Students of elementary algebra are often told to simplify their fractions in the final answer to a test question. For example, \[ \frac{2n+2}{4}\] ought to be simplified to \[ \frac{n+1}{2}.\]

Should students question why they have to do that, our answer—perhaps like yours—would be a combination of the following arguments:

- Math is complicated enough; we always want an answer to be as simple as possible;
- If there is another step in the problem, it will be easier for you to have a simpler starting point for the next step;
- In case you need to remember the answer, it's easier to memorize a less complicated formula.

All of the above lines of reasoning are generally sound advice; however, there are cases in which these lines of reasoning are actually faulty. The purpose of this short note is to give such an example, and we additionally will point out a surprising connection of this theme with medieval Arabic mathematics.

Consider the formula for the sum of the first \(n\) integers, \[\sum_{i=1}^n i = \frac{n(n+1)}{2}.\]

This formula is both well-known, and easy to remember: we take the product of the two consecutive integers that start at \(n\), and divide by \(2\).

On the other hand, the formula for the sum of the squares of the first \(n\) integers, \[ \sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6},\]

may also be well-known, but it is not quite as easy to remember. But we can remedy that as follows: \[ \sum_{i=1}^n i^2=\frac{2n(2n+1)(2n+2)}{24}.\]

If we *do not* simplify this fraction, we can memorize it as the product of the three consecutive integers that start at \(2n\), divided by \(24\) (or \(4!\)).

A similar phenomenon occurs if we look at the **sum of the squares of odd integers**: the formula

\begin{equation}\sum_{i=0}^n (2i+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3} (1) \end{equation}

is more easily memorized as

\begin{equation} \sum_{i=0}^n (2i+1)^2 = \frac{(2n+1)(2n+2)(2n+3)}{6}, (2)\end{equation}

the product of the three consecutive integers starting at \(2n+1\), and divided by \(3!\). Similarly, the formula for the **sum of squares of even integers**,

\begin{equation} \sum_{i=1}^n (2i)^2 = \frac{2n(n+1)(2n+1)}{3}, (3)\end{equation}

is easier to remember as

\begin{equation} \sum_{i=1}^n (2i)^2 = \frac{2n(2n+1)(2n+2)}{6}, (4)\end{equation}

the product of the three consecutive integers starting at \(2n\), divided by \(3!\).

It should not go unnoticed that the verbal explanations of these last two reformulations are pretty much the same, and that they are more easily grasped than the notational versions. Taking the verbal explanations as a guide, if we call \(n\) the highest number, then the formulas become identical. If \(n\) is odd, then

\begin{equation} 1^2+3^2+5^2+\ldots+n^2=\frac{n(n+1)(n+2)}{6}, (5)\end{equation}

and if \(n\) is even, then^{1}

\begin{equation}2^2+4^2+6^2+\ldots+n^2=\frac{n(n+1)(n+2)}{6}. (6)\end{equation}

To express the sums this way we had to abandon our handy sigma notation, which students initially resist anyway. And it is here, having distanced ourselves a little from the confines of our notation, that the essential nature of the rules becomes apparent.

Before the advent of modern algebraic notation, rules like these were routinely expressed verbally. Taking a large step back in time to the latter 13th century, the Moroccan polymath Ibn al-Banna̅ ͗ explained the sum of squares of odd integers and the sum of squares of even integers with wording that reflects formulas (5) and (6). He wrote in his *Condensed [Book] on the Operations of Arithmetic* (*Talki̅ṣ a ͑ma̅l al-ḥisa̅b*) that adding the odd squares "is given by multiplying a sixth of the upper extreme by the product of the two numbers that come after it," and the rule for the even squares is identical: "multiply a sixth of the upper extreme by the product of the two numbers that come after it."^{2}

It can be argued that a student with limited algebraic skills or notational experience would find the rules as described by Banna̅ ͗ easier to understand and remember than the two rules (1) and (2). The instructions "multiply a sixth of the upper extreme by the product of the two numbers that come after it" is more accessible than our \({1\over 6}n(n+1)(n+2)\) to a person who struggles with basic algebra. In fact, in Ibn al-Banna̅ ͗ 's time, these rules were considered to be part of arithmetic, not *al-jabr wa l-muq*a̅*bala*, as algebra was called then. Ibn al-Banna̅ ͗ thus gives the rules in his chapter on addition, not in his later chapter on algebra. We may view the medieval rules as identical with \({1\over 6}n(n+1)(n+2)\), but in the latter we have taken the step of naming the upper extreme \(n\), and forming an algebraic expression to replace Ibn al-Banna̅ ͗ 's sequence of operations. Arabic algebra was a specific technique of arithmetical problem-solving, and was not used to state arithmetical rules like these.

**1. This formula, of course, is also the sum of the first \(n\) triangular numbers. By noting that consecutive triangular numbers add up to a square, one can compose simple proofs for the rules.**

**2. See [2, 43.4,9] (i.e., lines 4 and 9 of page 43), our translation. These rules are also found in several other medieval Arabic books.**

But verbal expositions are not always so clear. The topic of this section is a rule for finding the "side," or cube root, of a perfect cube. The rule was inserted by a later hand as a kind of appendix to some manuscripts of the 10th-century *Epistles* of the Brethren of Purity (Ikhwa̅n a̅l-Ṣafa ͑). The *Epistles*, which were intended as a compendium of all scientific and philosophical knowledge of the time, were written by an anonymous, secretive group of scholars centered in Basra, in southern Iraq. Think of them as a kind of medieval version of the Bourbaki group.

Below is a translation of the appendix. Because the verbal explanation may be difficult to follow, we follow it with two worked-out examples.

Abu̅ Ṭa̅lib Aḥmad bin Ja ͑far bin Ḥamma̅d recounted that he found a characteristic property of the side of the cubic number that is being mentioned, which is that when asked for the side of a cubic number, the way [to find] it is that you take a sixth of the number being investigated. You see if the number [being investigated] is even, [in which case] you add up the squares of the odd numbers in a series until [you reach] what resulted from the sixth, and you multiply the remainder of the sum of the squares by six, to get the sought-after side. If the number being sought is odd, you take the squares of the even numbers in a series and you add them up. Where the sum ends, you look at the remainder and you multiply six by it, which gives the side. If the number that is investigated is odd, take the even squares in a series and add them up to whereof the addition ends, then look into the remainder and multiply it by six, wherein the side comes out.^{3}

We illustrate the rule by examples for both the even and the odd case.

**Example 1**

Suppose we wish to compute the cube root of \(14^3=2744\). Dividing \(2744\) by \(6\), we get \(457{1\over 3}\).

Then we add up the squares of the odd integers and we stop just before exceeding the latter number: \[1^2+3^2+5^2+7^2+9^2+11^2+13^2=455.\] The difference \(457{1\over 3}-455\) is \(2{1\over 3}\), and multiplying by \(6\) we find the cube root of \(2744\): \(6\times 2{1\over 3}=14\).

**Example 2**

We compute the cube root of \(17^3=4913\). Dividing \( 4913\) by \(6\) gives \(818{5\over 6}\).

Adding the squares of the even numbers as long as we stay below that gives \[2^2+4^2+\cdots+16^2=816.\] The difference is \(818{5\over 6}-816=2{5\over 6}\), and so multiplying by \(6\) we get \(17\).

In the Appendix, we provide more examples of Abu̅ Ṭa̅lib's rule in the form of a set of student exercises.

We have not seen this rule in any other text, either in Arabic or in any other language. Its obscurity can be accounted for by the fact that it is not at all a practical means of finding the cube root of a perfect cube, and it fails to give an approximation to the cube root of other numbers. (See the second student exercise in the Appendix.) It is a curiosity of number theory, never intended as an alternative to the standard rules practiced at the time, and equivalent to our Ruffini-Horner algorithm.^{4}

Abu̅ Ṭa̅lib's rule is stated with no hint of how it was derived. But its derivation is not difficult to figure out. First, note that \[n^3 - n = (n-1)n(n+1)\]

is the product of three consecutive integers. Al-Karaji̅, for one, stated and proved this identity in the early 11th century as a lemma to his rule for finding the sum \(1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+\ldots+8\cdot 9\cdot 10\).^{5} Therefore, the cube root \(n\) of \(n^3\) can be expressed as the difference \[n = n^3 - (n-1)n(n+1).\]

Now, if we remember that the product of three consecutive numbers occurs in the rules for summing odd and even squares (provided we do not simplify the answer!), we obtain the method discovered by Abu̅ Ṭa̅lib. If \(n\) is even, then \(n-1\) is odd, and vice-versa, thus the requirement to add the odd squares for an even cube, and the even squares for an odd cube. For the case that \(n\) is even, we have \[n = n^3 - (n-1)n(n+1)=6\left({\rm\frac{1}{6}}n^3-(1^2+3^2+5^2+\ldots+(n-1)^2)\right),\]

while for \(n\) odd it is \[n = n^3 - (n-1)n(n+1)=6\left({\rm\frac{1}{6}}n^3-(2^2+4^2+6^2+\ldots+(n-1)^2)\right).\]

In an age when modern algebraic formulas for numerical identities were still in the distant future, mathematicians could naturally conceive of the sums of consecutive squares of odd numbers and of even numbers as a sixth of the upper extreme multiplied by the product of the next two integers. With this understanding of the rules, the discovery of Abu̅ Ṭa̅lib's method to calculate cube roots of perfect cubes, described above, becomes plausible and easy to understand. On the other hand, modern-day mathematicians, accustomed to the simplified formulas \[\sum_{i=0}^n (2i+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}\] and \[\sum_{i=1}^n (2i)^2 = \frac{2n(n+1)(2n+1)}{3}\] afforded by our algebraic notation, may have to work harder to understand why such a method works. This is in fact what happened to one of us, who understood the rule only after he re-wrote (1) and (3) in the non-simplified forms (2) and (4), and from this experience conceived the idea for writing this note.

3. Translation adjusted by the authors from [1, p. 68]. This Abu̅ Ṭa̅lib is otherwise unknown.

4. See [3] for such examples of cube root extraction from China, the Muslim world, and medieval Europe.

Following are student exercises that build upon the historical material described on the previous page. A pdf version suitable for handing out and working through in a classroom is also available.

The goal of these exercises is to explore a curious rule for finding the "side," or cube root, of a perfect cube. The rule was inserted by a later author as a kind of appendix to some manuscript copies of the 10th-century *Epistles* of the Brethren of Purity (Ikhwa̅n a̅l-Ṣafa ͑). The *Epistles*, which were intended as a compendium of all scientific and philosophical knowledge of the time, were written by an anonymous, secretive group of scholars centered in Basra, in southern Iraq.

Below is a translation of the appendix. Before the advent of modern algebraic notation, rules like these were routinely expressed verbally. Because the verbal explanation may be difficult to decipher, we follow it with two worked-out examples.

Abu̅ Ṭa̅lib Aḥmad bin Ja ͑far bin Ḥamma̅d recounted that he found a characteristic property of the side of the cubic number that is being mentioned, which is that when asked for the side of a cubic number, the way [to find] it is that you take a sixth of the number being investigated. You see if the number [being investigated] is even, [in which case] you add up the squares of the odd numbers in a series until [you reach] what resulted from the sixth, and you multiply the remainder of the sum of the squares by six, to get the sought-after side. If the number being sought is odd, you take the squares of the even numbers in a series and you add them up. Where the sum ends, you look at the remainder and you multiply six by it, which gives the side. If the number that is investigated is odd, take the even squares in a series and add them up to whereof the addition ends, then look into the remainder and multiply it by six, wherein the side comes out.^{1}

We illustrate the rule by examples for both the even case and the odd case.

**Example 1**

Suppose we wish to compute the cube root of the even number \(2744\).

- Dividing \(2744\) by \(6\), we get \(457{1\over 3}\).
- Then we add up the squares of the odd integers, starting with \(1^2\) and stopping just before the sum exceeds the number we computed in the first step: \[1^2+3^2+5^2+7^2+9^2+11^2+13^2=455.\]
- The difference \(457{1\over 3}-455\) is \(2{1\over 3}\), and multiplying this by \(6\), we find the cube root of \(2744\): \[6\times 2{1\over 3}=14.\]
- To check that this is correct, we simply cube the result: \(14^3 = 2744\).

**Example 2**

Suppose we wish to compute the cube root of the odd number \(4913\). Dividing \( 4913\) by \(6\) gives \(818{5\over 6}\).

Adding the squares of the even integers as long as we stay below that value gives \(2^2+4^2+\cdots+16^2=816\).

The difference is \(818{5\over 6}-816=2{5\over 6}\), and multiplying this by \(6\), we find the cube root of \(4913\): \(6\times 2{5\over 6}=17\).

Checking this by cubing gives: \(17^3 = 4913\).

**Exercise 1**

In this exercise you will compute the cube root of a perfect cube using Abu̅ Ṭa̅lib's method. You may use a standard calculator to add, subtract, multiply and divide numbers, but not for any other operation.

a. Fill in the following tables with the squares, and the sums of the squares.

The first table is for odd numbers and the second for even numbers.

The first two entries of each table are already filled.

\[ \begin{array}{r|c|c|c|c|c|c|c|c|c|}

n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 & 17\\

\hline

\rule{0mm}{4mm} n^2 & 1 & 9 & & & & & & &\\

\hline

\rule{0mm}{4mm} 1^2+3^2+\cdots+n^2 &1 & 10 & & & & & & &

\end{array}\]

\[ \begin{array}{r|c|c|c|c|c|c|c|c|c|}

n & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18\\

\hline

\rule{0mm}{4mm} n^2 & 4 & 16 & & & & & & &\\

\hline

\rule{0mm}{4mm} 2^2+4^2+\cdots+n^2 &4 & 20 & & & & & & &

\end{array}\]

b. All the numbers in the table below are perfect cubes. Choose the one that corresponds to today's day of the week. We will call this number \(n^3\), and we need to find \(n\).

\[

\begin{array}{c|c|c|c|c|c|c}

Mon & Tue & Wed & Thu & Fri & Sat & Sun\\

\hline

\rule{0mm}{4mm} 3375 & 1728 & 2197 & 9261 & 4096 & 6859 & 5832

\end{array}\]

c. Divide the number you have chosen by 6. So it will now be \(n^3/6\). Enter both \(n^3\) and \(n^3/6\) in the table below. But **do not** use decimals. Use mixed numbers instead, as in Example 1, where \( 4913/6 = 818\frac{5}{6}\). Also enter in the table the largest sum of squares from the appropriate table that is not larger than \(n^3/6\).

\[

\begin{array}{c|c|c}

n^3 & n^3/6 & \mbox{Sum of squares}\\

\hline

\rule{0mm}{4mm} & &

\end{array}\]

d. Find the difference \(n^3/6- (\mbox{Sum of squares}) \) from the last table, and multiply it by \(6\). Your answer will be the cube root of \(n^3\), or \(n\). Check by cubing your answer to see if you get the number you originally chose.

**Exercise 2 **

The numbers used in Exercise 1 are all perfect cubes, and Abu̅ Ṭa̅lib's method works fine in that case—even though it is not a very practical means of finding the cube root of a perfect cube! In this exercise, we will show that Abu̅ Ṭa̅libb's method does not even give a good approximation of the cube root if the initial number is not itself a perfect cube.

a. Choose a number in the following table according to the last digit of your birthday. So if your birthday is August 17 you would choose the number for 7.

\[

\begin{array}{c|c|c|c|c|c|c|c|c|c}

0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\

\hline

\rule{0mm}{4mm} 5132 & 4501 & 1955 & 3222 & 4997 & 2538 & 6047 & 7001 & 2988 & 4001

\end{array}\]

b. Go through the same steps as before, and cube the final answer. Check that what you get is not even close to the original number.

**Exercise 3** This exercise is for more advanced students.

A brute force method for finding the cube root of a number \(n\) is to compute successively all the cubes \(1^3, 2^3, 3^3, \ldots\) until we find some \(k\) such that \(k^3=n\). This method can be stated as follows: \[\sqrt[3]{n}=\max\{k: n-k^3\geq 0\}.\] Show that Abu̅ Ṭa̅lib's method can be re-stated as the following variation of the brute force method:

If \(n\) is a perfect cube, then \[\sqrt[3]{n} = \left \{\begin{array}{cl} 1+2\max\{m: n-2m(2m+1)(2m+2)\geq 0\} & \mbox{ if \(n\) is odd}\\

2\max\{m: n-(2m-1)(2m)(2m+1)\geq 0\} & \mbox{ if \(n\) is even}

\end{array}\right . \] In other words, instead of using the perfect cubes \(k^3\) to determine the required maximum value, we use the products of three consecutive integers \(k-1, k, k+1\).

*Hint for Exercise 3 *

Medieval Islamic mathematicians were very familiar with summation formulas, which they also stated in verbal form. Here are modern symbolic translations of their verbal descriptions for two such formulas that will be useful for this exercise:

\[ \mbox{If \(n\) is even: }\,\,\, 1^2+3^2+5^2+\ldots+(n-1)^2=\frac{(n-1)n(n+1)}{6}\]

\[ \mbox{If \(n\) is odd: } \,\,\, 2^2+4^2+6^2+\ldots+(n-1)^2=\frac{(n-1)n(n+1)}{6}\]

1. This translation of the primary source excerpt is based on El Bizri's translation of *Epistles of the Brethren of Purity: On Composition and the Arts, Epistles 6-8*, Oxford University Press, 2018, p. 68, with some adjustments by the authors of these exercises. This Abu̅ Ṭa̅lib is otherwise unknown.

[1] Nader El-Bizri and Godefroid de Callataÿ (eds). *Epistles of the Brethren of Purity: On Composition and the Arts, Epistles 6-8*, Oxford University Press, 2018.

[2] Ibn al-Bannāʾ. *Talkhīṣ aʿmāl al-ḥisāb*. Texte établi, annoté et traduit par Mohamed Souissi. Tunis: Université de Tunis, 1969.

[3] Bo Göran Johansson. Cube root extraction in medieval mathematics. *Historia Mathematica* 38 (2011), 338-367.

[4] A. S. Saidan. *Tārīkh ʿilm al-jabr fī l-ʿālam al-ʿArabī *(*History of Algebra in Medieval Islam*), 2 vols. Kuwait: Al-Majlis al-Waṭanī lil-Thaqāfah wa’l-Funūn wa’l-Adāb, Qism al-Turāth al-ʿArabī, 1986.

Valerio De Angelis received his B.Sc. in Mathematics from Imperial College of Science and Technology, London, in 1982. He then spent some time at the University of Rochester (where he received an MS degree and met Jeff Oaks) before completing his Ph.D. studies in mathematics at the University of Washington, Seattle, in 1992, under the guidance of Selim Tuncel. He joined the mathematics faculty at Xavier University of Louisiana in New Orleans in 1998. He has published articles in dynamical systems, number theory and combinatorics.

Jeffrey A. Oaks received his Ph.D. in mathematics from the University of Rochester in 1991, and he has taught mathematics at the University of Indianapolis since 1992. He began the study of history of mathematics in 1999, and has since published many studies in history of algebra, particularly in medieval Arabic algebr*a.*