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{{Short description|Cubic equation unsolvable in real radicals}}
{{Italic title}}

'''{{lang|la|Casus irreducibilis}}''' ({{ety|la||the irreducible case}}) is the name given by mathematicians of the 16th century to [[cubic equation]]s that cannot be solved in terms of [[radical expression|real radicals]], that is to those equations such that the computation of the solutions cannot be ''reduced'' to the computation of [[square root|square]] and [[cube root]]s.

[[Cardano's formula]] for [[solution in radicals]] of a cubic equation was discovered at this time. It applies in the ''casus irreducibilis'', but, in this case, requires the computation of the square root of a negative number, which involves knowledge of [[complex number]]s, unknown at the time.

The casus irreducibilis occurs when the three solutions are real and distinct, or, equivalently, when the [[discriminant]] is positive.

It is only in 1843 that [[Pierre Wantzel]] proved that there cannot exist any solution in real radicals in the ''casus irreducibilis''.<ref name ="Wantzel">{{Citation|last = Wantzel|first = Pierre|author-link = Pierre Wantzel|url = http://archive.numdam.org/ARCHIVE/NAM/NAM_1843_1_2_/NAM_1843_1_2__117_1/NAM_1843_1_2__117_1.pdf|title = Classification des nombres incommensurables d'origine algébrique|journal = [[Nouvelles Annales de Mathématiques]]|year = 1843|volume = 2|pages = 117–127|language = french}}</ref>

== The three cases of the discriminant ==
Let
: <math>ax^3+bx^2+cx+d=0</math>
be a cubic equation with <math>a\ne0</math>. Then the discriminant is given by
: <math>D := \bigl((x_1-x_2)(x_1-x_3)(x_2-x_3)\bigr)^2 = 18abcd - 4ac^3 - 27a^2d^2 + b^2c^2 -4b^3d~.</math>
It appears in the algebraic solution and is the square of the product
: <math>\Delta := \prod_{j<k}(x_j-x_k) = (x_1-x_2)(x_1-x_3)(x_2-x_3) \qquad \qquad \bigl(\!= \pm\sqrt{D}\bigr)</math>
of the <math>\tbinom32 = 3</math> differences of the 3 roots <math>x_1,x_2,x_3</math>.<ref><math>\Delta</math> is closely related to the [[Vandermonde polynomial]].</ref>

# If {{math|''D'' < 0}}, then the polynomial has one real root and two complex non-real roots. <math>\Delta\in i\R^\times</math> is purely imaginary.<br />Although there are cubic polynomials with negative discriminant which are irreducible in the modern sense, ''casus irreducibilis'' does not apply.<ref>The polynomial <math>x^3+x+1</math> with discriminant {{math|''D'' {{=}} –31 < 0}} is easily seen to be irreducible, because according the [[Rational root theorem#Cubic equation|rational root theorem]] it would have to have roots <math>\in\{-1,1\}</math> which it does not have.</ref>
# If {{math|''D'' {{=}} 0}}, then <math>\Delta=0</math> and two of the roots  are equal. These two roots are also a root of the [[derivative]] of the polyomial. So, they are also a root of the [[greatest common divisor]] of the polynomial and its derivative, which can be computed with the [[Euclidean algorithm for polynomials]]. It follows that the three roots are real, and if the coefficients are rational numbers, the same is true for the roots. <br>That is, all the roots are expressible without radicals.
# If {{math|''D'' > 0}}, then <math>\Delta\in\R^\times</math> is non-zero and real, and there are three distinct real roots which are expressed by [[Cardano's formula]] as sums of two [[complex conjugate]]s cube roots.<br />Because complex numbers were not known in the 16th century, this case has been termed ''casus irreducibilis'', because the computation of the roots could not, at that time, be ''reduced'' to the computation of square and cube roots.<ref>[[James Pierpont (mathematician)|James Pierpont]] in [https://archive.org/details/sim_annals-of-mathematics_1900-10_2_1/page/38/mode/2up ''Annals of Mathematics 1900-1901''] on p.&thinsp;38: "To Cardan and his contemporaries who had no idea how such cube roots could be found this case was highly paradoxical. Since that time mathematicians have attempted to present these real roots as sums of real radicals. As their efforts were unsuccessful, the case when D > 0 was known as the casus irreducibilis."<br />[[Artur Ekert]] [https://archive.org/details/greatartorru00card/page/110/mode/2up ''Complex and unpredictable Cardano''] takes Cardano’s example <math>x^3-15x-4=0</math> having <math display="inline">q^2/4 + p^3/27 = (-4)^2/4 + (-15)^3/27 = -121 < 0 </math> and writes on p.&thinsp;9: "Cardano knew that <math>x = 4</math> was one of the solutions and yet it was a ''casus irreducibilis''". This shows that in the 16th century "irreducibilis" must have meant something like "not reducible to real radicals".<br />On the other hand, Cardano’s example may be used to show how real roots can arise from cube roots of non-real numbers:
:{|
|-
|We have || <math>p=-15, \; q=-4</math>,
|-
|which yields &nbsp; &nbsp; &nbsp; || <math>d:={q^2\over 4}+{p^3\over 27}=-121</math>,
|-
|from which || &nbsp; &nbsp; &nbsp; &nbsp; <math>-{q\over 2} \pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}} = -{q\over 2} \pm \sqrt{d}=2 \pm 11\,i</math>.
|-
|colspan=2| In the 16th century it was difficult („verè sophistica“) to find that
|-
| || &nbsp; &nbsp; &nbsp; &nbsp; <math>\sqrt[3]{2 + 11^{\color{white};} i} = 2 + i =: u</math>
|-
|and || &nbsp; &nbsp; &nbsp; &nbsp; <math>\sqrt[3]{2 - 11^{\color{white};} i} = 2 - i =: v</math>,
|-
|so that || <math>t_k = 
\omega_k   \sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} + 
\omega_k^2 \sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}</math>
|-
| || &nbsp; &nbsp; <math>= \omega_k u + \omega_k^2 v </math>.
|-
|colspan=2| This means in detail:
|-
|1st root || <math>t_1 = \omega_1 u + \omega_1^2 v </math>
|-
| || &nbsp; &nbsp; <math>= 1 \cdot (2 + i) + 1 \cdot (2 - i)</math>
|-
| || &nbsp; &nbsp; <math>= 4</math>,
|-
|2nd root || <math>t_2 = \omega_2 u + \omega_2^2 v </math>
|-
| || &nbsp; &nbsp; <math>= \biggl(\!\!-\frac{1}{2} + \frac{\sqrt{3}}{2}i\biggr) \cdot (2 + i) + \biggl(\!\!-\frac{1}{2} - \frac{\sqrt{3}}{2}i\biggr) \cdot (2 - i)</math>
|-
| || &nbsp; &nbsp; <math>= -2 - \sqrt{3}</math>,
|-
|3rd root || <math>t_3 = \omega_3 u + \omega_3^2 v </math>
|-
| || &nbsp; &nbsp; <math>= \biggl(\!\!-\frac{1}{2} - \frac{\sqrt{3}}{2}i\biggr) \cdot (2 + i) + \biggl(\!\!- \frac{1}{2} + \frac{\sqrt{3}}{2}i\biggr) \cdot (2 - i)</math>
|-
| || &nbsp; &nbsp; <math>= -2 + \sqrt{3}</math>.
|}
It may be noticed that <math display="inline">d := q^2/4 + p^3/27 = -121</math> is not the discriminant <math>D</math>; it is <math>D = -108 \, d = 13068 = 2^2 3^3 11^2 </math> with the sign inverted. Interestingly <math display="inline">\sqrt{d} = i\sqrt{D/108}</math> occurs in Cardano’s formula (as well as the primitive 3rd roots of unity <math>\omega_{2,3}</math> with their {{nowrap|<math>i \sqrt{3}</math>),}} although <math>\sqrt{D}~,</math> and not <math>\sqrt{d}~,</math> is necessarily an element of the splitting field.
</ref>

== Formal statement and proof ==
More generally, suppose that {{math|''F''}} is a [[formally real field]], and that {{math|''p''(''x'') &isin; ''F''[''x'']}} is a cubic polynomial, irreducible over {{math|''F''}}, but having three real roots (roots in the [[real closure]] of {{math|''F''}}). Then ''casus irreducibilis'' states that it is impossible to express a solution of {{math|''p''(''x'') {{=}} 0}} by radicals with real radicands.

To prove this,<ref>B.L. van der Waerden, ''Modern Algebra'' (translated from German by Fred Blum), Frederick Ungar Publ. Co., 1949, p.&thinsp;180.</ref> note that the discriminant {{math|''D''}} is positive. Form the [[field extension]] {{math|''F''({{sqrt|''D''}}) {{=}} ''F''(∆)}}. Since this is {{math|''F''}} or a [[quadratic extension]] of {{math|''F''}} (depending in whether or not {{math|''D''}} is a square in {{math|''F''}}), {{math|''p''(''x'')}} remains irreducible in it. Consequently, the [[Galois group]] of {{math|''p''(''x'')}} over {{math|''F''({{sqrt|''D''}})}} is the cyclic group {{math|''C''<sub>3</sub>}}. Suppose that {{math|''p''(''x'') {{=}} 0}} can be solved by real radicals. Then {{math|''p''(''x'')}} can be [[splitting field|split]] by a tower of [[cyclic extension]]s
:<math> F\sub F(\sqrt{D})\sub F(\sqrt{D}, \sqrt[p_1]{\alpha_1}) \sub\cdots \sub K\sub K(\sqrt[3]{\alpha})</math>

At the final step of the tower, {{math|''p''(''x'')}} is irreducible in the penultimate field {{math|''K''}}, but splits in {{math|''K''({{radic|''α''|3}})}} for some {{math|''α''}}. But this is a cyclic field extension, and so must contain a [[Conjugate element (field theory)|conjugate]] of {{math|{{radic|''α''|3}}}} and therefore a [[primitive root of unity|primitive 3rd root of unity]].

However, there are no primitive 3rd roots of unity in a real closed field. Suppose that &omega; is a primitive 3rd root of unity. Then, by the axioms defining an [[ordered field]], &omega; and &omega;<sup>2</sup> are both positive, because otherwise their cube (=1) would be negative. But if &omega;<sup>2</sup>>&omega;, then cubing both sides gives 1>1, a contradiction; similarly if &omega;>&omega;<sup>2</sup>.

== Solution in non-real radicals ==

=== Cardano's solution ===
{{Further|Cubic equation#Cardano's formula}}

The equation  {{math|''ax''<sup>3</sup> + ''bx''<sup>2</sup> + ''cx'' + ''d'' {{=}} 0}} can be depressed to a [[monic polynomial|monic]] [[trinomial]] by dividing by <math>a</math> and substituting {{math|''x'' {{=}} ''t'' − {{sfrac|''b''|3''a''}}}} (the [[Tschirnhaus transformation]]), giving the equation {{math|''t''<sup>3</sup> + ''pt'' + ''q'' {{=}} 0}} where
:<math>p=\frac{3ac-b^2}{3a^2}</math>
:<math>q=\frac{2b^3-9abc+27a^2d}{27a^3}.</math>

Then  regardless of the number of real roots, by [[Cubic equation#Cardano's formula|Cardano's solution]] the three roots are given by

:<math> t_k = \omega_k \sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} + \omega_k^2  \sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}</math>

where <math> \omega_k</math> (''k''=1, 2, 3) is a [[Root of unity#Trigonometric expression|cube root of 1]] (<math>\omega_1 = 1</math>, <math>\omega_2 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i</math>, and <math>\omega_3 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i</math>, where {{math|''i''}} is the [[imaginary unit]]). Here if the [[radicand]]s under the cube roots are non-real, the cube roots expressed by radicals are defined to be any pair of complex conjugate cube roots, while if they are real these cube roots are defined to be the real cube roots.

''Casus irreducibilis'' occurs when none of the roots are rational and when all three roots are distinct and real; the case of three distinct real roots occurs if and only if {{math|{{sfrac|''q''<sup>2</sup>|4}} + {{sfrac|''p''<sup>3</sup>|27}} < 0}}, in which case Cardano's formula involves first taking the square root of a negative number, which is [[Imaginary number|imaginary]], and then taking the cube root of a complex number (the cube root cannot itself be placed in the form {{math|''α'' + ''βi''}} with specifically given expressions in real [[nth root|radicals]] for {{math|''α''}} and {{math|''β''}}, since doing so would require independently solving the original cubic). Even in the reducible case in which one of three real roots is rational and hence can be factored out by [[polynomial long division]], Cardano's formula (unnecessarily in this case) expresses that root (and the others) in terms of non-real radicals.

=== Example ===
The cubic equation

:<math>2x^3-9x^2-6x+3=0</math>

is irreducible, because if it could be factored there would be a linear factor giving a rational solution, while none of the possible roots given by the [[rational root test]] are actually roots. Since its discriminant is positive, it has three real roots, so it is an example of ''casus irreducibilis.'' These roots can be expressed as

:<math>t_k=\frac{3-\omega_k\sqrt[3]{39-26i}-\omega_k^2\sqrt[3]{39+26i}}{2}</math>

for <math>k\in\left\{1, 2, 3\right\}</math>. The solutions are in radicals and involve the cube roots of [[complex conjugate]] numbers.

== Trigonometric solution in terms of real quantities ==
{{Main|Cubic equation#Trigonometric and hyperbolic solutions}}

While ''casus irreducibilis'' cannot be [[Algebraic solution|solved in radicals]] in terms of real quantities, it ''can'' be solved [[Trigonometry|trigonometrically]] in terms of real quantities.<ref>{{harvtxt|Cox|2012}}, Section 1.3B Trigonometric Solution of the Cubic, pp.&thinsp;18–19.</ref> Specifically, the depressed monic cubic equation <math>t^3+pt+q=0 </math> is solved by

:<math>t_k=2\sqrt{-\frac{p}{3}}\cos\left[\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right] \quad \text{for} \quad k=0,1,2 \,.</math>

These solutions are in terms of real quantities if and only if <math>{q^{2}\over 4}+{p^{3}\over 27} < 0</math> — i.e., if and only if there are three real roots. The formula involves starting with an angle whose cosine is known, trisecting the angle by multiplying it by 1/3, and taking the cosine of the resulting angle and adjusting for scale.

Although cosine and its inverse function (arccosine) are [[transcendental functions]], this solution is algebraic in the sense that <math>\cos\left[\arccos\left(x\right)/3\right]</math> is an [[algebraic function]], equivalent to [[angle trisection]].

==Relation to angle trisection==
The distinction between the reducible and irreducible cubic cases with three real roots is related to the issue of whether or not an angle is [[Angle trisection|trisectible]] by the classical means of [[compass and straightedge constructions|compass and unmarked straightedge]]. For any angle {{math|''θ''}}, one-third of this angle has a cosine that is one of the three solutions to

:<math>4x^3-3x-\cos(\theta)=0.</math>

Likewise, {{math|{{frac|''θ''|3}}}} has a sine that is one of the three real solutions to

:<math>4y^3-3y+\sin(\theta)=0.</math>

In either case, if the rational root test reveals a rational solution, {{mvar|x}} or {{mvar|y}} minus that root can be factored out of the polynomial on the left side, leaving a quadratic that can be solved for the remaining two roots in terms of a square root; then all of these roots are classically constructible since they are expressible in no higher than square roots, so in particular {{math|cos({{frac|''θ''|3}})}} or {{math|sin({{frac|''θ''|3}})}} is constructible and so is the associated angle {{math|{{frac|''θ''|3}}}}. On the other hand, if the rational root test shows that there is no rational root, then ''casus irreducibilis'' applies, {{math|cos({{frac|''θ''|3}})}} or {{math|sin({{frac|''θ''|3}})}} is not constructible, the angle {{math|{{frac|''θ''|3}}}} is not constructible, and the angle {{math|''θ''}} is not classically trisectible.

As an example, while a 180° angle can be trisected into three 60° angles, a 60° angle cannot be trisected with only compass and straightedge. Using [[List of trigonometric identities#Multiple-angle formulae|triple-angle formulae]] one can see that {{math|cos {{sfrac|π|3}} {{=}} 4''x''<sup>3</sup> − 3''x''}} where {{math|''x'' {{=}} cos(20°)}}. Rearranging gives {{math|8''x''<sup>3</sup> − 6''x'' − 1 {{=}} 0}}, which fails the rational root test as none of the rational numbers suggested by the theorem is actually a root. Therefore, the minimal polynomial of {{math|cos(20°)}} has degree 3, whereas the degree of the minimal polynomial of any constructible number must be a power of two.

Expressing {{math|cos(20°)}} in radicals results in

:<math>\cos\left(\frac{\pi}{9}\right)=\frac{\sqrt[3]{1-i\sqrt{3}}+\sqrt[3]{1+i\sqrt{3}}}{2\sqrt[3]{2}}</math>

which involves taking the cube root of complex numbers. Note the similarity to {{math|''e''<sup>''iπ''/3</sup> {{=}} {{sfrac|1+''i''{{sqrt|3}}|2}}}} and {{math|''e''<sup>''−iπ''/3</sup> {{=}} {{sfrac|1−''i''{{sqrt|3}}|2}}}}.

The connection between rational roots and trisectability can also be extended to some cases where the sine and cosine of the given angle is irrational. Consider as an example the case where the given angle {{math|''θ''}} is a vertex angle of a regular pentagon, a polygon that can be constructed classically. For this angle {{math|''5θ/3''}} is 180°, and standard trigonometric identities then give

:<math> \cos(\theta)+\cos(\theta/3) = 2\cos(\theta/3)\cos(2\theta/3)
=-2\cos(\theta/3)\cos(\theta)</math>

thus

:<math> \cos(\theta/3) = -\cos(\theta)/(1+2\cos(\theta)).</math>

The cosine of the trisected angle is rendered as a rational expression in terms of the cosine of the given angle, so the vertex angle of a regular pentagon can be trisected (mechanically, by simply drawing a diagonal).

== Generalization ==
''Casus irreducibilis'' can be generalized to higher degree polynomials as follows. Let {{math|''p''&nbsp;&isin;&nbsp;''F''[''x'']}} be an irreducible polynomial which splits in a formally real extension {{math|''R''}} of {{math|''F''}} (i.e., {{math|''p''}} has only real roots). Assume that {{math|''p''}} has a root in <math>K\subseteq R</math> which is an extension of {{math|''F''}} by radicals. Then the degree of {{math|''p''}} is a power of 2, and its splitting field is an iterated quadratic extension of {{math|''F''}}.<ref>{{harvtxt|Cox|2012}}, Theorem 8.6.5, p.&thinsp;222.</ref><ref name=Isaacs>I. M. Isaacs, "Solution of polynomials by real radicals", ''[[American Mathematical Monthly]]'' 92 (8), October 1985, 571–575,</ref>{{rp|571–572}}

Thus for any irreducible polynomial whose degree is not a power of 2 and which has all roots real, no root can be expressed purely in terms of real radicals, i.e. it is a ''casus irreducibilis'' in the (16th century) sense of this article. Moreover, if the polynomial degree ''is'' a power of 2 and the roots are all real, then if there is a root that can be expressed in real radicals it can be expressed in terms of square roots and no higher-degree roots, as can the other roots, and so the roots are [[constructible number|classically constructible]].

''Casus irreducibilis'' for [[Quintic function|quintic polynomials]] is discussed by Dummit.<ref>David S. Dummit [http://www.emba.uvm.edu/~dummit/quintics/solvable.pdf Solving Solvable Quintics] {{Webarchive|url=https://web.archive.org/web/20120307030156/http://www.emba.uvm.edu/~dummit/quintics/solvable.pdf |date=2012-03-07 }}</ref>{{rp|17}}

=== Relation to angle pentasection (quintisection) and higher ===
The distinction between the reducible and irreducible quintic cases with five real roots is related to the issue of whether or not an angle with rational cosine or rational sine is pentasectible (able to be split into five equal parts) by the classical means of compass and unmarked straightedge. For any angle {{math|''θ''}}, one-fifth of this angle has a cosine that is one of the five real roots of the equation

:<math>16x^5-20x^3+5x-\cos(\theta)=0.</math>

Likewise, {{math|{{sfrac|''θ''|5}}}} has a sine that is one of the five real roots of the equation

:<math>16y^5-20y^3+5y-\sin(\theta)=0.</math>

In either case, if the rational root test yields a rational root ''x<sub>1</sub>'', then the quintic is reducible since it can be written as a factor (''x—x''<sub>1</sub>) times a [[quartic polynomial]]. But if the test shows that there is no rational root, then the polynomial may be irreducible, in which case ''casus irreducibilis'' applies, {{math|cos({{frac|''θ''|5}})}} and {{math|sin({{frac|''θ''|5}})}} are not constructible, the angle {{math|{{frac|''θ''|5}}}} is not constructible, and the angle {{math|''θ''}} is not classically pentasectible. An example of this is when one attempts to construct a 25-gon (icosipentagon) with compass and straightedge. While a pentagon is relatively easy to construct, a 25-gon requires an angle pentasector as the minimal polynomial for {{math|cos(14.4°)}} has degree 10:

:<math>\begin{align}
\cos\left(\frac{2\pi}{5}\right) &= \frac{\sqrt{5}-1}{4} \\
16x^5-20x^3+5x+\frac{1-\sqrt{5}}{4} &= 0 \qquad\qquad x=\cos\left(\frac{2\pi}{25}\right) \\
4\left(16x^5-20x^3+5x+\frac{1-\sqrt{5}}{4}\right)\left(16x^5-20x^3+5x+\frac{1+\sqrt{5}}{4}\right) &= 0 \\
4\left(16x^5-20x^3+5x\right)^2+2\left(16x^5-20x^3+5x\right)-1 &= 0 \\
1024x^{10}-2560x^8+2240 x^6+32x^5-800 x^4-40x^3+100x^2+10x-1 &= 0.
\end{align}</math>

Thus,

:<math>\begin{align}
e^{2\pi i/5} &= \frac{-1+\sqrt{5}}{4}+\frac{\sqrt{10+2\sqrt{5}}}{4}i \\
e^{-2\pi i/5} &= \frac{-1+\sqrt{5}}{4}-\frac{\sqrt{10+2\sqrt{5}}}{4}i \\
\cos\left(\frac{2\pi}{25}\right) &= \frac{\sqrt[5]{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}+\sqrt[5]{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}}{2\sqrt[5]{4}}.
\end{align}</math>

== Notes ==
{{reflist}}

== References ==
* {{citation
 | last = Cox | first = David A. | author-link = David A. Cox
 | doi = 10.1002/9781118218457
 | edition = 2nd
 | isbn = 978-1-118-07205-9
 | publisher = John Wiley & Sons
 | series = Pure and Applied Mathematics
 | title = Galois Theory
 | year = 2012}}. See in particular Section 1.3 Cubic Equations over the Real Numbers (pp.&thinsp;15–22) and Section 8.6 The Casus Irreducibilis (pp.&thinsp;220–227).
* {{citation|first=Bartel Leendert|last=van der Waerden|author-link=Bartel Leendert van der Waerden|title=Modern Algebra I|isbn=978-0-387-40624-4|publisher=Springer|year=2003|others= F. Blum, J.R. Schulenberg}}

== External links ==
*{{PlanetMath|urlname=CasusIrreducibilis|title=casus irreducibilis}}

[[Category:Polynomials]]
[[Category:Algebra]]