Mark Kac (pronounced “cots”, 1914–1985) was a pioneer in the development of the modern theory of probability.¹ He was a star among the constellation of great Polish mathematicians who emerged in the early part of the twentieth century.² Kac’s first language was Russian, since he was born in an area then under the control of the Russian Empire. As a child he learned to speak Hebrew, the language of instruction at his primary school (his father was the principal), and French. He was eleven before he learned Polish, which was required by his new school, the Krzemieniec Lyceum of his home town. Intending to study engineering at Jan Kazimierz University (Lwów), he was diverted into mathematics by the events related below, and he earned a doctorate (1937) there under Hugo Steinhaus (a student of D. Hilbert’s, 1911). Thanks to a postgraduate scholarship at Johns Hopkins awarded in December 1938, Kac acquired his fifth language, English, and escaped the Nazis’ murder of his immediate family among the other Jewish citizens of Krzemieniec. His career was spent almost entirely in the United States: Hopkins, Cornell, war work at the Radiation Lab of MIT, Rockefeller University, and finally the University of Southern California. He pioneered the application of probability measures to topics in unexpected areas: number theory with P. Erdös [Erdös and Kac 1940]; the physics of ideal gases with G. Uhlenbeck and P. Hemmer [Kac, Uhlenbeck, and Hemmer 1963], and, closely related to work by R. Feynman, the “Feynman-Kac formula” [Kac 1949] within quantum theory (in particular, quantum field theory).³

Figure 1. Photograph of Mark Kac taken by Paul R. Halmos in 1962.
Who's That Mathematician? Images from the Paul R. Halmos Photograph Collection, page 26.

Kac’s willingness to look at things in a new way, and the simplicity and elegance of his expression, serve as models for all who work in the mathematical sciences.⁴ On two occasions (1950 and 1968) Kac won the Chauvenet Prize, the MAA’s highest award for expository writing; the second for the justly famous essay “Can one hear the shape of a drum?” [Kac 1966]. A master of the well-chosen example, Kac preferred the specific and concrete to the general and abstract. It would be contrary to the spirit of this good man not to provide such examples. Here is his oft-quoted distinction between types of genius:

There are two kinds of geniuses: the “ordinary”, and the “magicians”. An ordinary genius is a fellow that you and I would be just as good as, if we were only many times better. There is no mystery as to how his mind works. Once we understand what he has done, we feel certain that we, too, could have done it. It is different with the magicians. They are, to use mathematical jargon, in the orthogonal complement of where we are and the working of their minds is for all intents and purposes incomprehensible. Even after we understand what they have done, the process by which they have done it is completely dark. . . . Richard Feynman is a magician of the highest caliber [Kac 1985, xxv.].

Beyond his mathematical achievements, Kac was admired and loved for his quick wit and his genial, kind personality; he didn’t take himself too seriously. Concerning his swift tongue, a friend related this anecdote from 1952:

Kac went to Pasadena to lecture at the California Institute of Technology. Richard Feynman was in the audience. After the lecture, Feynman got up and announced: “If all mathematics disappeared, it would set physics back precisely one week.” Without a pause Kac responded: “Precisely the week in which God created the world” [Cohen 1986, 1147].

Illustrative of his kindness and lack of pretension, Kac told the story of examining a “not terribly good” doctoral candidate. Kac asked him to describe the behavior of the function \(\frac{1}{z}\) in the complex plane.

The student replied, “The function is analytic, sir, in the whole plane except at \(z = 0\), where it has a singularity,” he answered, and it was perfectly correct. “What is this singularity called?” I continued. The student stopped in his tracks. “Look at me,” I said. “What am I?” His face lit up. “A simple Pole, sir,” which is in fact the correct answer [Kac 1985, 126].

To this can be added an anecdote of Ulam’s:

I remembered him in Poland as very slim and slight, but here he became rather rotund. I asked him, a couple of years after his arrival [in America], how it had happened. With his characteristic good humor, he replied: “Prosperity!” [Ulam 1991, 269].

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Notes

[1] Most of the biographical details which follow come from Kac’s autobiography [Kac 1985]. Also helpful are [Cohen 1986], [McKean 1990], [Niss 2018], and [Stroock 2015].

[2] Consider only these few names (in order of birth): J. Łukasiewicz (1878–1956), W. Sierpiński (1882–1969), S. Banach (1882–1945), J. Schauder (1899–1943), A. Zygmund (1899–1992), A. Tarski (1901–1993), S. Ulam (1909–1984), and S. Eilenberg (1913–1998).

[3] Kac and Feynman overlapped for a few years at Cornell. Kac was working with the ideas of N. Wiener on Brownian motion and other stochastic processes when he attended a talk by Feynman on his doctoral work. He realized that their methods, involving a generalized form of integration, had much in common; for example, the diffusion equation describing Brownian motion and the Schrödinger equation of quantum mechanics turn into each other when the time variable \(t\) is mapped to \(\pm it\).

[4] A lecture series given to undergraduates at Haverford and Bryn Mawr in 1958 provides a charming and undergraduate-accessible introduction to Kac’s work; it was written up as an MAA Carus monograph [Kac 1959a]. An overview for professionals in the physical sciences is given in [Kac 1959b]. For a survey of mathematics aimed at a more general audience, see [Kac and Ulam 1968], written with his friend and fellow Polish-American, Stanislaw Ulam. This was originally commissioned to be an appendix in the Encyclopaedia Britannica [Ulam 1991, 268].

Kac related the story of how he became a mathematician in the prologue⁵ of his autobiography [Kac 1985]. In the spring of 1930, he was fifteen, a year from graduation, and had become obsessed with the cubic equation. His teachers dismissed his questions about the cubic as too difficult for a high school student. He decided that he would study it on his own and soon found in a textbook Cardano’s formula⁶ for a root of the cubic \(x^3 + px +q = 0\),

\[
x = \sqrt[3]{-\dfrac{q}{2} + \sqrt{\left(\dfrac{q}{2}\right)^{2} + \left(\dfrac{p}{3}\right)^{3}}} +  \sqrt[3]{-\dfrac{q}{2} - \sqrt{\left(\dfrac{q}{2}\right)^{2} + \left(\dfrac{p}{3}\right)^{3}}}
\hspace{.6in}(*)\]

 The book’s derivation began with, “Set \(x = u + v\).” The young student thought this “grossly unfair,” presupposing knowledge of the solution (that a root x would be obtained as the sum of two cube roots), and vowed to work it out for himself. He announced this lofty goal to his skeptical father, who promised to pay his son five Polish złotys (“in those days a lot of money”)⁷ should he succeed.

All summer he worked at the problem, “filling reams of paper with formulas before I collapsed into bed at night,” everything else put aside, even dating. At last the formulas were before him. He showed his pieced-together work to his father, who looked it over and promptly paid up. When school reopened, he presented “a neatly written manuscript” to his teacher, who submitted it on the young man’s behalf to a national journal for school mathematics, Młody Matematyk (The Young Mathematician). “That seemed to be the end of it,” writes Kac, “because the receipt was not acknowledged, and for months nothing was heard from distant Warsaw,” 487 km from Kac’s home in Krzemieniec.

In early May, a few weeks before his graduation, he was caught in the hall by the school’s principal. Fearing a bad outcome, he was surprised to be addressed with something like respect. The principal announced that “His Excellency, the Counselor of the Ministry of Education, Antoni Marian Rusiecki, who is visiting our institution, would like to see you at 2:30 this afternoon.” It took Mark Kac only a few seconds to recall that His Excellency was also the editor-in-chief of Młody Matematyk.⁸

Kac made himself presentable, in his Sabbath best, and met the visitor exactly at 2:30. Rusiecki told him that they had indeed received his paper and explained the lengthy delay in responding. They had been reluctant to accept his derivation, because they felt that it must have been found previously. But a careful search of the literature indicated that it was new, and so they were going to publish it. “And so they did. It appeared a few months after my graduation under the name of Katz because I thought that the German spelling was fancier than the Slavic Kac.”⁹ What happened next is so dramatic that only the author’s words will suffice.

Before my brief visit was over Mr. Rusiecki asked me my plans for the future. I told him that my family thought that I should study engineering. “No,” he said, “you must study mathematics; you clearly have a gift for it.” I followed this advice and it saved my life. In mathematics, as it turned out, I was good enough (and lucky enough) to win a post-doctoral fellowship to go abroad in 1938. The fellowship was endowed by the wealthy, thoroughly assimilated Polish-Jewish family Parnas, and by the terms of the endowment, one of the two yearly endowments had to be given to a Jewish applicant. I came to Johns Hopkins University in December of 1938 and the war caught me there. Had I gone into engineering, I would unquestionably have shared the fate of my family and six million others [Kac 1985, 4].

Work good enough to launch such a career and save a life should be known. But what was Mark Kac’s derivation? It was not given nor even sketched in the autobiography. A search for copies of this volume of Młody Matematyk in the United States initially proved fruitless (in part because it was listed under Parametr), nor did this publication merit either inclusion in a collection of Kac’s selected papers [Kac 1979] or mention in his bibliography [McKean 1990, 226–235]. Perhaps, I thought, it could be found.¹⁰

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Notes

[5] The prologue of [Kac 1985], “How I Became a Mathematician”, originally appeared as an essay in the journal of the Weizmann Institute of Science, Israel: Rehovot, vol. 9, no. 2 (1981/82).

[6] The story of Cardano's formula is told entertainingly in [Livio 2005, 63–73]. Gerolamo Cardano (or Jerome Cardan) (1501–1576), the first to publish it, was the third to obtain it, after substantial help from the notebook of Scipione del Ferro (1465–1526), and from conversations with Niccolò Fontana (1499–1557), known to history as Niccolò Tartaglia, “the stammerer.” (Fontana acquired his nickname as a result of the 1512 French invasion of his home town, Brescia, during which he was slashed across his jaw and throat by a saber; the wound left him with a lifelong speech impediment and a perpetual beard to hide the scar.) Del Ferro never made his methods public, but he had taught a method for solving cubics to his students Annibale della Nave (ca 1500–1558), who was married to del Ferro’s daughter Filippa, and Antonio Maria del Fiore (dates unknown, fl. 1520–1535). Fiore briefly made a career out of public challenges (standard procedure for Italian Renaissance mathematicians) to solve a class of cubics. Knowing that a method existed, Tartaglia accepted Fiore’s challenge (1535) and worked out more general rules shortly before the contest, winning it easily. Cardano asked the now-famous Tartaglia to show him the method, which Tartaglia did only under a promise of secrecy, but he withheld the derivation; in those times discoveries afforded competitive advantages in contests and so were concealed like buried treasure. A rumor reached Cardano that della Nave also knew the secret. Cardano traveled with his student Lodovico de Ferrari (1522–1565) to Bologna to meet with della Nave, who allowed the visitors to read his late father-in-law’s notebook, where del Ferro had recorded his solution many years before Tartaglia. Given del Ferro’s prior claim, Cardano felt no longer bound by his promise to Tartaglia and published the formula in his landmark book on algebra, Ars Magna (The Great Art). Though Cardano gave due credit to del Ferro and Tartaglia [Cardano 1545, 8, 96], he did not mention his promise to Tartaglia.

[7] Kac cited [1985, 31] a half-stipend for a college student of 60 złotys a month, which he called “about twelve dollars,” a rate in 1931 of five złotys to 1 US dollar. That suggests five 1931 złotys would be roughly equivalent to $15–$20 in 2021 dollars.

[8] The magazine Młody Matematyk was an offshoot of the magazine Parametr (in English, Parameter), published from 1930 to 1932 and then reappearing in 1939 for one year, its publication ceasing with the German invasion of Poland. Though Polish mathematics had attained world-class status, the level of mathematics taught in the schools was in need of improvement. A. Tarski (1901–1983), at the time a secondary school teacher as well as a university professor, had promoted (in 1929) the publication of a magazine devoted to the teaching of mathematics, and Parametr was the response. The magazine was founded and edited by Antoni M. Rusiecki (1892–1956), a mathematics instructor and government official in the Ministry of Religious Denominations and Public Education (Warsaw), and Stefan Straszewicz (1889–1983), the chair of that Ministry’s mathematics committee. Straszewicz earned a PhD (in 1914) under E. Zermelo (1871–1953) at the University of Zürich. Both men taught in underground schools during the German occupation, and after the war worked to develop the Polish Mathematical Olympiad. From the beginning Parametr had included material for secondary school teachers and students, but in the 1931/1932 volume, this material was taken out of Parametr and published as a new magazine, Młody Matematyk, distributed for free with Parametr. Rusiecki and Straszewicz not only served as editors but also contributed many articles to the two magazines. As part of his governmental duties, Rusiecki visited schools throughout Poland to provide teacher training, often for twenty days a month, and the extra work of editing two magazines was unsustainable. Besides Kac, among the contributors were H. Steinhaus, (1887–1972), W. Sierpiński (1882–1969) and Tarski. See [Dabkowska 2019, 69–130] and [McFarland, McFarland, and Smith 2014, 204–213].

[9] The Polish pronunciation of “Marek Kac” and the German pronunciation of “Mark Katz” are nearly identical.

[10] As far as can be determined, the only physical copies of Młody Matematyk outside of Poland are in Brown University’s John Hay Library, Providence, RI, bound with the Library’s copies of Parametr.

A clue to the whereabouts of the article appears in a book by Daniel Kalman [2009, 79–82], which presents Kac’s solution. Clearly, the author had seen the original article, or something very close to it. In email he told me that his discussion had been based on an unpublished paper [Roy n.d.] by Ranjan Roy of Beloit College in Wisconsin. Correspondence with Prof. Roy confirmed that he had once had a copy of the Kac article, but that he could not find either it or even his own paper cited by Prof. Kalman! He added that he had obtained a scan of the original article from Prof. Joel E. Cohen of Rockefeller University, one of the world’s experts on population biology and a close friend of Kac’s. The reminiscences that became Kac’s autobiography [Kac 1985] were largely drawn from stories he told Prof. Cohen, tape recorder in hand, as they walked together through the halls at Rockefeller.

Prof. Cohen very kindly responded to email, though the pandemic had largely shut down normal life and he was not in New York City. A week or two later, he braved the virus to return to his office, and sent a scan, drawing attention to two small penciled remarks to ensure they were not overlooked. On the magazine’s title page, Kac had written, “See how bright I was when I was 15–16! M K”. And at the start of his article, there is an addendum, “\(m\neq n\)!”.

    

Figure 2. The pages of Kac’s article, from the scan by Cohen.

What follows is my translation of Mark Kac’s article¹¹ from Polish to English. Though ignorant of Polish, I have some Russian; that, with the help of Google Translate, was enough for a reasonable rendering.¹² Note that there is one alteration from the original. A third handwritten notation in Kac’s copy crosses out “(4)” cited in the last sentence of Kac’s work, before the Editor’s Note, and substitutes “(6)”. This substitution is incorporated in the English translation and in the transcribed Polish version of Kac's original article, "O nowym sposobie rozwiązywania równań stopnia trzeciego'' (pdf).

 

MAREK KATZ
(Krzemieniec)
On a new way of solving equations of the third degree

To solve the equation \(z^{3} + az^{2} + bz + c = 0\), we substitute

\[z = x - \frac{1}{3}a\]

and we obtain an equation without the unknown in the second power:

\[\hspace{3in}x^{3} + px + q = 0, \hspace{2.1in}(1)\]

where \(p\) and \(q\) are terms dependent on \(a\), \(b\), and \(c\). Here I want to give—as it seems to me—a new way of solving equations of type (1).

My way is to find certain numbers \(A\), \(B\), \(m\), \(n\), such that for every value of \(x\) this equality holds:

\[\hspace{2in}x^{3} + px + q = A(x + m)^{3} - B(x + n)^{3}. \hspace{1.45in}(2)\]

Then we get the equation in the form

\[\hspace{2.5in}A(x + m)^{3} - B(x + n)^{3} = 0 \hspace{1.8in}(3)\]

and it will be easy to solve this; for we observe that the left side of the equation breaks down into factors:

\(A(x + m)^{3} - B(x + n)^{3} \)
\(\hspace{.1in}= \left[\sqrt[3]{\!A}(x + m) - \sqrt[3]{B}(x + n)\right]\cdot\left[\sqrt[3]{\!A^{2}}(x + m)^{2} + \sqrt[3]{\!AB}(x + m)(x + n) + \sqrt[3]{B^{2}}(x + n)\right]\!.\)

Setting this product equal to zero, we have two possibilities:

\[\hspace{1.8in}\begin{array}{c}\hspace{.8in}\sqrt[3]{A}(x + m) - \sqrt[3]{B}(x + n) = 0, \hspace{1.4in}(4)\\ \sqrt[3]{\!A^{2}}(x + m)^{2} + \sqrt[3]{\!AB}(x + m)(x + n) + \sqrt[3]{B^{2}}(x + n) = 0.\hspace{.5in}(5)\end{array}\]

As we can see from equation (4), it will be easy to find one of the roots of the given equation (1).

So let's find the numbers \(A\), \(B\), \(m\), and \(n\). Expanding the right side of the equality (2), we obtain:

\[x^{3} + px + q = (A - B)x^{3} + 3(Am - Bn)x^{2} + 3(Am^{2} - Bn^{2})x + (Am^{3} - Bn^{3}).\]

If two polynomials of one variable have equal values for every value of this variable, then the coefficients of correspondingly equal powers of the variable must be equal:

\[\begin{array}{c}A - B = 1,\\ Am - Bn = 0,\\ Am^{2} - Bn^{2} = \frac{1}{3}p,\\ Am^{3} - Bn^{3} = q.\end{array}\]

 We are to solve the system of 4 equations in 4 unknowns. From the first equation we find \(A = B + 1\) and substitute into the remaining equations:

\[\begin{array}{c}B(m - n) + m = 0,\\ B(m^{2} - n^{2}) + m^{2} = \frac{1}{3}p,\\ B(m^{3} - n^{3}) + m^{3} = q.\end{array}\]

Now we find from the group \(B(m-n) = -m\) and substitute into the two remaining equations:

\[\begin{array}{c}-m(m + n) + m^{2} = \frac{1}{3}p,\\ -m(m^{2} + mn + n^{2}) + m^{3} = q.\end{array}\]

After the simplifications we get:

\[\begin{array}{c}mn = -\frac{1}{3}p,\\ m + n = \frac{3q}{p},\end{array}\]

we assume that \(p \neq 0\).  (The case when \(p = 0\) gives the equation \(x^{3} + q = 0\), whose solution we will not find difficult.)

We find the values of \(m\) and \(n\) as the roots of the quadratic equation

\[\hspace{2.5in}u^{2} - \frac{3q}{p}u - \frac{1}{3}p = 0. \hspace{2.2in}(6)\]

If it turns out that this equation has its discriminant positive, then it has two roots which are not equal. By assuming one of these values for \(m\), and the other for \(n\), we can easily calculate the values of \(A\) and \(B\), namely

\[B = \frac{-m}{m-n}, \quad A = \frac{- n}{m-n}.\]

Substituting these values into (4) and multiplying both sides of this equation by \(\sqrt[3]{m-n}\), we obtain the equation

\[- \sqrt[3]{n}(x + m) + \sqrt[3]{m}(x + n) = 0.\]

Hence, after rearranging, we have the following equation:

\[\left(\!\sqrt[3]{m} - \sqrt[3]{n}\,\right)x = m\sqrt[3]{n} - n\sqrt[3]{m}.\]

By transforming the right side, we will have

\[\left(\!\sqrt[3]{m} - \sqrt[3]{n}\,\right)x = \sqrt[3]{mn}\left(\!\sqrt[3]{m} - \sqrt[3]{n}\,\right)\left(\!\sqrt[3]{m} + \sqrt[3]{n}\,\right),\]

so finally we will have

\[\hspace{2.1in}x = \sqrt[3]{mn}\left(\!\sqrt[3]{m} + \sqrt[3]{n}\,\right). \hspace{2.2in}(7)\]

Substituting the values of \(m\) and \(n\), determined from equation (6), we will obtain the formula known as Cardano's formula.

 

Młody Matematyk Editor's Note.

Granting room to the method presented by Mr. Marek Katz, who is a student of the 8th grade of the Krzemieniec Lyceum preparatory school, the Editorial Board must add a few comments.

The author of the article limited himself to considering the case when the discriminant of  equation (6) is positive, i.e., the case when the expression

\[\hspace{2.2in}\left(\frac{1}{2}q\right)^{2} + \left(\frac{1}{3}p\right)^{3} \hspace{2.3in}(8)\]

has a positive value.

From the theory of the third degree equation, it is known that in this case equation (1) has one real root. The method described in the article relates to finding this root.

In the case when expression (8) is zero, the discriminant of equation (6) is zero, so this equation has a double root:

\[m = n = \frac{3q}{2p}.\]

But in this case, formula (7) takes the form \(x = \frac{3q}{p}\), and it is not difficult to verify by direct substitution that this formula gives the root of equation (1).

In the event that expression (8) has a negative value, \(m\) and \(n\) are not real numbers. The case here is known as the casus irreducibilis—peculiar in that in this case the third degree equation has three real roots, but they cannot be determined via an algebraic path. Clearly, the particular method described by Mr. M. Katz, fails in this case. Indeed, the equation

\[(x - 1)(x - 2)(x + 3) = 0\]

has three roots: \(+1,\) \(+2\), \(-3\). Expanding the left side, we get the equation

\[x^{3} - 7x + 6 = 0,\]

so we have these values: \(p = -7\), \(q = 6\). Equation (6) takes the form¹³

\[u^{2} + \frac{18}{7}u + \frac{7}{3} = 0.\]

It is easy to check that this equation does not have real roots, so the real values of \(m\) and \(n\) cannot be determined from it; but it does not follow that the given third-degree equation has no roots.

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Notes

[11] Originally published as “O nowym sposobie rozwiązywania równań stopnia trzeciego,” Młody Matematyk, Rok 1, Kwiecień-Maj 1931, Nr. 4–5, pp. 69–71 ("On a new way of solving equations of the third degree," Young Mathematician 1(4–5) (April–May 1931): 69–71.)

[12] The translation was checked by Zbigniew Kantorosinski, a scholar at the Library of Congress and a native speaker of Polish. He had not only the translation, but my retyped Polish of the original. Editors’ Note: During the blind refereeing process for this article, the author’s Polish-to-English translation was also vetted by a mathematician fluent in both English and Polish, based on the published version of Kac’s article in Młody Matematyk.

[13] In the original article, the fraction \(\frac{18}{7}\) was erroneously written as \(\frac{6}{7}\).

 

The comment by Rusiecki, though correct, seems off-target. In the casus irreducibilis, or “irreducible case”, the three real roots are usually given in terms of a trigonometric function (see below); a transcendental expression, not algebraic. The “failure” however lies not with Kac’s derivation, but with Cardano’s formula. Kac set out to find a derivation of Cardano’s formula that did not assume the form of the answer, and he succeeded. Even in the “irreducible case”, Cardano’s formula provides a correct expression for a real root; the purported “failure” is that this is not explicitly done in terms of real numbers alone—roots of complex numbers arise. The argument is made (particularly by Tignol [2001, 16–20] and Nahin [1998, 25]) that, contrary to what legions of high school students are taught, imaginary numbers found a home in algebra not because of quadratics, but because of cubics. But this invites a longer discussion.

In Cardano’s time, not merely the square roots of negative numbers, but negative numbers themselves were viewed as “fictitious”, or, if encountered as the root of a quadratic, “impossible” [Kline 1972, 252–253], not least because most algebraic questions were couched in geometric terms: a geometric problem requiring a negative answer violated common sense. When at last Cardano derived the del Ferro-Tartaglia expressions, he did so via a geometric argument [Cardano 1545, 96–98].¹⁴ But when the root of a cubic known to have a real root was expressed by Cardano’s formula in terms of the square roots of negative numbers, there had to be a reckoning. Indeed, Rafael Bombelli (1526–1572)—in the case of the cubic¹⁵—and Cardano himself—in the case of a quadratic—were able to show through manipulation of complex quantities that a sum or a product of complex quantities could in fact equal a real number:

If it should be said, Divide \(10\) into two parts, the product of which is \(30\) or \(40\), it is clear that this case is impossible. Nevertheless, we will work thus: We divide \(10\) into two equal parts, making each \(5\). These we square, making \(25\). Subtract \(40\), if you will, from the \(25\) thus produced,  . . . leaving a remainder of \(-15\), the square root of which added to or subtracted from \(5\) gives parts the product of which is \(40\). These will be \(5 + \sqrt{-15}\) and \(5 - \sqrt{-15}\). . . . Putting aside the mental tortures involved,¹⁶ multiply \(5 + \sqrt{-15}\) by \(5 - \sqrt{-15}\), making \(25 - (-15)\), which is \(+40\). Hence this product is \(40\). . . . So progresses arithmetic subtlety, the end of which, as is said, is as refined as it is useless [Cardano 1545, 219–220].

It is likely that Rusiecki would have described the extraction of complex roots of a quadratic equation as algebraic. His disqualification of Cardano’s formula as algebraic can only be due to its inclusion, in the irreducible case, of a radical of a number not in the base field \(R\) of the equation’s coefficients. Well, is there perhaps some method of writing, in general, the complex expressions occurring in the casus irreducibilis as \(a+i\sqrt{b}\), where a and b are real? That would apparently render Cardano’s formula “algebraic”, in accord with the quadratic formula.

Whether the roots of a cubic or any polynomial with real coefficients can be expressed in terms of real roots only (perhaps multiplied by \(i\)) has been answered. One of the first investigations is due to Mollame [1890], who found that, given the casus irreducibilis, if the coefficients of the polynomial satisfied the relation (10) in his article,

\[\left(\frac{q}{2}\right)^{2} = -\frac{1}{2}\left(\frac{p}{3}\right)^{3},\]

then a real root was expressed simply as

\[x = \sqrt[3]{2q}.\]

For the general irreducible case, however, Mollame was skeptical that real radicals alone could be used.

Van der Waerden [1970, 189–190] stated unequivocally that “it is actually impossible to solve the equation [in the irreducible case] by real radicals unless the equation is reducible in the base field K” (in which the coefficients lie), and provided the proof. More recently, Isaacs [1985] has shown that, given an irreducible polynomial with rational real coefficients, and which has only real roots, then if it has any root expressible in terms of real radicals, the degree of the polynomial must be a power of 2. That clearly rules out the cubic. In short, what Rusiecki found lacking in the Cardano formula is simply not to be found at all, at least in terms of real roots.

But, as Rusiecki implied, the expression of a cubic’s three real roots in the casus irreducibilis as real expressions can be accomplished, by means of a trigonometric function. This was accomplished by François Viète (1540–1603), in a posthumous publication [Viète 1615, 174–175]. In modern terms, the argument is this [Nahin 1998, 22–23]. From (8)’s being negative and the reality of \(p\), \(q\), it follows that \(p < 0\). Then we can say

\[p = -3a^{2}, \quad q = - a^{2}b,\]

where we can assume \(a > 0\) because \(a\) appears only as \(a^{2}\), and the original cubic becomes

\[y^{3} - 3a^{2}y - a^{2}b = 0. \hspace{.2in}(†)\]

The irreducible case, in which expression (8) must be negative, leads to

\[\frac{|b|}{2a} < 1\]

so that we can set, without loss of generality,

\[\frac{b}{2a} = \cos \theta\]

for some angle \(\theta\), \(0 < \theta < \pi\). Viète, a master of trigonometry,¹⁷ knew the identity

\[\cos 3\phi = 4\cos^{3}\phi - 3\cos \phi. \hspace{.2in} (‡)\]

This follows easily from the addition and double angle formulas, expanding \(\cos 3\phi\) as \(\cos (2\phi + \phi)\). Multiplying both sides of (‡) by \(2a^{3}\) gives

\[(2a\cos \phi)^{3} - 3a^{2}(2a\cos \phi) - a^{2}(2a\cos 3\phi) = 0,\]

which is identical to (†), with \(y = 2a\cos \phi\), provided the angle \(3\phi\) is such that

\[\cos 3\phi = \frac{b}{2a} = \sqrt{\frac{(\tfrac{1}{2}q)^{2}}{(-\frac{1}{3}p)^{3}}}.\]

Thus, we obtain the three roots of the cubic (†) as

\[y_{k} = 2a\cos \phi_{k}, \quad \phi_{k} = \tfrac{1}{3}\cos^{-1}(b/2a) + \tfrac{2}{3}\pi(k-1),\; k = \{1, 2, 3\}.\]

Incidentally, the restriction “\(m\neq n!\)” penciled in by Kac should be taken not as a requirement, but simply as a warning to the reader that these are not necessarily equal. They can be, as Rusiecki pointed out, but in that case, there is a double root.

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Notes

[14] Very clear discussions of the geometric argument cast into algebraic language are given by Struik [1969, 64–65, see notes 3 and 5 in particular] and Bewersdorff [2006, 5–6]. According to Kline [1972, 253], the first to give both positive and negative roots of a quadratic was Albert Girard (1595–1632) in his L’Invention nouvelle en l’algèbre (1629).

[15] See [Fauvel and Gray 1987, 265].

[16] The Latin original is dismissis incruciationibus, literally “the incruciations having been dismissed”. Tignol [2001, 20] suggests Cardano may have been punning on the word crux (cross), i.e., not only “ignoring the (ex)cruciating pain,” but “dismissing the cross products” involved in the multiplication of the complex binomials.

[17] In 1593, Adriaen van Roomen (1561–1615), a Belgian visitor to the court of Henry IV, challenged the mathematicians of France to solve a 45th degree equation. Viète did so at once, recognizing a term from the expansion of sin(45θ). Viète, a wealthy man, hosted van Roomen at his home for a month, and the two men became friends [Boyer 1985, 341].

Even world-class mathematicians start from modest beginnings, and were once high school students themselves, perhaps even among those made uncomfortable when approached by the school’s principal. Here is a calculation well within the grasp of most algebra students, and yet it was good enough to spark a major career. It is hoped that some young people will find this an inspiring story and Mark Kac an appealing person. Maybe there are other old problems out there for which they can find a new solution all their own, or even a first solution. Mathematics is a deep subject with a rich history. It invites exploration.

The cubic itself is a fascinating topic. An argument can be made for its being the springboard for the whole of today’s algebra. Its historical and cultural importance is well known. For example, Feynman wrote to his family while touring Greece that he was disheartened by the locals’ excessive respect for the great accomplishments of the past:

They were very upset when I said that the thing of greatest importance to mathematics in Europe was the discovery by Tartaglia that you can solve a cubic equation—which, altho [sic] it is very little used, must have been psychologically wonderful because it showed that a modern man could do something no ancient Greek could do, and therefore helped in the renaissance which was the freeing of man from the intimidation of the ancients—what they are learning in school is to be intimidated into thinking they have fallen so far below their super ancestors [Feynman 2006, 327].

Considering how simply the Cardano formula can be derived, as a consequence of the quadratic—as shown by Viète—it’s surprising that it is not often taught in the schools. Here’s how that might be done.

First, it should be pointed out that the general cubic equation,

\[ Ax^{3} + Bx^{2} + Cx + D = 0 \]

can be reduced to the monic

\[x^{3} + bx^{2} + cx + d = 0\]

simply by dividing by \(A\). Then, use the Tschirnhaus¹⁸ transformation [Tignol 2001, 68]: let

\[x = y - \frac{1}{3}b\]

and, substituting this expression for \(x\) into the monic polynomial, rewrite the equation in terms of \(y\). This removes the quadratic term (the same trick can be used for any polynomial, e.g., let \(x=y-\frac{1}{2}b\) for a quadratic). One returns to the standard form (∗) considered by Kac,

\[y^3 + py + q = 0,\]

 where \(p = c - \frac{1}{3}b\), \(q = d - \frac{1}{3}bc + \frac{2}{27}b^{3}\). Thus any cubic can be turned into the standard form for which the Cardano formula holds. Now comes the master stroke given in [Viète 1615, 287]: let

\[y = u - \frac{p}{3u}\]

Plugging this into the previous equation results in the wonderfully simple

\[u^3 + q - \frac{p^3}{27u^3} = 0\]

Multiply both sides by \(u^3\), and obtain a quadratic in \(u^3\):

\[u^6 + qu^3 – \frac{p^3}{27} = 0\]

for which the roots are given by the familiar quadratic formula,

\[u^{3} = \frac{-q \pm \sqrt{q^{2} + \frac{4p^{3}}{27}}}{2} = -\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3}\right)^{3}}\]

Let \(u_1\) correspond to the real cube root of the quantity with the plus sign, and \(u_2\) to that of the minus. That means

\[u^{6} + qu^{3} - \frac{p^{3}}{27} = (u^{3} - u_{1}^{3})(u^{3} - u_{2}^{3})\]

so that \(u_{1}^{3}u_{2}^{3} = -(p/3)^{3}\), or \(u_{2} = - (p/3u_{1})\). Finally a root of the cubic is given by

\[y = u_{1} - \frac{p}{3u_{1}} = u_{1} + u_{2} = \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3}\right)^{3}}} +  \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^{2} +\left(\frac{p}{3}\right)^{3}}},\]

exactly the Cardano formula.¹⁹

Once a student has learned to solve the cubic, Kac’s method should be presented. Many students are presented with only one way to do something; it’s worthwhile to show two or three ways to reach a result. For students in calculus, a cute exercise is to show that a function of the form (∗) will have three distinct real roots only if the product of its local maximum and its local minimum is negative; only then will the graph cross the axis three times. And of course this is just the condition that the expression (8) is negative.

It’s true that cubic equations do not arise in the sciences as often as quadratics, but they do crop up occasionally, for example in chemistry problems involving a particular type of equilibrium reaction, or in classical mechanical problems involving rotations and eigenvectors of 3 × 3 matrices. Even so, the cubic should be taught, if only to highlight the creativity of great mathematicians who recast these equations into forms more easily solved. Beyond the inventive algebra, the dramatic human story behind the cubic has few peers in mathematical history for intrigue and conflict.

Finally, should a student become interested in the theory of equations generally, she might begin by retracing Ferrari’s path to the quartic, and perhaps even look at Abel’s proof (1824) that the general fifth-degree equation cannot be solved with radicals. Resources suitable for talented and motivated high school students include [Livio 2005], [Pesic 2003], and [Alekseev 2004]. That might in turn lead her to Galois theory and its creator, Évariste Galois (1811–1832), whose tragic biography is perhaps the most romantic story in mathematics. In addition to [Livio 2005], [van der Waerden 1970], [Bewersdorff 2006], and [Tignol 2001], works on Galois theory that can be recommended highly to beginners include [Stewart 2015] and [Artin 2007]. And to come full circle, a beautiful introduction to the ideas of Galois theory is given by Kac and Ulam in [Kac and Ulam 1968, 56–60].

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Notes

[18] Ehrenfried W. von Tschirnhaus (1651–1708), mathematician and possible inventor of European porcelain.

[19] It should be noted that the cubic has two other roots, but they're easily found. There are three cube roots of \(u_{1}^{3}\): \(u_{1}\), \(\omega u_{1}\), and \(\omega^{2}u_{1}\), where \(\omega = \exp(\frac{2}{3}\pi i)\) is a primitive cube root of 1, and \(\frac{1}{\omega }= \omega^{2}\). Keeping the conditions \(u_{1}u_{2} = - p/3\) and \(y = u - \frac{p}{3u}\), but replacing \(u_{1}\) first by \(\omega u_{1}\) and then by \(\omega^{2}u_{1}\) gives \(y_{2} = \omega^{2}u_{1} + \omega u_{2}\) and  \(y_{3} = \omega u_{1} + \omega^{2}u_{2}\) respectively. Simple!

In our correspondence, I asked Prof. Roy why he had not published his paper on Kac’s derivation of Cardano’s formula. He replied that he had subsequently decided that Kac had independently rediscovered a derivation given nearly a century earlier by George Boole (1815–1864) in [Boole 1842], and he suggested that I consult his book [Roy 2011, 732–733]. Boole’s derivation of the cubic is also discussed by Wolfson [2008]. I have not had a chance to consult Prof. Roy’s book at length, but thanks to his friend and Beloit colleague Prof. Mehmet Dik, I was able to read a scan of the relevant pages; therein Prof. Roy suggested that it was fortunate that Rusiecki did not know Boole’s work. In my opinion the methods, though similar, are not the same, and Boole’s derivation in any case should not have prevented the publication of Kac’s work.

Interestingly, as a postscript to his prologue, Kac described being at a talk given by Gian-Carlo Rota (1932–1999) which involved a theorem of J. J. Sylvester (1814–1897). Sylvester’s work followed Boole’s in invariant theory. Kac concluded his postscript:

Almost en passant [Rota] said, “I’ll now show you how one can use Sylvester’s theorem to solve cubic equations.” After only a few words a feeling of dèjà vu came over me; it was the method I had discovered in the summer of 1930 [Kac 1985, 5].

Rota’s derivation, like Boole’s, was based on invariant theory. Mark Kac did not need any of that.

Alekseev, V. B. 2004. Abel’s Theorem in Problems and Solutions, based on the lectures of Professor V. I. Arnold. Translated by Francesca Aicardi. Dordrecht: Kluwer Academic Publishers (Springer).

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Acknowledgements

It’s a pleasure to thank Prof. Joel E. Cohen for sending me a scan of the original article, and for his encouragement to proceed with this project. I am happy to thank Prof. Dan Kalman for his many excellent suggestions to earlier versions of this paper, much to its betterment, and for not only providing the crucial clue as to where the original could be found, but also for sending a copy of Ranjan Roy’s paper to me. (Thanks to Prof. Kalman, I was able to provide Prof. Roy with a replacement copy of his own paper.) This paper could not have been written without their help. I am very grateful to Zbigniew Kantorosinski, who graciously checked the translation. I want to thank my friend Barbara Dash, of the Library of Congress, for introducing me to her colleague (and incidentally, four decades ago, also to my wife). I would also like to thank the referees for their valuable suggestions, and in particular one who suggested I look at [Mollame 1890] for an early attempt to handle the casus irreducibilis in real roots, and at [McFarland, McFarland, and Smith 2014] for background information on Parametr, Młody Matematyk, and their editors Antoni Marian Rusiecki and Stefan Straszewicz. While I was able to find the Mollame article online, the pandemic prevented my obtaining the book from the University of Chicago’s library. The referee kindly forwarded to me an electronic copy of the relevant chapter of the book.

I very much regret that I cannot here thank Prof. Roy, who died suddenly in August of 2020 while walking with his wife. But I did thank him many times during this project, not least for helping me contact Prof. Cohen.

I offer this article in memory of the tens of thousands in Mark Kac’s city of Krzemieniec, including his immediate family, killed in the Holocaust, and in memory of Ranjan Roy.

About the Author

David Derbes is a retired high school physics teacher. He earned an undergraduate degree in physics at Princeton in 1974, an ex post facto master’s degree from Part III of the Mathematical Tripos, Cambridge University, and a PhD in theoretical physics at the University of Edinburgh, Scotland, in 1979; he is Peter Higgs’s only American student. He taught mathematics and physics at the school from which he graduated, Isidore Newman in New Orleans, from 1979 to 1984, engineering physics at Tulane University between 1984 and 1986, and mostly physics at the University of Chicago’s Laboratory Schools from 1986 to 2019. His retirement occupation is manuscript salvage: translation and/or editing of important, never-published, or out of print lecture notes and articles, most recently as part of a team working on Sidney Coleman’s lectures on quantum field theory. He is married (fortieth anniversary in July 2021) and has one adult daughter in government service, and two very mischievous cats. He can be reached at derbes.physics@gmail.com.