# An elementary proof of the insolvability of the quintic

When I was in middle school, I remember my teacher telling me, after I learned the quadratic formula, that there was a general formula for solving cubic and quartic equations, but no such formula existed for solving the quintic. This was also when I first heard the infamous story of young Galois’s death from a duel.

Using my profound middle-school logic, I took this story as a challenge to devise my own formula for solving the quintic. Naturally, my efforts came up short.

When I was in high school, with this obsession still fully intact, I attempted to read through the wonderful monograph Field Theory and Its Classical Problems. Here’s the MAA review of this book:

Hadlock’s book sports one of the best prefaces I’ve ever read in a mathematics book. The rest of the book is even better: in 1984 it won the first MAA Edwin Beckenbach Book Prize for excellence in mathematical exposition.

Hadlock says in the preface that he wrote the book for himself, as a personal path through Galois theory as motivated by the three classical Greek geometric construction problems (doubling the cube, trisecting angles, and squaring the circle — all with just ruler and compass) and the classical problem of solving equations by radicals. Unlike what happens in most books on the subject, all three Greek problems are solved in the first chapter, with just the definition of field as a subfield of the real numbers, but without even defining degree of field extensions, much less proving its multiplicativity (this is done in chapter 2). Doubling the cube is proved to be impossible by proving that the cube root of 2 cannot be an element of a tower of quadratic extensions: if the cube root of 2 is in a quadratic extension, then it is actually in the base field. Repeating the argument, we conclude that it is not constructible because it is not rational. A similar argument works for proving that trisecting a 60 degree angle is impossible. Of course, proving that duplicating the cube is impossible needs a different argument: chapter 1 ends with Niven’s proof of the transcendence of π.

After this successful bare-hands attack at three important problems, Chapter 2 discusses in detail the construction of regular polygons and explains Gauss’s characterization of constructible regular polygons, including the construction of the regular 17-gon. Chapter 3 describes Galois theory and the solution of equations by radicals, including Abel’s theorem on the impossibility of solutions by radicals for equations of degree 5 or higher. Chapter 4, the last one, considers a special case of the inverse Galois problem and proves that there are polynomials with rational coefficients whose Galois group is the symmetric group, a result that is established via Hilbert’s irreducibility theorem.

Many examples, references, exercises, and complete solutions (taking up a third of the book!) are included and make this enjoyable book both an inspiration for teachers and a useful source for independent study or supplementary reading by students.

As I recall, I made it successfully through the first couple of chapters but started to get lost with the Galois theory somewhere in the middle of Chapter 3. Despite not completing the book, this was one of the most rewarding challenges of my young mathematical life. Perhaps one of these days I’ll undertake this challenge again.

Anyway, this year I came across the wonderful article The Abel–Ruffini Theorem: Complex but Not Complicated in the March issue of the American Mathematical Monthly. The article presents a completely different way of approaching the insolvability of the quintic that avoids Galois theory altogether.

The proof is elementary; I’m confident that I could have understood this proof had I seen it when I was in high school. That said, the word “elementary” in mathematics can be a bit loaded — this means that it is based on simple ideas that are perhaps used in a profound and surprising way. Perhaps my favorite quote along these lines was this understated gem from the book Three Pearls of Number Theory after the conclusion of a very complicated proof in Chapter 1:

You see how complicated an entirely elementary construction can sometimes be. And yet this is not an extreme case; in the next chapter you will encounter just as elementary a construction which is considerably more complicated.

I believe that a paid subscription to the Monthly is required to view the above link, but the main ideas of the proof can be found in the video below as well as this short PDF file by Leo Goldmakher.

# Engaging students: Factoring quadratic polynomials

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Chris Brown. His topic, from Algebra: factoring quadratic polynomials.

What interesting word problems using this topic can your students do now?

The ability to factor quadratic polynomials is at the essence of many two-dimensional kinematic word problems that students will encounter in the future physics courses. One specific word problem that students can now solve, is, “In a tied game between the Golden State Warriors and the Houston Rockets, Steph Curry has the ball for his team. If Steph Curry is 20ft away from the basketball hoop and throws the basketball up in the air at a velocity of 3 m/s, will he be able to make the shot if 3 seconds is left on the clock and win the game for his team? Consider this to be an isolated system.” This special type of problem gives them initial distance, final distance, initial velocity, and acceleration. He student then needs to solve for time, which turns this into a quadratic scenario that requires factoring. I feel like this problem situation is super relevant to the high school age group as it seems to be popular amongst that age group, and with this problem they can extend it to any real-world scenario that searches for time when given distance and velocity.

How does this topic extend what your students should have learned in previous courses?

When factoring quadratic equations, one of the universal methods of factoring is called factoring by grouping. Let’s identify a quadratic equation to be ax2 + bx + c = 0. When factoring by grouping, the students must first multiply ‘a’ and ‘c,’ and then find factors of the product which sum to ‘b’. Let’s call these specific factors ‘n’ and ‘m’. Thus far, this brings in students abilities to create factor trees from 3rd grade mathematics. The next step requires students to replace ‘b’ with the factors ‘n’ and ‘m,’ such that we now have ax2 + nx + mx + c = 0. Now the students have to group the ‘ax2’ term and ‘c’ with either the ‘nx’ and ‘mx’ terms in such a way that when the greatest common divisor is pulled away, what’s left is identical for each group. The ability to identify the greatest common divisor between two terms stems from what they learned in 5th grade mathematics. Then, the last step would be to factor out the common term. This entire process, which was not completed here, has used two very fundamental skills from elementary mathematics.

How can technology be used to effectively engage students with this topic?

I believe Symbolab is an amazing website, that the students can use to aid them in the understanding of the process of factoring quadratic polynomials. I chose this website, because it focuses on the process of factoring and uses common language to explain their steps which the students should be aware of. Lastly, I love this website because it gives students the option to hide the steps and just see the answer. With this, the students can type in random quadratics and work towards the solution, and if they get stuck, they can see all the steps. All in all, it is an amazing way to practice the skill of factoring quadratic equations for as long as they please!

Here is the link to Symbolab: https://www.symbolab.com/solver/factor-calculator/factor%20x%5E%7B2%7D-4x%2B3%3D0

# My Favorite One-Liners: Part 90

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a typical problem that arises in Algebra II or Precalculus:

Find all solutions of $2 x^4 + 3 x^3 - 7 x^2 - 35 x -75 =0$.

There is a formula for solving such quartic equations, but it’s very long and nasty and hence is not typically taught in high school. Instead, the one trick that’s typically taught is the Rational Root Test: if there’s a rational root of the above equation, then (when written in lowest terms) the numerator must be a factor of $-10$ (the constant term), while the denominator must be a factor of $2$ (the leading coefficient). So, using the rational root test, we conclude

Possible rational roots = $\displaystyle \frac{\pm 1, \pm 3, \pm 5, \pm 15, \pm 25, \pm 75}{\pm 1, \pm 2}$

$= \pm 1, \pm 3, \pm 5, \pm 15, \pm 25, \pm 75 \displaystyle \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2}, \pm \frac{25}{2}, \pm \frac{75}{2}$.

Before blindly using synthetic division to see if any of these actually work, I’ll try to address a few possible misconceptions that students might have. One misconception is that there’s some kind of guarantee that one of these possible rational roots will actually work. Here’s another: students might think that we haven’t made much progress toward finding the solutions… after all, we might have to try synthetic division 24 times before finding a rational root. So, to convince my students that we actually have made real progress toward finding the answer, I’ll tell them:

Yes, 24 is a lot\dots but it’s better than infinity.

# My Favorite One-Liners: Part 85

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s one-liner is one that I’ll use when I want to discourage students from using a logically correct and laboriously cumbersome method. For example:

Find a polynomial $q(x)$ and a constant $r$ so that $x^3 - 6x^2 + 11x + 6 = (x-1)q(x) + r$.

Hypothetically, this can be done by long division:

However, this takes a lot of time and space, and there are ample opportunities to make a careless mistake along the way (particularly when subtracting negative numbers). Since there’s an alternative method that could be used (we’re dividing by something of the form $x-c$ or $x+c$, I’ll tell my students:

Yes, you could use long division. You could also stick thumbtacks in your eyes; I don’t recommend it.

Instead, when possible, I guide students toward the quicker method of synthetic division:

# My Favorite One-Liners: Part 82

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In differential equations, we teach our students that to solve a homogeneous differential equation with constant coefficients, such as

$y'''+y''+3y'-5y = 0$,

the first step is to construct the characteristic equation

$r^3 + r^2 + 3r - 5 = 0$

by essentially replacing $y'$ with $r$, $y''$ with $r^2$, and so on. Standard techniques from Algebra II/Precalculus, like the rational root test and synthetic division, are then used to find the roots of this polynomial; in this case, the roots are $r=1$ and $r = -1\pm 2i$. Therefore, switching back to the realm of differential equations, the general solution of the differential equation is

$y(t) = c_1 e^{t} + c_2 e^{-t} \cos 2t + c_3 e^{-t} \sin 2t$.

As $t \to \infty$, this general solution blows up (unless, by some miracle, $c_1 = 0$). The last two terms decay to 0, but the first term dominates.

The moral of the story is: if any of the roots have a positive real part, then the solution will blow up to $\infty$ or $-\infty$. On the other hand, if all of the roots have a negative real part, then the solution will decay to 0 as $t \to \infty$.

This sets up the following awful math pun, which I first saw in the book Absolute Zero Gravity:

An Aeroflot plan en route to Warsaw ran into heavy turbulence and was in danger of crashing. In desparation, the pilot got on the intercom and asked, “Would everyone with a Polish passport please move to the left side of the aircraft.” The passengers changed seats, and the turbulence ended. Why? The pilot achieved stability by putting all the Poles in the left half-plane.

# My Favorite One-Liners: Part 81

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that hypothetically could appear in Algebra II or Precalculus:

Find the solutions of $x^4 + 2x^3 + 10 x^2 - 6x + 65 = 0$.

While there is a formula for solving quartic equations, it’s extremely long and hence is not typically taught to high school students. Instead, the techniques that are typically taught are the Rational Root Test and (sometimes, depending on the textbook) Descartes’ Rule of Signs. The Rational Root Test constructs a list of possible rational roots (in this case $\pm 1, \pm 5, \pm 13, \pm 65$) to test… usually with synthetic division to accomplish this as quickly as possible.

The only problem is that there’s no guarantee that any of these possible rational roots will actually work. Indeed, for this particular example, none of them work because all of the solutions are complex ($1 \pm 2i$ and $2 \pm 3i$). So the Rational Root Test is of no help for this example — and students have to somehow try to find the complex roots.

So here’s the wisecrack that I use. This wisecrack really only works in Texas and other states in which the state legislature has seen the wisdom of allowing anyone to bring a handgun to class:

What do you do if a problem like this appears on the test? [Murmurs and various suggestions]

Shoot the professor. [Nervous laughter]

It’s OK; campus carry is now in effect. [Full-throated laughter.]

# My Favorite One-Liners: Part 59

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Often I’ll cover a topic in class that students really should have learned in a previous class but just didn’t. For example, in my experience, a significant fraction of my senior math majors have significant gaps in their backgrounds from Precalculus:

• About a third have no memory of ever learning the Rational Root Test.
• About a third have no memory of ever learning synthetic division.
• About half have no memory of ever learning Descartes’ Rule of Signs.
• Almost none have learned the Conjugate Root Theorem.

Often, these students will feel somewhat crestfallen about these gaps in their background knowledge… they’re about to graduate from college with a degree in mathematics and are now discovering that they’re missing some pretty basic things that they really should have learned in high school. And I don’t want them to feel crestfallen. Certainly, these gaps need to be addressed, but I don’t want them to feel discouraged.

Hence one of my favorite motivational one-liners:

It’s not your fault if you don’t know what you’ve never been taught.

I think this strikes the appropriate balance between acknowledging that there’s a gap that needs to be addressed and assuring the students that I don’t think they’re stupid for having this gap.

# My Favorite One-Liners: Part 49

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s post is certainly not a one-liner but instead is my pseudohistory for how the roots of polynomials were found.

When I teach Algebra II or Precalculus (or train my  future high school teachers to teach these subjects), we eventually land on the Rational Root Test and Descartes’ Rule of Signs as an aid for finding the roots of cubic equations or higher. Before I get too deep into this subject, however, I like to give a 10-15 minute pseudohistory about the discovery of how polynomial equations can be solved. Historians of mathematics will certain take issue with some of this “history.” However, the main purpose of the story is not complete accuracy but engaging students with the history of mathematics. I think the story I tell engages students while remaining reasonably accurate… and I always refer students to various resources if they want to get the real history.

To begin, I write down the easiest two equations to solve (in all cases, $a \ne 0$:

$ax + b = 0 \qquad$ and $\qquad ax^2 + bx + c = 0$

These are pretty easy to solve, with solutions well known to students:

$x = -\displaystyle \frac{b}{a} \qquad$ and $\qquad x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In other words, there are formulas that you can just stick in the coefficients and get the answer out without thinking too hard. Sure, there are alternate ways of solving for $x$ that could be easier, like factoring, but the worst-case scenario is just plugging into the formula.

These formulas were known to Babylonian mathematicians around 2000 B.C. (When I teach this in class, I write the date, and all other dates and discoverers, next to the equations for dramatic pedagogical effect.) Though not written in these modern terms, basically every ancient culture on the globe that did mathematics had some version of these formulas: for example, the ancient Egyptians, Greeks, Chinese, and Mayans.

Naturally, this leads to a simple question: is there a formula for the cubic:

$ax^3 + bx^2 + cx + d = 0$

Is there some formula that we can just plug $a$, $b$, $c$, and $d$ to just get the answer?  The answer is, Yes, there is a formula. But it’s nasty. The formula was not discovered until 1535 A.D., and it was discovered by a man named Tartaglia. During the 1500s, the study of mathematics was less about the dispassionate pursuit of truth and more about exercising machismo. One mathematician would challenge another: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” Then the second mathematician would solve it and challenge the first: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” And so on. Well, Tartaglia came up with a formula that would solve every cubic equation. By plugging in $a$, $b$, $c$, and $d$, you get the answer out.

Tartaglia’s discovery was arguably the first triumph of the European Renaissance. The solution of the cubic was perhaps the first thing known to European mathematicians in the Middle Ages that was unknown to the ancient Greeks.

In 1535, Tartaglia was a relatively unknown mathematician, and so he told a more famous mathematician, Cardano, about his formula. Cardano told Tartaglia, why yes, that is very interesting, and then published the formula under his own name, taking credit without mention of Tartaglia. To this day, the formula is called Cardano’s formula.

So there is a formula. But it would take an entire chalkboard to write down the formula. That’s why we typically don’t make students learn this formula in high school; it’s out there, but it’s simply too complicated to expect students to memorize and use.

$ax^4 + bx^3 + cx^2 + dx + e = 0$

The solution of the quartic was discovered less than five years later by an Italian mathematician named Ferrari. Ferrari found out that there is a formula that you can just plug in $a$, $b$, $c$, $d$, and $e$, turn the crank, and get the answers out. Writing out this formula would take two chalkboards. So there is a formula, but it’s also very, very complicated.

Of course, Ferrari had some famous descendants in the automotive industry.

So now we move onto my favorite equation, the quintic. (If you don’t understand why it’s my favorite, think about my last name.)

$ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0$

After solving the cubic and quartic in rapid succession, surely there should also be a formula for the quintic. So they tried, and they tried, and they tried, and they got nowhere fast. Finally, the problem was solved nearly 300 years later, in 1832 (for the sake telling a good story, I don’t mention Abel) by a French kid named Evariste Galois. Galois showed that there is no formula. That takes some real moxie. There is no formula. No matter how hard you try, you will not find a formula that can work for every quintic. Sure, there are some quintics that can be solved, like $x^5 = 0$. But there is no formula that will work for every single quintic.

Galois made this discovery when he was 19 years old… in other words, approximately the same age as my students. In fact, we know when wrote down his discovery, because it happened the night before he died. You see, he was living in France in 1832. What was going on in France in 1832? I ask my class, have they seen Les Miserables?

France was torn upside-down in 1832 in the aftermath of the French Revolution, and young Galois got into a heated argument with someone over politics; Galois was a republican, while the other guy was a royalist. More importantly, both men were competing for the hand of the same young woman. So they decided to settle their differences like honorable Frenchmen, with a duel. So Galois wrote up his mathematical notes one night, and the next day, he fought the duel, he lost the duel, and he died.

Thus giving complete and total proof that tremendous mathematical genius does not prevent somebody from being a complete idiot.

For the present, there are formulas for cubic and quartic equations, but they’re long and impractical. And for quintic equations and higher, there is no formula. So that’s why we teach these indirect methods like the Rational Root Test and Descartes’ Rule of Signs, as they give tools to use to guess at the roots of higher-order polynomials without using something like the quadratic formula.

Real references:

http://mathworld.wolfram.com/CubicFormula.html

http://mathworld.wolfram.com/QuarticEquation.html

http://mathworld.wolfram.com/AbelsImpossibilityTheorem.html

http://mathworld.wolfram.com/QuinticEquation.html

http://library.wolfram.com/examples/quintic/

http://library.wolfram.com/examples/quintic/timeline.html

# My Favorite One-Liners: Part 48

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

One of the techniques commonly taught in Algebra II or Precalculus is the Rational Root Test, which is a way of making a list of candidates of rational numbers that might (emphasis, might) be roots of the polynomial. This is a commonly taught method for finding the roots of polynomials whose degree is higher than 3. (Other techniques that are typically taught to students are Descartes’ Rule of Signs and (less commonly) the Upper and Lower Bound Rules.) For example, for the polynomial $f(x) = 2x^3 + 5 x^2 - 2x - 15$.

• The factors of the constant term are $\pm 1, \pm 3, \pm 5$ and $\pm 15$, and so the numerator of any rational root must be one of these numbers.
• The factors of the leading coefficient are $\pm 1$ and $\pm 2$, and so the denominator of any rational root must be one of these numbers.
• In conclusion, if there’s a rational root, then it’s $\pm 1, \pm 3, \pm 5, \pm 15, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}$ and $\pm \frac{15}{2}$. In other words, we have a list of 16 possible rational roots. Not all of them will be roots, of course, since the cubic polynomial only has at most three distinct roots. Also, there’s no guarantee that any of them will be roots. The only way to find out if any of them work is by testing them, usually using synthetic division.

So, after a practice problem or two, I’ll ask my students,

What guarantee do you have that at least one of the possible rational roots will actually work?

After letting them think for a few seconds, I give them the answer: