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Chapter 26 of *Ars Magna* starts with a rule: "If the fourth and first powers are equal to the square and constant and if, having divided the coefficient of *x* and the constant by the coefficient of *x*^{2}, half the result of dividing the coefficient of *x* is [equal to] the square root of that which came out of the division of the constant of the equation, then take the square root of the constant of the original equation, add to it one-fourth the coefficient of *x*^{2}, take the square root of the sum, from this subtract the square root of the same one fourth the coefficient of *x*^{2}, and the remainder is the value of *x*." Translated into modern symbols by T.R. Witmer ([1], p. 171), the rule establishes that if *x*^{4 }+ *ax *= *bx*^{2 }+ *N*, and if *a*/2*b *= √(*N*/*b)*, then $$x=\sqrt{\sqrt{N}+{b\over 4}} - \sqrt{b\over 4}.$$ We must stress that when an equation to be solved appears in the statement of a problem in chapter 26 of *Ars Magna*, without exception the coefficients are positive.

Cardano does not present a proof, so we might as well provide one. Since *N *= *a*^{2}/4*b*, we will have $$x^4=bx^2-ax+N=bx^2-ax + {a^2\over 4b} = b\left(x^2-{a\over b}x+{a^2\over 4b^2}\right) = b\left(x-{a\over 2b}\right)^2,$$ hence $$x^2=\pm\sqrt{b}\left(x-{a\over 2b}\right).$$ The solutions of these two quadratic equations are $$x=\sqrt{b\over 4} \pm \sqrt{{b\over 4}-\sqrt{N}} \qquad {\rm and} \qquad x = -\sqrt{b\over 4}\pm \sqrt{{b\over 4} +\sqrt{N}}.$$ Cardano's solution, namely $$x = -\sqrt{b\over 4} + \sqrt{{b\over 4}+\sqrt{N}},$$ is one of them (obviously it is positive). The solutions of the first quadratic might be complex if *b*/4 - √*N* < 0.

It should come as no surprise that algebraic proofs, as we understand them nowadays, are absent in *Ars Magna*; in particular in chapter 26. Modern symbolic algebra came into being in the 17th century after the pioneering works of François Viète (1540-1603).

The first problem of chapter 26 has to do with money deposits, which leads to the equation *x*^{4 }+ 10*x *= 5*x*^{2 }+ 5. We observe that 5 = 10^{2}/(4 x 5), thus the equation is of the type just discussed. Rather than applying the rule, as Cardano does, it might be advisable to follow the procedure that led to a proof: *x*^{4 }= -10*x *+ 5*x*^{2 }+ 5 = 5(*x*^{2 }- 2*x *+ 1) = 5(*x *- 1)^{2}, whence *x*^{2 }= ±√5(*x *- 1). The equation *x*^{2 }= -√5(*x *- 1) has the solutions $$x= -\sqrt{5\over 4} \pm \sqrt{{5\over 4}+\sqrt{5}},$$ while the equation *x*^{2 }= √5(*x *- 1) has the complex solutions $$x=\sqrt{5\over 4} \pm i\sqrt{\sqrt{5}-{5\over 4}}.$$ As expected, Cardano considers only the positive solution $$x= -\sqrt{5\over 4} + \sqrt{{5\over 4}+\sqrt{5}}.$$

Figure 2: First problem from chapter 26

The second rule of chapter 26 is very similar to the first. In modern symbols it establishes that if *x*^{4 }= *bx*^{2 }+ *ax *+ *N* and *a*/2*b *= √(*N*/*b*) then $$x=\sqrt{{b\over 4}+\sqrt{N}} + \sqrt{b\over 4}.$$ It can be proven, in an almost identical way as the first rule. Indeed, the equation has the solutions $$x=\sqrt{b\over 4} \pm \sqrt{{b\over 4}+\sqrt{N}}, \quad {\rm and} \quad x=-\sqrt{b\over 4} \pm \sqrt{{b\over 4}-\sqrt{N}}$$ and the latter will be complex if *b/*4 - √*N* < 0. For instance, *x*^{4 }= 5*x*^{2 }+ 10*x *+ 5 ([1], p. 172) has the solutions $$x = \sqrt{5\over 4} \pm \sqrt{{5\over 4}+\sqrt{5}} \quad {\rm and} \quad x = -\sqrt{5\over 4} \pm i\sqrt{\sqrt{5}-{5\over 4}}.$$ Cardano writes only the positive solution $$x = \sqrt{5\over 4} + \sqrt{{5\over 4}+\sqrt{5}}.$$

The third rule, in Witmer's translation into modern symbols ([1], p.173), reads as follows: "If *x*^{4 }+ *bx*^{2 }+ *ax *= *cx*^{3 }+ *N*, and if *N *= *b*+2, and if *a *= *c*, and if *a*/2 = √*N*, then $$x = {a\over 4} \pm \sqrt{\left({a\over 4}\right)^2+1-\sqrt{\left({a\over 2}\right)^2+1}}."$$ That is to say, we have to solve the equation *x*^{4}+(*a*^{2}/4 - 2)*x*^{2 }+ *ax *= *ax*^{3 }+ *a*^{2}/4. This problem looks more challenging since there is a cubic term in the equation. Can we provide a proof?

We start by observing that the given equation is equivalent to *x*^{4 }- *ax*^{3}+(*a*^{2}/4)*x*^{2 }= 2*x*^{2 }- *ax *+ *a*^{2}/4. Thus *x*^{2}(*x*^{2 }- *ax *+ *a*^{2}/4) = 2*x*(*x *- *a*/2) + *a*^{2}/4, which in turn leads to *x*^{2}(*x *- *a*/2)^{2 }- 2*x*(*x *- *a*/2) + 1 = *a*^{2}/4 + 1. Therefore, $$\left(x\left(x-{a\over 2}\right) - 1\right)^2 = \left(\sqrt{{a^2\over 4}+1}\right)^2 \quad {\rm or} \quad x\left(x-{a\over 2}\right)-1 = \pm\sqrt{{a^2\over 4}+1}.$$ The equation $$x^2-{a\over 2}x - \left(1+\sqrt{1+{a^2\over 4}}\right) = 0$$ has the two solutions $$x = {a\over 4} \pm \sqrt{\left({a\over 4}\right)^2+1+\sqrt{\left({a\over 2}\right)^2+1}},$$ while the equation $$x^2-{a\over 2}x - \left(1-\sqrt{1+{a^2\over 4}}\right) = 0$$ has the two solutions $$x = {a\over 4} \pm \sqrt{\left({a\over 4}\right)^2+1-\sqrt{\left({a\over 2}\right)^2+1}},$$ Only the latter appear in the third rule.

As an example, Cardano puts forward the equation *x*^{4 }+ 34*x*^{2 }+ 12*x *= 12*x*^{3 }+ 36. We will then have the solutions 3 ± √(10-√37) (both positive) and the solutions 3 ± √(10+√37) (one positive and the other negative). Cardano does not consider the solutions 3 ± √(10+√37), although one of them is positive.

Harald Helfgott (University of Bristol) and Michel Helfgott (East Tennessee State University), "A Modern Vision of the Work of Cardano and Ferrari on Quartics - Some Special Quartics," *Loci* (February 2010), DOI:10.4169/loci003312