# Polynomials with degree $5$ solvable in elementary functions?

Quadratic, cubic and quartic polynomials are solvable in radicals, so there is no question here.

What about the polynomials of degree $5$ (quintic)? Do we know all such polynomials (classes of polynomials) with all (or some) of the roots expressible as elementary functions of the coefficients?

If we do, I would be glad for a reference. If we don't (or it hasn't been proven that we do) then it might be possible to find a new class of solvable quintics?

Elementary functions are understood in the usual sense: algebraic functions, exponents, logarithms, trigonometric and inverse trigonometric functions and some finite combination of the above.

Which means that I'm not interested in the solution using Bring radicals, Hypergeometric functions etc.

A related question

Information on polynomials of higher degree would be appreciated as well.

The other (more practical) question - is there a complete list of known quintics solvable in elemetary functions? (Not necessarily in radicals).

Edit

The importaint point: If at least one root of the quintic can be expressed as elementary function, then all its roots can be expressed as elementary functions, since the general quartic is solved in radicals.

I think this greatly simplifies the problem.

For example, the following quintic has a root expressed as elementary function:

$$x^5-5ax^4-10x^3+10ax^2+5x-a=0 \tag{1}$$

$$x_1=\tan \left( \frac{1}{5} \arctan a \right)$$

Then we can divide $(1)$ by $(x-x_1)$ and obtain a quartic equation with all the roots expressible in radicals. (The example is not ideal though, since this quintic can most likely be solved in radicals).

• As of 1999 this was not known -- see Conjecture 2 at the bottom of p. 442 of What is a closed-form number? by Timothy Y. Chow [American Mathematical Monthly 106 #5 (May 1999), 440-448]. I suspect it's still not known, since I believe such a result would have become relatively well known given its intrinsic interest and the fact that understanding the issue at hand doesn't require a lot of advanced mathematical training. – Dave L. Renfro Jun 16 '16 at 14:27
• "If at least one root of the quintic can be expressed as elementary function, then all its roots can be expressed as elementary functions" Nice observation! – Dave L. Renfro Jun 16 '16 at 16:43
• A very relevant answer - but misleading, since special functions are used – Yuriy S Jun 16 '16 at 20:10
• Other relevant questions: 1 and 2 – Yuriy S Jun 16 '16 at 20:22

These two are relatively well-known.

I. For the DeMoivre quintic: $$x^5+5ax^3+5a^2x+b = 0\tag1$$

$$x = \left(\frac{-b+\sqrt{D}}{2}\right)^{1/5}-a\left(\frac{-b+\sqrt{D}}{2}\right)^{-1/5},\quad D=b^2+4a^5\\ \color{blue}{\text{or}}\\ \\ x_k = 2\sqrt{-a}\;\sin\left(\tfrac{1}{5}\,\arcsin\big(\tfrac{-b}{2\sqrt{-a^5}}\big)-\tfrac{2\pi\,k}{5}\right)\\ \color{blue}{\text{or}}\\ \\ x_k = 2\sqrt{-a}\;\cos\left(\tfrac{1}{5}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^5}}\big))-\tfrac{2\pi\,k}{5}\right)$$

for all five roots $x_k$ with $k =0,1,2,3,4$. Note that all quintics can be reduced, in radicals (using a quadratic Tschirnhausen transformation), to the form,

$$x^5+5ax^3+5bx+c=0\tag2$$

so the general quintic is tantalizingly close to, but not quite, solvable.

P.S. This is directly analogous to the soln of the depressed cubic which all cubics can be reduced to,

$$x^3+3ax+b = 0\tag3$$

where,

$$x_k = 2\sqrt{-a}\;\cos\left(\tfrac{1}{3}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^3}}\big))-\tfrac{2\pi\,k}{3}\right)$$

II. For the tangent quintic: $$x^5 + 5a x^4 + 10 b x^3 + 10a b x^2 + 5b^2 x + a b^2 = 0\tag4$$

$$x = \sqrt{b}\;\frac{1+R^{1/5}}{1-R^{1/5}},\quad R=\frac{a+\sqrt{b}}{a-\sqrt{b}}\\ \color{blue}{\text{or}}\\ \\ x_k =\sqrt{-b}\,\tan\left(\tfrac{1}{5}\,\arctan\big(\tfrac{-a}{\sqrt{-b}}\big)-\tfrac{2\pi\,k}{5}\right)$$

There does not seem to be a commonly known quintic with roots that can be expressible in terms of elementary functions (trigonometric, logarithmic, hyperbolic, etc) that cannot be expressed in terms of radicals as well.

• Thanks, I've already learned how to obtain this simple kind of quintics. (Actually, almost any quintic with simple trigonometric roots). – Yuriy S Jul 8 '16 at 20:00
• Or, indeed, hyperbolic roots – Yuriy S Jul 8 '16 at 20:07
• @You'reInMyEye: I'm trying to figure out if radicals can indeed express all quintic roots expressible in terms of the other elementary functions. – Tito Piezas III Jul 8 '16 at 20:10
• @ Tito, according to the paper linked in Dave's comment above, it's still an open conjecture – Yuriy S Jul 8 '16 at 20:14
• @You'reInMyEye: Actually, its different. The author states: "Conjecture 2. The roots $r_i$ of $2x^5-10x+5 = 0$ are not in $E$." This quintic does not have a solvable Galois group so are not expressible in radicals. – Tito Piezas III Jul 8 '16 at 20:43