# Impossible Equations: Algebraic Equations in the Renaissance

### Mathematics Department, King's College London

The present account of the history of cubic and quartic equations comes from the following sources.

## References

[1]
Katz, V. J; A History of Mathematics-An Introduction, Addison-Wesley, 1998.
[2]
Calinger, C., A Contextual History of Mathematics, Prentice-Hall, 1999.

## 1  Completing the Square

The Greeks knew how to solve some quadratic equations by completing the square, for example
 x2+10x = 39.
(1)

The left-hand side of the equation is the shaded area of the diagram. The missing square has area 25, so

 (x+5)2 = x2+10x+25 = 39+25 = 64.
(2)
Hence
 x+5 = ±8,
(3)
so x=3 or -13. Negative numbers or even zero were not considered to be solutions, as numbers were lengths, so: x=3. In modern notation, x=unknown, and we solve the general equation
 x2+bx+c = 0
(4)
by writing
 x2+bx = (x+b/2)2-b4/4
(5)
so if (4) is true, then
 (x+b/2)2
 -
 b2/4+c = 0 so
 (x+b/2)2
 =
 b2/4-c so
 x+b/2
 =
 ±Ö(b2/4-c), so

 x = -b±Ö(b2-4c) 2 .
(6)
This is the modern formula with a = 1, which is no loss of generality. Naturally, if b2 < 4c we would say that there are complex solutions
 x = -b±(Ö(4c-b2)i 2
(7)
but to the Greeks there were no such numbers.

In 1494 Luca Pacioli concluded that (in his opinion) the general cubic (quartic, quintic and higher) could not be solved by a formula. The historical development of the study of these equations is as follows:

Cubic.
solved by Scipione in 1515:
 x3+bx2+cx+d = 0
(8)
Quartic.
solved by Ferrari in 1545:
 x4+bx3+cx2+d = 0
(9)
Quintic.
showed to be unsolvable by Abel in 1824:
 x5+bx4+cx3+dx2+ex+f = 0
(10)

Giralamo Cardano (1501-1576) invented, or at least popularised, the method of ``linear substitution''. Write any equation as an equivalent equation for the unknown y, where y = x+p, with p a parameter to be chosen conveniently. Then x = y-p and the quadratic equation becomes

 (y-p)2+b(y-p)+c = 0,
(11)
which can be expanded to give a quadratic equation for y:
 y2+(b-2p)y+p2-bp+c = 0.
(12)
Seems more complicated than the original equation, but we can choose p = b/2, so second term is zero, and we see that y2 =    a known quantity. When this method is applied to the cubic equation, we can choose p = b/3 to remove the coefficient of y2, or by solving a quadratic equation for p, we can remove the coefficient of y; but unless we are lucky, we cannot remove both the y2 and the y terms, to ``complete the cube''. This happens only when a special relation exists between b and c.

## 2  Solving by factorisation

If we can write
 x2+bx+c = (x-a)(x-b)
(13)
then eq. (4) is true if x = a or if x = b. Conversely, if eq. (4) is true, then either x-a = 0 or x-b = 0. The right-hand side of eq. (13) is x2-(a+b)x+ab, and is true for all x, so we can identify
 a+b = -b; ab = c.
(14)
``When the coefficient of x2 is 1, then the sum of the roots of a quadratic is -b and their product is c''. This helps us solve the quadratic for the second solution, if we are given the first.

Exercise. Show that if ac ¹ 0, then the two solutions to ax2+bx+c = 0 can be written

 x = 2c -b±Ö(b2-4ac) .
(15)
Similar things occur for cubics, quartics etc. Write
 x3+bx2+cx+d = (x-a)(x-b)(x-g)
 =
 x3-(a+b+g)x2+(ab+bg+ga)x-abg.
Hence
 b
 =
 -sum of roots
(16)
 c
 =
 sum of products of roots, two at a time
(17)
 d
 =
 -product of the three roots.
(18)
Similarly, if a,b,g,d are the roots of a quartic,
 x4+bx3+cx2+dx+e = (x-a)(x-b)(x-g)(x-d)
 =
 x4-(a+b+g+d)x3
 +
 -
 (abg+bgd+gda+dab)x+abgd.
Therefore, for the quartic with unit coefficient of x4:
 b
 =
 - sum of roots
(19)
 c
 =
 sum of products of roots two at a time
(20)
 d
 =
 - sum of products of roots 3 at a time
(21)
 e
 =
 product of the four roots.
(22)
These relations were known for positive roots to Viète 1591, and for complex roots to Albert Girard (1629).

## 3  Occurence of cubics etc in the theory of finance

From about 1344, Dardi could solve special cubics arising in finance.
Question At what % compound interest will an investor gain half his capital after 3 years?

Answer. Interest is compound, so we get the equation

 æç è 1+ x 100 ö÷ ø 3 = 1+ 1 2 .
(23)
Hence we find y: = 1+x/100
 y: = 1+x/100 = (3/2)1/3
(24)
and hence x. When we expand eq. (23) we get a cubic equation, but of the special form that can be solved by ``completing the cube''. The general cubic was not really needed for finance, but became a challenge after 1494 and Pacioli's remark. By posing problems as challenges, the scolars competed for students and the few paid positions.

## 4  The solution of the cubic

Around 1500, cubics were classified into 13 different types, depending on whether a coefficient was positive or not, rather than on whether there were real or complex solutions. The cubic of the form
 x3+cx = d
(25)
with any positive c,d, was solved within 25 years of Pacioli's remarks by Scipione dal Ferro, probably before 1515, in Bologna. He kept it secret, except from his student Antonio Fiore, who challenged Niccolo Tartaglia (1499-1557) of Venice to a contest, because Tartaglia was claiming he could solve x3+bx2 = d, with b and d as usual, positive. During the contest, Tartaglia worked out how to solve Fiore's problems, of the form (25), and won the challenge, as Fiore could not solve equations with a quadratic term.

At school, I read a book by Sir J. H. Jeans on astronomy, written in about 1920. In it, Sir James remarks in a three-line footnote that the solution to the cubic was hardly a great discovery; one just needs to write x = p+q, where p and q are unknowns, and compare the identity

 (p+q)3-3pq(p+q)-(p3+q3) = 0
(26)
with the equation x3+cx+d = 0 to get
 -(3pq)3
 =
 c3
(27)
 p3+q3
 =
 -d.
(28)
Then, knowing p3+q3 and p3q3 (sum and product of unknowns), we see that p3 and q3 are the solutions to a known quadratic equation, and hence can be solved. This is Viète's later explanation. It is probable that Scipione hit on a version of this identity.

## 5  Arab and Medieval notation

Jeans's comments are unfair. Notation in algebra used words and some shorthand, and was very cumbersome.

+ was written piu or p
- was written meno or m
x was written cosa, or co, meaning ``thing''
x2 was written censo or ce
x3 was written co ce
x4 was written cece, or æ
= was written - -
Ö was written radix or R
(...)1/3 was written cub R
and a typical equation was
co ce p 3ce p 3 co - -7

Notation changed slowly. Bombelli (1526-1572) introduced powers:

x3 was written

 3
 È
 1

2x2 was written

 2
 È
 2

3x was written

 1
 È
 3

Bombelli introduced brackets, but continued to use Rad, cub Rad, while Ö had been introduced in Germany in 1525 by Rudolff, with ÖÖ as cube-root and ÖÖÖ as fourth root (only the first survived). Also in Germany, Stifel used letters A,B,C,D,F and also x as unknowns, and in France, Viète writes A,E,I,O,U,Y as unknowns, and B,C,D,...as given numbers. He also introduced

 A B
for division of A by B. Even 50 years after Pacioli, it must have been hard to manipulate equations. Descartes later introduced a,b,c¼ for knowns and x,y,z for unknowns.

## 6  Tartaglia's quarrel with Cardano

Tartaglia had won the contest with Scipioni's student Fiore, but had kept the method secret. Cardano persuaded Tartaglia to tell him, on oath not to put it in his, Cardano's, forthcoming book on arithmetic. He kept his oath, but then Cardano and his student Ferrari (1522-1565) visited Bologna where Scipione's son-in-law, Annibale della Nave, was in charge of Scipione's papers, which they were allowed to see. Cardano confirmed that Scipione had the solution 20 years before Tartaglia, and that Scipione's special case could be used to solve Tartaglia's cases, by linear substitution. So he felt under no further obligation to keep the secret, and he published the solutions in Ars Magna (The Great Work) in 1545, which included a solution to the quartic, by then solved by Ferrari. Tartaglia was furious, though Cardano had given him credit for independently solving the cubic) and challenged Ferrari to a contest, of which Ferrari was deemed the winner.

To this day, the formula for the cubic is known as Cardano's formula. He never claimed that he found it himself, but he wrote a good book about it.

## 7  Cardano's Formula for the Cubic

Suppose
 x3+Cx+D = 0.
Write x = p+q. Then
 p3q3 = -C3/27; p3+q3 = -D.
So, p3,q3 are the two solutions to
 y2+Dy-C3/27 = 0, so
 y = {-D±Ö(D2+4C3/27)}/2
One solution is p3, the other is q3. Therefore
 x = ìí î -D+Ö(D2+4C3/27) 2 üý þ 1/3 + ìí î -D-Ö(D2+4C3/27) 2 üý þ 1/3 .
(29)

## 8  Ferrari's solution to the quartic

First reduce the equation by removing the x3 term by linear substitution: put y=x+b/4 in eq. (9). Then rearrange, to get
 x4 = -(cx2+dx+e).
(30)
Now add 2px2+p2 to each side, where p is a parameter to be conveniently chosen later. We get
 x4+2px2+p2 = (2p-c)x2-dx+(p2-e)
(31)
The left-hand side is a perfect square, whatever we choose for p: it is (x2+p)2. Choose p so that the right-hand side is a quadratic expression with two equal roots, say a. This means ``b2 = 4ac'', that is
 d2 = 4(2p-c)(p2-e).
This is a cubic in p with known coefficients, so can be solved by Cardano's formula. Also we then know a = d/(2(2p-c)), from the quadratic with equal roots. Then we may take the square root of each side of eq. (31), to give
 x2+p = ±Ö(2p-c)(x-a);
a quadratic equation to find x.

## 9  My method of solution

Following Jeans's note, I tried for a year to solve quartics, and succeeded in November 1953. We write x = p+q+r and compare the identity below with the equation:
 x4
 -
 2(p2+q2+r2)x2-8pqr x+(p2+q2+r2)2-4(p2q2+p2r2+q2r2) = 0
 x4
 +
 cx2+dx+e = 0.
This gives the symmetric polynomials of p2,q2,r2:
 p2+q2+r2
 =
 -c/2
 p2q2r2
 =
 d2/64
 -4(p2q2+p2r2+q2r2)+c2/4
 =
 e.
This gives a cubic, whose three solutions are p2,q2,r2. I spent most of the year with another, less useful identity, which led nowhere. Then I found this one, and got a cubic for p4,q4,r4 instead of the squares. I had it stuck in my mind that the analogue of Viète's method should require fourth roots in the answer.

## 10  Bombelli's (1526-1572) Complex Numbers

Cardano often listed negative solutions, but had doubts as to their meaning. He also met complex solutions to quadratics, but concluded that they were useless and that algebra had run away to a point beyond its usefulness. He also noticed that the solution to x3+6x = 20 is given (by the famous formula) as

 x = {(Ö(108)+10)}1/3-{(Ö(108)-10)}1/3
This is the solution 2, but how can we see it? Worse, for the equation x3 = 15x+4, the ``Cardano'' formula gives
 x = {(2+Ö(-121))}1/3+{(2-Ö(-121))}1/3
though the solution is 4, a real number. Bombelli realised that he had to go via complex numbers; he argued that (if the solution is an integer) it must be 4, by introducing complex numbers and the rules for their manipulation. He had the first notation for complex numbers:
 2-3i was written 2m di m3
This was done at a time when even negative numbers were suspect.

## 11  Viète

We have mentioned Viète's better notation. He also explained the Cardano formula using the algebra in Jeans's footnote. He showed how to use trigonometry to solve the cube-root problem of complex numbers. Suppose that we can use linear substitution y = x+p and scaling z = qy to transform the cubic to
 z3-(3/4)z = A.
Then complex numbers will be needed in the solution. But we can compare with
 cos3q- 3 4 cosq = 1 4 cos(3q).
Thus put z = cosq, and we find q from cos(3q) = 4A, from tables of cosines, which were then becoming available. This paved the way for de Moivre's formula. Another 280 years were to pass without any progress on the quintic. Abel in 1824 proved that there exists NO formula involving a finite number of surds giving the solution to the general quintic. Galois (aged 20) gave a general theory of when a quintic has a solution using surds, and invented group theory and class-field theory in his thesis. He wrote it up the day before dying in a duel.

File translated from TEX by TTH, version 2.01.
On 3 Jul 2001, 21:00.