Editing Quintic function (section)

==Solvable quintics==

Some quintic equations can be solved in terms of radicals. These include the quintic equations defined by a polynomial that is [[irreducible polynomial|reducible]], such as {{math|''x''<sup>5</sup> − ''x''<sup>4</sup> − ''x'' + 1 {{=}} (''x''<sup>2</sup> + 1)(''x'' + 1)(''x'' − 1)<sup>2</sup>}}. For example, it has been shown<ref>{{cite journal |last1=Elia |first1=M. |last2=Filipponi |first2=P. |title=Equations of the Bring–Jerrard Form, the Golden Section, and Square Fibonacci Numbers |journal=The Fibonacci Quarterly |date=1998 |volume=36 |issue=3 |pages=282–286 |url=https://www.fq.math.ca/Scanned/36-3/elia.pdf}}</ref> that

:<math>x^5-x-r=0</math>

has solutions in radicals [[if and only if]] it has an [[integer]] solution or ''r'' is one of ±15, ±22440, or ±2759640, in which cases the polynomial is reducible.

As solving reducible quintic equations reduces immediately to solving polynomials of lower degree, only irreducible quintic equations are considered in the remainder of this section, and the term "quintic" will refer only to irreducible quintics. A '''solvable quintic''' is thus an irreducible quintic polynomial whose roots may be expressed in terms of radicals.

To characterize solvable quintics, and more generally solvable polynomials of higher degree, [[Évariste Galois]] developed techniques which gave rise to [[group theory]] and [[Galois theory]]. Applying these techniques, [[Arthur Cayley]] found a general criterion for determining whether any given quintic is solvable.<ref>A. Cayley, "On a new auxiliary equation in the theory of equation of the fifth order", ''Philosophical Transactions of the Royal Society of London'' '''151''':263-276 (1861) {{doi|10.1098/rstl.1861.0014}}</ref> This criterion is the following.<ref>This formulation of Cayley's result is extracted from Lazard (2004) paper.</ref>

Given the equation
:<math> ax^5+bx^4+cx^3+dx^2+ex+f=0,</math>
the [[Tschirnhaus transformation]] {{math|''x'' {{=}} ''y'' − {{sfrac|''b''|5''a''}}}}, which depresses the quintic (that is, removes the term of degree four), gives the equation

:<math> y^5+p y^3+q y^2+r y+s=0,</math>

where

:<math>\begin{align}p &= \frac{5ac-2b^2}{5a^2}\\[4pt]
q &= \frac{25a^2d-15abc+4b^3}{25a^3}\\[4pt]
r &= \frac{125a^3e-50a^2bd+15ab^2c-3b^4}{125a^4}\\[4pt]
s &= \frac{3125 a^4f-625a^3 be+125a^2b^2 d-25ab^3 c+4 b^5}{3125a^5}\end{align}</math>

Both quintics are solvable by radicals if and only if either they are factorisable in equations of lower degrees with rational coefficients or the polynomial {{math|''P''<sup>2</sup> − 1024 ''z'' Δ}}, named '''{{vanchor|Cayley's resolvent}}''', has a rational root in {{mvar|z}}, where

:<math>\begin{align} 
P = {} &z^3-z^2(20r+3p^2)- z(8p^2r - 16pq^2- 240r^2 + 400sq - 3p^4)\\[4pt]
&- p^6 + 28p^4r- 16p^3q^2- 176p^2r^2- 80p^2sq + 224prq^2- 64q^4 \\[4pt]
&+ 4000ps^2 + 320r^3- 1600rsq
\end{align}
</math>

and

:<math>\begin{align}
\Delta = {} &-128p^2r^4+3125s^4-72p^4qrs+560p^2qr^2s+16p^4r^3+256r^5+108p^5s^2 \\[4pt]
&-1600qr^3s+144pq^2r^3-900p^3rs^2+2000pr^2s^2-3750pqs^3+825p^2q^2s^2 \\[4pt]
&+2250q^2rs^2+108q^5s-27q^4r^2-630pq^3rs+16p^3q^3s-4p^3q^2r^2. \end{align}
</math>
Cayley's result allows us to test if a quintic is solvable. If it is the case, finding its roots is a more difficult problem, which consists of expressing the roots in terms of radicals involving the coefficients of the quintic and the rational root of Cayley's resolvent.

In 1888, [[George Paxton Young]] described how to solve a solvable quintic equation, without providing an explicit formula;<ref>George Paxton Young, "Solvable Quintic Equations with Commensurable Coefficients", ''American Journal of Mathematics'' '''10''':99–130 (1888), {{JSTOR|2369502}}</ref> in 2004, [[Daniel Lazard]] wrote out a three-page formula.<ref>{{harvtxt|Lazard|2004|p=207}}</ref>

===Quintics in Bring–Jerrard form===

There are several parametric representations of solvable quintics of the form {{math|''x''<sup>5</sup> + ''ax'' + ''b'' {{=}} 0}}, called the [[Bring–Jerrard form]].

During the second half of the 19th century, John Stuart Glashan, George Paxton Young, and [[Carl Runge]] gave such a parameterization: an [[irreducible polynomial|irreducible]] quintic with rational coefficients in Bring–Jerrard form
is solvable if and only if either {{math|''a'' {{=}} 0}} or it may be written
:<math>x^5 + \frac{5\mu^4(4\nu + 3)}{\nu^2 + 1}x + \frac{4\mu^5(2\nu + 1)(4\nu + 3)}{\nu^2 + 1} = 0</math>
where {{math|''μ''}} and {{math|''ν''}} are rational.

In 1994, Blair Spearman and Kenneth S. Williams gave an alternative,
:<math>x^5 + \frac{5e^4( 4c + 3)}{c^2 + 1}x + \frac{-4e^5(2c-11)}{c^2 + 1} = 0.</math>

The relationship between the 1885 and 1994 parameterizations can be seen by defining the expression
:<math>b = \frac{4}{5} \left(a+20 \pm 2\sqrt{(20-a)(5+a)}\right)</math>
where {{tmath|1=a=5\tfrac{4\nu+3}{\nu^2+1} }}. Using the negative case of the square root yields, after scaling variables, the first parametrization while the positive case gives the second.

The substitution {{tmath|1=c = - \tfrac{m}{\ell^5}, }} {{tmath|1=e = \tfrac{1}{\ell} }} in the Spearman–Williams parameterization allows one to not exclude the special case {{math|''a'' {{=}} 0}}, giving the following result:

If {{mvar|a}} and {{mvar|b}} are rational numbers, the equation {{math|''x''<sup>5</sup> + ''ax'' + ''b'' {{=}} 0}} is solvable by radicals if either its left-hand side is a product of polynomials of degree less than 5 with rational coefficients or there exist two rational numbers ''ℓ'' and {{mvar|m}} such that
:<math>a=\frac{5 \ell (3 \ell^5-4 m)}{m^2+\ell^{10}}\qquad b=\frac{4(11 \ell^5+2 m)}{m^2+\ell^{10}}.</math>

===Roots of a solvable quintic===
A polynomial equation is solvable by radicals if its [[Galois group]] is a [[solvable group]]. In the case of irreducible quintics, the Galois group is a subgroup of the [[symmetric group]] {{math|''S''<sub>5</sub>}} of all permutations of a five element set, which is solvable if and only if it is a subgroup of the group {{math|''F''<sub>5</sub>}}, of order {{math|20}}, generated by the cyclic permutations {{math|(1 2 3 4 5)}} and {{math|(1 2 4 3)}}.

If the quintic is solvable, one of the solutions may be represented by an [[algebraic expression]] involving a fifth root and at most two square roots, generally [[nested radical|nested]]. The other solutions may then be obtained either by changing the fifth root or by multiplying all the occurrences of the fifth root by the same power of a [[root of unity|primitive 5th root of unity]], such as
:<math>\frac{\sqrt{-10-2\sqrt{5}}+\sqrt{5}-1}{4}.</math>

In fact, all four primitive fifth roots of unity may be obtained by changing the signs of the square roots appropriately; namely, the expression
:<math>\frac{\alpha\sqrt{-10-2\beta\sqrt{5}}+\beta\sqrt{5}-1}{4},</math>
where <math> \alpha, \beta \in \{-1,1\}</math>, yields the four distinct primitive fifth roots of unity.

It follows that one may need four different square roots for writing all the roots of a solvable quintic. Even for the first root that involves at most two square roots, the expression of the solutions in terms of radicals is usually highly complicated. However, when no square root is needed, the form of the first solution may be rather simple, as for the equation {{math|''x''<sup>5</sup> − 5''x''<sup>4</sup> + 30''x''<sup>3</sup> − 50''x''<sup>2</sup> + 55''x'' − 21 {{=}} 0}}, for which the only real solution is

: <math>x=1+\sqrt[5]{2}-\left(\sqrt[5]{2}\right)^2+\left(\sqrt[5]{2}\right)^3-\left(\sqrt[5]{2}\right)^4.</math>

An example of a more complicated (although small enough to be written here) solution is the unique real root of {{math|''x''<sup>5</sup> − 5''x'' + 12 {{=}} 0}}. Let {{math|''a'' {{=}} {{sqrt|2''φ''<sup>−1</sup>}}}}, {{math|''b'' {{=}} {{sqrt|2''φ''}}}}, and {{math|''c'' {{=}} {{radic|5|4}}}}, where {{math|''φ'' {{=}} {{sfrac|1+{{sqrt|5}}|2}}}} is the [[golden ratio]]. Then the only real solution {{math|''x'' {{=}} −1.84208...}} is given by

: <math>-cx = \sqrt[5]{(a+c)^2(b-c)} + \sqrt[5]{(-a+c)(b-c)^2} + \sqrt[5]{(a+c)(b+c)^2} - \sqrt[5]{(-a+c)^2(b+c)} \,,</math>

or, equivalently, by

:<math>x = \sqrt[5]{y_1}+\sqrt[5]{y_2}+\sqrt[5]{y_3}+\sqrt[5]{y_4}\,,</math>

where the {{math|''y<sub>i</sub>''}} are the four roots of the [[quartic equation]]

:<math>y^4+4y^3+\frac{4}{5}y^2-\frac{8}{5^3}y-\frac{1}{5^5}=0\,.</math>

More generally, if an equation {{math|1=''P''(''x'') = 0}} of prime degree {{math|''p''}} with rational coefficients is solvable in radicals, then one can define an auxiliary equation {{math|1=''Q''(''y'') = 0}} of degree {{math|''p'' − 1}}, also with rational coefficients, such that each root of {{math|''P''}} is the sum of {{math|''p''}}-th roots of the roots of {{math|''Q''}}. These {{math|''p''}}-th roots were introduced by [[Joseph-Louis Lagrange]], and their products by {{math|''p''}} are commonly called [[Lagrange resolvent]]s. The computation of {{math|''Q''}} and its roots can be used to solve {{math|1=''P''(''x'') = 0}}. However these {{math|''p''}}-th roots may not be computed independently (this would provide {{math|''p''<sup>''p''−1</sup>}} roots instead of {{math|''p''}}). Thus a correct solution needs to express all these {{math|''p''}}-roots in term of one of them. Galois theory shows that this is always theoretically possible, even if the resulting formula may be too large to be of any use.

It is possible that some of the roots of {{math|''Q''}} are rational (as in the first example of this section) or some are zero. In these cases, the formula for the roots is much simpler, as for the solvable [[de Moivre]] quintic{{anchor|de Moivre quintic}}

:<math>x^5+5ax^3+5a^2x+b = 0\,,</math>

where the auxiliary equation has two zero roots and reduces, by factoring them out, to the [[quadratic equation]]

:<math>y^2+by-a^5 = 0\,,</math>

such that the five roots of the de Moivre quintic are given by

:<math>x_k = \omega^k\sqrt[5]{y_i} -\frac{a}{\omega^k\sqrt[5]{y_i}},</math>

where ''y<sub>i</sub>'' is any root of the auxiliary quadratic equation and ''ω'' is any of the four [[primitive root of unity|primitive 5th roots of unity]].  This can be easily generalized to construct a solvable [[septic equation|septic]] and other odd degrees, not necessarily prime.

===Other solvable quintics===

There are infinitely many solvable quintics in Bring–Jerrard form which have been parameterized in a preceding section.

Up to the scaling of the variable, there are exactly five solvable quintics of the shape <math>x^5+ax^2+b</math>, which are<ref>{{cite web |first=Noam |last=Elkies |title=Trinomials {{nobr|a x{{sup|n}} + b x + c}} with interesting Galois groups |url=https://www.math.harvard.edu/~elkies/trinomial.html |publisher=[[Harvard University]]}}</ref> (where ''s'' is a scaling factor):
:<math>x^5-2s^3x^2-\frac{s^5}{5} </math>
:<math> x^5-100s^3x^2-1000s^5</math>
:<math>x^5-5s^3x^2-3s^5 </math>
:<math>x^5-5s^3x^2+15s^5 </math>
:<math> x^5-25s^3x^2-300s^5</math>

Paxton Young (1888) gave a number of examples of solvable quintics:
:{| <math>x^5+3x^2+2x-1 </math> ||
|-
| <math> x^5-10x^3-20x^2-1505x-7412</math> ||
|-
| <math>x^5+\frac{625}{4}x+3750 </math> ||
|-
| <math>x^5-\frac{22}{5}x^3-\frac{11}{25}x^2+\frac{462}{125}x+\frac{979}{3125} </math> ||
|-
| <math>x^5+20x^3+20x^2+30x+10 </math> || <math>~\qquad ~</math> Root: <math> \sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4</math>
|-
|<math>x^5-20x^3+250x-400 </math> ||
|-
| <math>x^5-5x^3+\frac{85}{8}x-\frac{13}{2} </math> ||
|-
|<math> x^5+\frac{20}{17}x+\frac{21}{17}</math> ||
|-
|<math>x^5-\frac{4}{13}x+\frac{29}{65}</math> ||
|-
|<math> x^5+\frac{10}{13}x+\frac{3}{13}	</math> ||
|-
| <math> x^5+110(5x^3+60x^2+800x+8320)</math> ||
|-
| <math>x^5-20 x^3 -80x^2 -150x -656</math>||
|-
| <math> x^5 -40x^3 +160x^2 +1000x -5888</math> ||
|-
|<math> x^5-50x^3-600x^2-2000x-11200</math> ||
|-
| <math>x^5+110(5x^3 + 20x^2 -360x +800)</math> ||
|-
| <math> x^5 -20x^3 +170x + 208</math> ||
|}

An infinite sequence of solvable quintics may be constructed, whose roots are sums of {{mvar|n}}th [[roots of unity]], with {{nobr|{{math|''n'' {{=}} 10''k'' + 1}}}} being a [[prime number]]:

:{|
|-
| <math>x^5+x^4-4x^3-3x^2+3x+1</math> || || Roots: <math>2\cos\left(\frac{2k\pi}{11}\right)</math>
|-
| <math> x^5+x^4-12x^3-21x^2+x+5</math> || || Root: <math> \sum_{k=0}^5 e^\frac{2i\pi 6^k }{31}</math>
|-
| <math>x^5+x^4-16x^3+5x^2+21x-9</math> || || Root: <math>\sum_{k=0}^7 e^\frac{2i\pi 3^k }{41}</math>
|-
| <math>x^5+x^4-24x^3-17x^2+41x-13</math> || <math>~\qquad ~</math> || {{nowrap|1= Root: <math>\sum_{k=0}^{11} e^\frac{2i\pi (21)^k }{61}</math>}}
|-
| <math>x^5+x^4 - 28x^3 + 37x^2 + 25x + 1</math> || || {{nowrap|1= Root: <math>\sum_{k=0}^{13} e^\frac{2i\pi (23)^k }{71}</math>}}
|}

There are also two parameterized families of solvable quintics:
The Kondo–Brumer quintic,

:<math> x^5 + (a-3)\,x^4 + (-a+b+3)\,x^3 + (a^2-a-1-2b)\,x^2 + b\,x + a = 0 </math>

and the family depending on the parameters <math>a, \ell, m</math>
:<math>	x^5 - 5\,p \left( 2\,x^3 + a\,x^2 + b\,x \right) - p\,c = 0 </math>

where

::<math> p = \tfrac{1}{4} \left[\, \ell^2 (4m^2 + a^2) - m^2 \,\right] \;,</math>
:
::<math> b = \ell \, ( 4m^2 + a^2 ) - 5p - 2m^2 \;,</math>
:
::<math> c = \tfrac{1}{2} \left[\, b(a + 4m) - p(a - 4m) - a^2m  \,\right] \;.</math>

===''Casus irreducibilis''===

Analogously to [[cubic equation]]s, there are solvable quintics which have five real roots all of whose solutions in radicals involve roots of complex numbers. This is ''[[casus irreducibilis]]'' for the quintic, which is discussed in Dummit.<ref>David S. Dummit [http://www.emba.uvm.edu/~dummit/quintics/solvable.pdf Solving Solvable Quintics]</ref>{{rp|p.17}} Indeed, if an irreducible quintic has all roots real, no root can be expressed purely in terms of real radicals (as is true for all polynomial degrees that are not powers of 2).