

Solving it gives N = 6 and S = 6.

Checking an inheritance of 36 bezants

My concern is that, although the reasoning is correct, it does not begin with the complete hypothesis: there are not merely two sons. Moreover, why do the third, fourth, fifth, and sixth sons also receive the same amount?

A solution that uses all the data

Let us begin with the correct hypothesis. Let T be the entire inheritance and k the amount received by each son. Before son j takes his turn, (j − 1)k coins have already been distributed. Son j therefore receives:

j + [T − (j − 1)k − j] / 7

Because every son receives the same amount, for j = 1, 2, 3, … we can write:

j + [T − (j − 1)k − j] / 7 = k

Multiplying both sides by 7 and expanding:

7j + T − (j − 1)k − j = 7k
7j + T − jk + k − j = 7k
T + k − 7k = jk − 7j + j
T − 6k = jk − 6j
T − 6k = j(k − 6)

This equality must hold for every j. That happens only if T − 6k = 0 and k − 6 = 0. Thus k = 6 and T = 36.

The first generalisation

In this problem, 7 = k + 1. Let us try replacing the denominator 7 with k + 1 and see what happens. This is an exploratory attempt: the coincidence alone does not guarantee that the formula works in other cases.

j + [T − (j − 1)k − j] / (k + 1) = k

Multiplying both sides by k + 1 gives:

(k + 1)j + T − (j − 1)k − j = (k + 1)k
(k + 1)j + T − jk + k − j = (k + 1)k
kj + j + T − jk + k − j = (k + 1)k
T + k = k² + k
T = k²

Why it always works — a note by ChatGPT

Explanatory note developed by ChatGPT in dialogue with Salvatore Mosaico, added after the original 2010 article. This argument does not start from the answer: it starts only from the requirement of equal shares.

Consider two consecutive sons. The next takes one extra fixed bezant. To receive the same total share, he must therefore receive one bezant less from the fractional part.

The fraction is one seventh: for that part to fall by one bezant, its calculation base must fall by seven. Where does this decrease come from? From the still unknown share paid to the previous son and the extra fixed bezant deducted for the next.

Call the common share k. Without knowing its value beforehand, we obtain:

k + 1 = 7 → k = 6

The share of six bezants follows from compensation; it was not our starting point. The first son must receive six: one fixed bezant and five from the fraction. After deducting the first bezant, 5 × 7 = 35 must remain, so the initial inheritance is 36. Dividing it into shares of six gives six sons.

Why does the mechanism continue? After a share of six is paid, the next fractional base falls by 6 + 1 = 7. One seventh falls by one while the fixed part rises by one: the share stays six. The first turn works, and every turn reproduces the same compensation until the last:

1 + 5 = 2 + 4 = 3 + 3 = 4 + 2 = 5 + 1 = 6 + 0

Generally, with denominator d and common share k, the base falls by k + 1. Compensation requires (k + 1)/d = 1, so k = d − 1. For the first son, 1 + (T − 1)/d = d − 1 gives T = (d − 1)² and n = T/k = d − 1. For at least two sons and the same rule through the last turn, take an integer d ≥ 3. Compensation preserves the share at each turn until the inheritance is exhausted.

The mechanism is easy to check once we identify what to compare; discovering it is not necessarily immediate. The following proof formalises this explanation.

Proof of the generalisation — contribution by ChatGPT

This proof was developed by ChatGPT in dialogue with Salvatore Mosaico as a later addition to the 2010 article. The preceding exploratory attempt and generalisation belong to the author; this supplement makes their necessity, sufficiency and compensation mechanism explicit.

1. An independent denominator

Do not assume the denominator is k + 1: call it d. Let T be the inheritance, k > 0 the common share and n ≥ 2 the number of sons; d > 1. We adopt the uniform rule used in the checks: the last son also follows the same procedure, exhausting the inheritance. If “the last takes everything” were instead an exception, the conditions would need separate clarification.

Before son j, the preceding j − 1 sons have received (j − 1)k. Son j takes j bezants, then 1/d of the remainder. Equal shares require:

j + [T − (j − 1)k − j]/d = k
dj + T − (j − 1)k − j = dk
dj + T − jk + k − j = dk
T − (d − 1)k = j[k − (d − 1)]

2. Why d must equal k + 1

The final right-hand side must give the same value for every j. Write it for j = 1 and j = 2, then subtract:

j = 1:  T − (d − 1)k = k − (d − 1)
j = 2:  T − (d − 1)k = 2[k − (d − 1)]
0 = k − (d − 1)
k = d − 1;  d = k + 1

The relation d = k + 1 is no longer an assumption: it is necessary. Substituting it and using T = nk gives:

T = (d − 1)k = k²
T = nk;  nk = k²;  n = k
d = n + 1;  k = n;  T = n²

3. Why it works for every son

Necessary conditions are not enough: we must establish sufficiency. Choose an integer n ≥ 2, an inheritance n² and denominator n + 1. Suppose the first j − 1 sons have each received n. After assigning the fixed j bezants to son j, the remaining amount is:

n² − (j − 1)n − j
= n² − jn + n − j
= (n + 1)(n − j)
j + [(n + 1)(n − j)]/(n + 1) = j + (n − j) = n

The fractional part is n − j. The total share is therefore j + (n − j) = n. For j = 1 there are no previous shares to assume, establishing the first step. The argument then passes from each son to the next. For 1 ≤ j ≤ n the remainder is nonnegative; the last share is n + 0, leaving n² − n × n = 0. This is proof by induction, not merely a check of selected examples.

4. The compensation mechanism

From one son to the next, the fixed part increases by 1. The base used to calculate the fraction falls by k + 1: k already paid to the previous son, plus one extra fixed bezant for the next. The fractional part therefore falls by (k + 1)/d. To offset the fixed increase exactly, (k + 1)/d must equal 1, so d = k + 1. This explains the mechanism.

5. Comparison with the second version

In the second version, described later, the fraction of the available money is calculated first, and j bezants are then added. Instead we obtain:

j + [T − (j − 1)k]/d = k
T − (d − 1)k = j(k − d)
k = d;  T = d(d − 1);  n = d − 1
d = n + 1;  k = n + 1;  T = n(n + 1)

The fractional base now falls only by k. Compensation requires k/d = 1. Sufficiency also follows: with T = n(n + 1), if each previous son received n + 1, son j receives (n − j + 1) + j = n + 1, until the inheritance is exhausted.

6. A clarification about the first two equations

Using the first two sons’ equations does not assume there are only two sons: these are necessary conditions even with many sons. After finding candidate values, however, we must show that they work for everyone and exhaust the inheritance. The formula involving j makes that step explicit. Exploration describes the discovery; the independent denominator proves necessity; induction proves sufficiency.

Because the total inheritance is also T = N · k, the equality T = k² implies N = k (for k > 0). The situation therefore occurs when the inheritance is the square of the number of sons.

Example with 8 sons and 64 bezants

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/9 of what remains.”

To the second son he said: “You shall take 2 bezants and 1/9 of what remains.”

To the third son he said: “You shall take 3 bezants and 1/9 of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/9 of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Example with 9 sons and 81 bezants

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/10 of what remains.”

To the second son he said: “You shall take 2 bezants and 1/10 of what remains.”

To the third son he said: “You shall take 3 bezants and 1/10 of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/10 of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Concise check. Before son j takes his turn, the remaining inheritance is 81 − (j − 1) · 9 = 9(10 − j). His share is therefore:

j + [9(10 − j) − j] / 10 = 9

This identity holds for every j = 1, …, 9; consequently, all nine sons receive 9 bezants.

Conclusions for the first form

In general, if there are n sons and an inheritance of n² bezants, the problem becomes:

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/(n + 1) of what remains.”

To the second son he said: “You shall take 2 bezants and 1/(n + 1) of what remains.”

To the third son he said: “You shall take 3 bezants and 1/(n + 1) of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/(n + 1) of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Answer: there were n sons, the inheritance was n² bezants, and every son received n bezants.

A second form of the problem

The same method also generalises the following problem, in which the fraction of the available money is calculated first and the fixed number of bezants is then added.

Shortly before his death, a man called his eldest son and said: “My sons, divide my modest possessions among yourselves. You shall keep 1/7 of all the bezants, plus 1 bezant.”

To the second son he said: “You shall take 1/7 of the remaining bezants, plus another 2 bezants.”

He called all his sons in decreasing order of age in this way, giving each one 1/7 of the remaining bezants and one bezant more than the preceding son.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Note. This problem also appears in part seven of chapter XII of the Liber Abaci, under the title “The inheritance of a man’s property”, on folio 70 recto of the manuscript.

The proposed solution again lets N be the number of sons and S the share of each:

First son:  (N · S) / 7 + 1 = S
Second son: ((N − 1) · S) / 7 + 2 = S

The solutions are N = 6 and S = 7.

Checking an inheritance of 42 bezants

By the same reasoning used above, it is easy to prove that, in general, if n is the number of sons, the inheritance is n(n + 1) and each son receives n + 1 bezants. The statement becomes:

Shortly before his death, a man called his eldest son and said: “My sons, divide my modest possessions among yourselves. You shall keep 1/(n + 1) of all the bezants, plus 1 bezant.”

To the second son he said: “You shall take 1/(n + 1) of the remaining bezants, plus another 2 bezants.”

He called all his sons in decreasing order of age in this way, giving each one 1/(n + 1) of the remaining bezants and one bezant more than the preceding son.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Proof of the second generalisation

Let T = n(n + 1) be the inheritance and k = n + 1 the common share. Before son j takes his turn, the amount left is:

T − (j − 1)k = (n − j + 1)(n + 1)

The amount received by son j is therefore:

[T − (j − 1)k] / (n + 1) + j
= n − j + 1 + j
= n + 1

The calculation is valid for every j = 1, …, n, so all n sons receive the same share.

Answer: there are n sons, each receives n + 1 bezants, and the inheritance is n(n + 1) bezants.

Prepared by Salvatore Mosaico.

From 2010 to 2026: the inheritance problem generalized