In 1901 the Swiss historian of mathematics and Arabist Heinrich Suter (1848-1922) published in the journal *Bibliotheca* *mathematica* the German translation of an Arabic mathematical treatise composed by (as the author himself wrote) “Master Abu Zakariya Muhammad ibn Abd Allah known under the name of el-Hassar,” in which we find the earliest known description of an iterative process for extracting square roots yielding closer and closer approximations to the real value. Although this ingenious yet basically simple way to calculate square roots – completely different from our method today – was taught later by the outstanding Italian mathematicians Luca Pacioli (*ca.* 1445-1517), Hieronimo Cardano (1501-1576), Nicolo Tartaglia (1499/1500-1557), and Pietro Antonio Cataldi (1548-1626), it is largely unknown today. But it is worthy of study, and we present it in this article.

In 1901 the Swiss historian of mathematics and Arabist Heinrich Suter (1848-1922) published in the journal *Bibliotheca* *mathematica* the German translation of an Arabic mathematical treatise composed by (as the author himself wrote) “Master Abu Zakariya Muhammad ibn Abd Allah known under the name of el-Hassar,” in which we find the earliest known description of an iterative process for extracting square roots yielding closer and closer approximations to the real value. Although this ingenious yet basically simple way to calculate square roots – completely different from our method today – was taught later by the outstanding Italian mathematicians Luca Pacioli (*ca.* 1445-1517), Hieronimo Cardano (1501-1576), Nicolo Tartaglia (1499/1500-1557), and Pietro Antonio Cataldi (1548-1626), it is largely unknown today. But it is worthy of study, and we present it in this article.

All we know about Abu Zakariya (in Hebrew manuscripts, Abu Bakr) Muhammad ibn Abu Abd Allah Ayyash el- or al-Hassar (the rush mat maker) is that he lived in the 12th century in the Islamic West, in the Maghreb (today's Morocco, Algeria and Tunisia) or in Andalus (southern Spain). As often happens, no original manuscript by the author has survived, but five later copies do survive:

- the one in the ducal library in Gotha, Germany, written in 1432, and translated by Suter in 1901;
- the oldest, dated 1194, in the Lawrence J. Schoenberg Collection at the University of Pennsylvania (Kunitzsch, 2007), which has made the manuscript available digitally;
- two in Morocco; and
- one in Syria.

The chapter on the square roots is not contained in all five. Some copies bear the title *Book of Proof and* *Recall.* The copy in Gotha has no title. It has 128 leaves.

In the chapter on fractions we read: “but if you have to represent a fraction, then write the denominator under a [horizontal]* line and above each its mentioned part.” This is the first known use of the fraction bar. A little later Leonardo Pisano (“Fibonacci”) became the first European mathematician to employ it in his *Liber ab(b)aci *of 1202.

Al-Hassar's book is supposed to be the oldest mathematical text of the Maghreb and Moslem Spain. He wrote a second larger work with the title, *The large book* or *The complete book on the art of number,* of which the first part, Book I, was found in* *1986, the first pages missing, in the Library Ibn Yusuf in the* *Moroccan city of Marrakesh. Its contents were described in 1987 by* *Mohamed Aballagh and Ahmed Djebbar. The second part, Book II,* *probably is lost.

Al-Hassar does not claim to be the originator of the new arithmetic propositions he expounds. On the contrary, he makes it clear at the outset that the opposite is true:

And everything that I have put together, described and explained in this book, stems from the statements of older scholars, and I have drawn it from the books of the ancestors, I have collected it, commented on it, and found and derived it thanks to their sound reasonings.

Unfortunately most of the “books of the ancestors” (or, rather, manuscripts of the ancestors) have been lost.

Only a few decades later, in his 1202 *Liber Ab(b)aci, *Leonardo Pisano used al-Hassar’s method to calculate the first and second approximations of the square root of 10 and of larger numbers. Pisano did not mention al-Hassar’s name, and probably had learned the method from someone else.

In 1271 the Jewish physician, author, and outstanding translator Moses ben Samuel ibn Tibbon, who lived in Montpellier in southern France, translated al-Hassar's text into Hebrew under the title *Sefer ha-heshbon *(*Book of arithmetic*), of which there is one manuscript in the Vatican and one in Oxford.

Later, al-Hassar's method was taken up again in the hand-written *Treatise of Arithmetic* by Aboûl Hasan Ali ben Mohammed Alkalçadi, who was born in Spain in 1412 and died in Tunisia in 1486 (Woepcke, 1859).

The first *printed* book that contained all the mathematical knowledge of its time was the *Summa de* *Arithmetica Geometria Proportioni et Proportionalita* of Luca Pacioli of 1494 (reprinted in 1523), and of course you find al-Hassar's procedure in it (Woepcke, 1874). From 1556 to 1560 Nicolo Tartaglia published his mathematical encyclopedia, the *General Trattato di Numeri et* *Misure* (*General Treatise of Numbers and Measurements*), and he wrote in the second volume in the chapter “Regola di saper sempre approssimarsi piu nelle radici sorde” (“Rule for knowing how to approach always closer the irrational roots”) that “those ancient Arabs, most expert investigators of these practices, ... sought yet another rule for an increasingly close approximation of the true value, & ad infinitum.” This means Tartaglia knew that the rule did not originate in Europe.

Al-Hassar’s method was treated also in 1539 in the *Practica arithmetice* of Hieronimo Cardano, in 1559 in the *Logistica* (*Arithmetic*) of the Frenchman Ioannes (Jean) Buteo, and in 1613 in the *Trattato del modo brevissimo di trouare la radice quadra* *delli numeri* (*Treatise on the shortest method to find the square* *root of numbers*) (Woepcke, 1874), in which the author Pietro Antonio Cataldi (1548-1626) published his discovery of continued fractions.

Beginning in the 16th century, arithmetic books in most countries adopted the present-day method for the extraction of square roots, the one that is still currently taught in schools. When more accurate results were needed, pairs of zeros (\(00\) or \(0000,\) etc.) were added to the radicand. This is still done today. Things were different in Italy, however. Italian arithmetic books applied the present-day method only to square numbers. For non-square numbers, the method described by the great Italian mathematicians Pacioli, Cardano, and Tartaglia was retained in textbooks and reprints until the very end of the 18th century. Thereafter, the method we trace back to al-Hassar fell into oblivion even in Italy.

_______________________

* Throughout this article, material [in brackets] has been inserted by the author, in order to assist the reader.

Imagine you live in a time with no computers. If you want to approximate the square root of a non-square number \(N,\) there is a logical way to do it: Look for the square number \(a^2\) nearest to \( N.\) If \(N\) is greater than \(a^2,\) then \(N=a^2+r\) (where we use \( r\) for *remainder*); if \(N\) is smaller than \(a^2,\) then (using \( r\) again) \(N=a^2-r.\) Now think of the formula \(\left(a \pm b\right)^2=a^2\pm 2ab+b^2,\) and assume that \(b\) is so small that you may neglect its square \(b^2.\) This means \(r\approx 2ab\) and, from this, \(b\approx \frac{r}{2a}.\)

Thus the first approximation of the square root of \(N\) is \[\sqrt{N}=\sqrt{a^2 \pm r}\approx a\pm \frac{r}{2a}.\] Take, for example, \(N=3=2^2-1. \) That is \(a=2,\: r=1,\) and \[\sqrt{3}\approx 2-\frac{1}{4}=1\frac{3}{4}=1.75,\] a very good approximation because \[\left(1\frac{3}{4}\right)^2=\left(\frac{7}{4}\right)^2=\frac{49}{16}=3\frac{1}{16},\] with an excess of only \(\frac{1}{16}\) or \( 0.0625.\)

This method (but not, of course, the symbolic formula) was used in Greek antiquity, especially by the first century A.D. mathematician Heron of Alexandria. It was more accurate the nearer \(N\) was to \(a^2\); that is, the smaller \(r\) was. The square of \(a \pm \frac{r}{2a}\) is \(a^2 \pm r + \left(\frac{r}{2a}\right)^2.\) This means that, for both positive and negative \(r,\) there is always a surplus or excess of \(\left(\frac{r}{2a}\right)^2.\)

Of course, some ancient Greek mathematicians were able to calculate better approximations. Archimedes (287-212 B.C.), for example, used \(\sqrt{3}\approx 1\frac{571}{780}=\frac{1351}{780},\) or, in decimal notation, \(1.7320512\dots.\) The real value of \(\sqrt{3}\) is \(1.7320508\dots.\) (Because decimal fractions began spreading only at the beginning of the 17th century, square roots of non-square numbers were represented only in the form of common fractions prior to that time.)

But the Greeks had no general method to improve the accuracy of the above approximations. We find the earliest known description of such a procedure in the five pages of the “Chapter on the extraction of irrational roots by approximation” of al-Hassar’s book (beginning on the reverse of leaf 125 of the Gotha copy). It is typical that al-Hassar does not give general instructions, rules, or formulas, but only numerical examples that show how to proceed. And he does not derive his method or give a general proof, but shows only that his method gives the right answers in his examples.

Al-Hassar knew the Greek procedure for both the positive and the negative case, and he calculated the example \(\sqrt{5}=\sqrt{2^2+1}\approx 2\frac{1}{4}.\) His “Chapter on the extraction of irrational roots by approximation” begins with the words (English translation of Suter's German translation):

When it is said: which is the square root of \(5,\) so take the next square number to \(5,\) this is equal to \(4,\) subtract it from \(5,\) the remainder is \(1,\) divide this by \(4,\) that gives \(\frac{1}{4},\) and add this to the root of \(4,\) which is equal to \(2.\) That gives \(2\frac{1}{4}.\) And this is the approximate root of \(5.\) Namely, when you multiply \(2\frac{1}{4} \: \left[=\frac{9}{4} \right]\) by itself you get \(\left[\frac{81}{16}=\right]\: 5\frac{1}{16}.\) The deviation is an excess of \(\frac{1}{16}.\)

(Later al-Hassar obtained also \(\sqrt{10}=\sqrt{3^2+1}\approx 3 \frac{1}{6}\) with the square \(10\frac{1}{36}\) – an excess of \(\frac{1}{36}.\)) And now comes the important instruction:

But if you want a closer approximation, then double \(2\frac{1}{4},\) that gives \(4\frac{1}{2}.\) Divide \(\frac{1}{16}\) by \(4\frac{1}{2}.\) This gives \(\frac{1}{72}\) (an eighth of a ninth). Subtract this from \(2\frac{1}{4}\:\left[=2\frac{18}{72}\right].\) This leaves a remainder of \(2\frac{17}{72}\: \left[=\frac{161}{72}\right].\) When you multiply this by itself you get \(\left[\frac{25921}{5184}=\right]\: 5\frac{1}{5184},\) and this is nearer [to \(5\)] than \(5\frac{1}{16}.\)

This procedure, expressed in modern notation, is \[ \sqrt{a^2+r}\approx\left(a+\frac{r}{2a}\right)-\frac{\left(\frac{r}{2a}\right)^2}{2\left(a+\frac{r}{2a}\right)}.\]

Al-Hassar continues:

If you want a closer approximation yet, double \(2\frac{17}{72},\) divide \(\left(\frac{1}{72}\right)^2\: \left[=\frac{1}{5184}\right]\) by the result, and subtract what you get from \(2\frac{17}{72}.\) So the result will be an even closer approximation than the first and second roots. You may continue like this as far as you want.

In other words, apply this method several times. Of course, in the process the numerator and the denominator of the fractions get longer and longer, and the multiplications become more and more unwieldy. Today's calculator or computer spares us the multidigit calculations that had to be done by hand in the past.

A warning about calculating the first approximation: In both the positive case \(N=a^2+r\) and the negative case \(N=a^2-r,\) the \(r\) must not exceed \(a.\) Therefore, \(\sqrt{14}\) must not be calculated by \(\sqrt{3^2+5},\) but only by \(\sqrt{4^2-2}.\) With respect to the square root of \(20,\) al-Hassar remarked that you may take either \(16\) (positive \(r\)) or \(25\) (negative \(r\)) as the nearest square, that is, \(\sqrt{4^2+4}\) or \(\sqrt{5^2-5}.\) In both cases you get the same first approximation, namely \(4+\frac{1}{2}=5-\frac{1}{2}=4\frac{1}{2}.\) This is true for all numbers of the form \( a^2+a=(a+1)^2-(a+1)=a(a+1)\); that is, for \(2, 6, 12, 20, 30, 42, 56, \) and so on. The first approximation is always \(a+\frac{1}{2}.\)

In the case of \(\sqrt{5}=\sqrt{2^2+1}\) with \(a=2\) and \(r=1,\) \(\:\frac{2a}{r}=\frac{4}{1} = 4\) is an integer. Therefore, its reciprocal, \(\frac{r}{2a}=\frac{1}{4},\) is a unit fraction; that is, a fraction with numerator \(1.\) For all square roots *for which this is true, and only for these,* there is a simplification of al-Hassar's method.

First we see that \[ \left(a\pm \frac{1}{n}\right)^2= a^2\pm\frac{2a}{n}+\frac{1}{n^2}.\] This means that, for both the plus and the minus cases, the excess is \(\frac{1}{n^2},\) again a unit fraction. This is true for all further approximations. In all of these cases the first approximation is \(a\pm \frac{1}{n}=\frac{an\pm 1}{n}.\) We call this fraction \(\frac{p}{q}.\)

To get the next approximation according to al-Hassar's rule, the excess \(\frac{1}{n^2}=\frac{1}{q^2}\) has to be divided by double the preceding approximation, namely \(2\times{\frac{p}{q}}=\frac{2p}{q}.\) We have \[\frac{1}{q^2}\div \frac{2p}{q}=\frac{1}{q^2}\times\frac{q}{2p}=\frac{1}{2pq}.\] This has to be subtracted from \(\frac{p}{q}.\) But \[ \frac{p}{q}-\frac{1}{2pq}= {\frac{2p^2-1}{2pq}}.\] And with this simple formula – much easier to use than al-Hassar's – we obtain an increasingly accurate series of approximations. (This formula is not new. It was given in 1766 by the Italian-French mathematician Joseph-Louis Lagrange (1736-1813); see *Oeuvres de Lagrange,* vol. 1, p. 695.)

Let us apply the formula \[{\frac{2p^2-1}{2pq}}\] to get the third approximation of \(\sqrt{5}\) from al-Hassar's second approximation \(\frac{p}{q}=\frac{161}{72}.\) We have \[2p^2-1= 2\times 161^2-1=2\times 25921-1=51841\] and \[2pq=2\times 161\times 72=23184. \] Therefore, the third approximation is \(\frac{51841}{23184},\) or, in decimals, \(2.2360679779158\dots .\) Its square is \(5.000000001860\dots .\) The excess is \(1.860...\times 10^{-9}.\) This is equal to \(\left(\frac{1}{23184}\right)^2.\)

When we calculate the next approximations with our formula, we find that the excess of the fourth approximation is \(1.730...\times 10^{-19}\) and of the fifth approximation \(1.497...\times 10^{-39},\) really excellent results obtained so easily.

In February 1657 (he gave no exact day), the French co-founder of analytic geometry and of probability calculus, pioneer of infinitesimal calculus, and great number-theoretician Pierre de Fermat (1607**-1665) addressed a letter to the French amateur mathematician Bernard Frénicle de Bessy (1605-1675) that is worth translating from French into English:

Every non-square number is of such a nature that one can find an infinity of square numbers, by which you multiply the given number, and when you add the unit [that is, \(1\)] to the product, a square number is coming out.

Example: \(3\) is a non-square number, which multiplied by \(1,\) which is a square number, makes \(3,\) and, taking [adding] the unit, makes \(4,\) which is a square number.

The same \(3,\) multiplied by \(16,\) which is a square number, makes \(48\) and, taking the unit, makes \(49,\) which is a square number.

There are infinite numbers, which, multiplying \(3,\) adding the unit, make likewise a square number. [The next is \(15^2=225,\) for which \( 3\times 225+1=676=26^2. \)]

I demand from you a general rule, when a non-square number is given, to find square numbers, which, multiplied by the given number, adding the unit, produce square numbers.Which is, for example, the smallest square number, which, multiplying \(61,\) adding the unit, makes a square number?

[The answer is \( 226\: 153\: 980^2.\) Note that \( 61\times 226\: 153\: 980^2 +1 = 1\: 766\: 319\: 049^2.] \)

Equally, which is the smallest square number, which, multiplying \(109,\) and adding the unit, makes a square number?

[The answer is \( 109 \times 15\:140\:424\:455\:100^2+1=158\:070\:671\:986\:249^2.] \)

If you do not send me the general solution, send me the particular one of these two numbers, which I have chosen from the smallest ones in order not to give you too much pains.

After having received your answer I will propose you something else. It is clear, without saying, that my proposition is only for finding integers, which satisfy the question because in the case of fractions the most insignificant arithmetician would manage the problem.

Also in February 1657, Fermat addressed a challenge to English mathematicians about the same mathematical problem, this time in Latin. It was received by the Irish mathematician Viscount William Brouncker (1620-1684), the co-founder and first president of the Royal Society, in March 1657. Brouncker at first delivered solutions in the form of fractions, but, after Fermat's demand for integral solutions, he provided these as well.

The main proposition of the challenge, in Latin and English, was:

Dato quovis numero non quadrato, dantur infiniti quadrati qui, in datum numerum ducti, adscitâunitate, conficiant quadratum.Given any non-square number, there are infinite square numbers, which multiplied by the given number, [and] the unit added, yield a square number.

The challenge was to prove this theorem, or to find a square number, which multiplied by \(149,\) \(109,\) or \(433,\) plus \(1,\) produces a square number.

If the non-square number is called \(d,\) the multiplying square number \(y^2,\) and the produced square number \(x^2,\) the resulting equation can be expressed as \(dy^2+1=x^2.\) Today it is usually written in the form \(x^2-dy^2=1.\) (Caution: \(dy^2\) is *not* a differential!)

Although it would have been logical to call the problem of finding \(x\) and \(y\) for a given \(d\) the *Fermat equation,* through a misunderstanding it was named instead the *Pell equation* after the Englishman John Pell (1611-1685), despite Pell's having little to do with it. But it is utterly futile to try to rectify this well-established misnomer.

It was the most prominent mathematician of the 18th century, the Swiss mathematician Leonhard Euler (1707-1783), who wrote in his popular 1770 book *Vollst**ä**ndige Anleitung zur Algebra* (*Complete instruction in algebra; *better known in English as *Elements of Algebra*), in the second section of the second part, “On indeterminate analysis,” in Chapter 7, “About a special method to make the formula \(ann +1\) into a square in integral numbers”: “For this an erudite Englishman by the name of Pell has invented a very ingenious method...” This remark was the origin of the erroneous designation.

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**Most biographies of Fermat still give 1601 as his birth year. However, Klaus Barner, Professor Emeritus, University of Kassel, Germany, has found that Fermat’s father had a son named Piere (one "r"), born in 1601, who died shortly after his birth. A second son named Pierre was born in 1607, and he became the famous mathematician (Barner, 2001; 2007). For more information, see the appendix, "When Was Fermat Born?"

There is a very intimate relation between the approximations of square roots and the Pell equation. Thus the ancient and medieval Indian approximations for \(\sqrt{2},\) namely \(\frac{17}{12}\) and \(\frac{577}{408},\) and the approximations \(\frac{26}{15}\) and \(\frac{1351}{780}\) for \(\sqrt{3},\) all satisfy the Pell equation, when we take the radicand for \(d,\) the numerator for \(x,\) and the denominator for \(y.\) For example, \[ 1351^2-3 \times 780^2=1825201 - 3 \times 608400=1825201-1825200=1.\]

We detect that \(p\) and \(q\) of our approximations \(\frac{p}{q},\) but *only those with unit fraction excesses,* are solutions of the Pell equation if we set the radicand \(N=d,\: p=x, \) and \(q=y,\) which yields the Pell equation \(p^2-Nq^2=1.\) With the formula \[\frac{2p^2-1}{2pq},\] it is easy to obtain new and larger solutions \(x\) and \(y\) for Pell equations with a given \(d.\)

It is not difficult to prove why in the first approximation \(p\) and \(q\) fulfill the Pell equation: You can apply our formula only if \(N = a^2\pm \frac{2a}{n}.\) Then the first approximation is \[ \frac{p}{q}=a\pm \frac{1}{n}=\frac{an\pm 1}{n},\] with \(p=an\pm 1\) and \( q=n.\) Their squares are \(p^2=a^2n^2\pm 2an+1\) and \(q^2=n^2.\) Then the Pell equation is \[ p^2-Nq^2=a^2n^2\pm 2an+1-\left(a^2\pm \frac{2a}{n}\right)n^2=1.\]

It is important to note that the numerator and the denominator of square roots whose \(\frac{r}{2a}\) does not have the form \(\frac{1}{n},\) where \(n\) is an integer, are *not* solutions to Pell equations with the result \(1.\) This can be easily shown: If we reduce \(\frac{r}{2a}\) to the lowest numerator \(m\) and denominator \(n,\) the first approximation is \[a\pm \frac{m}{n}=\frac{an\pm m}{n}.\] Its square is \[\frac{a^2n^2\pm 2amn+m^2}{n^2},\] where \[N=\frac{a^2n^2\pm 2amn}{n^2}\] and the excess is \(\frac{m^2}{n^2}.\) The Pell equation is \[ a^2n^2\pm 2amn+m^2-Nn^2=a^2n^2\pm 2amn+m^2-\frac{a^2n^2\pm 2amn}{n^2}n^2=m^2.\] Hence \(m^2=1\) only if \(m=1.\)

From this we can draw the conclusion that \(x^2-dy^2 = 1\) only if the square of the calculated \(\sqrt{d}\:\) has an excess in the form of a unit fraction \(\frac{1}{n}.\)

For example, for \[ \sqrt{22}=\sqrt{5^2-3}\approx 5 - \frac{3}{10} = 4\frac{7}{10}=\frac{47}{10},\] where \(a=5,\: m=3,\: n=10,\) and \(\frac{3}{10}\) is *not* a unit fraction, we get the wrong Pell equation: \[47^2-22\times 10^2=2209-2200=9,\] not \(1,\) because \(m=3\) and \(m^2=9.\)

For \(d=22,\) the smallest, or the so-called fundamental, Pell solution, calculated with the help of continued fractions, is \(x=197\) and \(y=42,\) yielding \(197^2=38809,\) \(22\times 42^2=38808,\) and \(\left(\frac{197}{42}\right)^2=22\frac{1}{1764}.\) Here we have the expected unit fraction. Therefore, we can now apply our formula for the next approximation, \[\frac{2p^2-1}{2pq},\] with \(p=197\) and \(q=42.\) We get \(\frac{77617}{16548},\) and the Pell equation \[ 77617^2-22\times 16548^2=6024398689-6024398688=1.\]

It is especially simple to find the fundamental solution of the Pell equation for the numbers \(d=a^2+1,\, a^2+2,\, \) and \(a^2-2.\) The first approximations for the square roots of these numbers are respectively \[ a+\frac{1}{2a}=\underline{\frac{2a^2+1}{2a}},\, a+\frac{2}{2a}=a+\frac{1}{a}=\underline{\frac{a^2+1}{a}},\,\,{\rm and} \,\,a -\frac{2}{2a}=a-\frac{1}{a}=\underline{\frac{a^2-1}{a}}.\] The numerators and denominators of the underlined fractions are \(x\) and \(y\) for \(d.\) For \(d=a^2-1,\) the fundamental solution is always \(x=a,\) \(y=1.\)

**Conclusion**

Today you get square roots with many correct decimal places instantly with your calculator or computer. However, as we saw with the fourth and fifth approximations of \(\sqrt{5},\) far greater accuracy can be achieved by means of a few easy calculations invented more than 800 years ago.

**About the Author**

Friedrich Katscher was born in Vienna in 1923. He is a lifelong Viennese who studied theoretical physics and mathematics and was science editor of a newspaper. Since the 1960s, he has been active as a historian of mathematics, giving lectures at the Vienna Technological University (Technische Universität Wien) and specializing in Italian Renaissance mathematics. His ability to speak German, English, French, Italian, and Spanish aids him in his mathematics history research. He invites you to correspond with him about this article and about Italian Renaissance mathematics via email (dr.katscher.vienna@chello.at).

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Woepcke, F. (1874). “Intorno ad un metodo per la determinazione approssimativa degl'irrazionali di secondo grado,” *Bullettino di bibliografia e di storia delle scienze* *matematiche e fisiche*, vol. VII (1874), 255-262.

* Editor's note:* As of 2016, an expanded version of this appendix now appears as the separate article, When Was Pierre de Fermat Born? The expanded article includes a photograph of Fermat's gravestone with the inscription reading that Fermat died in 1665 at the age of 57 years.

Most biographies of Fermat still give 1601 as his birth year. However, Klaus Barner, Professor Emeritus of the University of Kassel, Germany, has discovered that this is incorrect (2001). Investigating in French archives, he found the following facts: In the baptismal register of the city of Beaumont, he read that Piere (with one “r”), son of Dominique Fermat, was baptized August 20, 1601. Other documents show that the name of Piere's mother was Françoise Cazenove. It seems that she, her son, and a daughter died after 1603. Sometime between 1603 and 1607, Dominique Fermat married a second time. His new wife, Claire de Long, bore him five children, among them a son named Pierre again, like his deceased half-brother. Unfortunately the baptismal registers from 1607 to 1611 are completely missing. But the mathematician's eldest son, Samuel, had written in the Latin epitaph of his father at the family grave in Toulouse:

OB.[iit] XII. IAN[uarii] .M.DC.LXV. AET[ate] .AN.[norum] .LVII.

That is, Pierre de Fermat died on January 12, 1665, at the age of 57 years. This means he was born between January 13, 1607, and January 12, 1608, and most probably in 1607. Additional research led Barner to conclude that Fermat was born between October 31 and December 6, 1607 (2007). In 1631 Fermat was conferred the right to add the title of nobility "de" to his name.