There’s the quadratic formula and the one for 3 and 4 factor ones (apologies if I’m wording this wrong), but I just heard that apparently there isn’t anything like a quintic(?) formula and so on. Why is this?? Googling gives me a bunch of confusing terminology that’s difficult to parse.
In: 75
The traditional way of proving this is with Galois Theory which is the kind of math that makes math grad students cry. We’re not going to have a good time trying to do that math in an ELI5 setting, even hand waving away a lot of the complexity.
Fortunately there’s a considerably simpler proof that is more graphical, then it goes into a bit of the “cups and balls” classic magic trick before finally dipping its toes into just a bit of higher level math to actually eek out the final result. There’s a lot of good intuition that comes earlier than that, though, so you can get a lot better grip of the partial result than with the Galois approach even if you get lost before having a full rigorous proof of the final result. I will warn, though, that while this is *more* accessible than Galois Theory it is still quite a lot to digest.
If you prefer your explanation in video form then [this](https://www.youtube.com/watch?v=BSHv9Elk1MU) video does a great job.
As an outline, we will:
1. Discuss complex numbers in polar form
2. Ask what order the roots of a polynomial come in
3. Prove that there is no rational equation for a quadratic
4. Prove that there is no equation for a cubic with only one nth root
5. Explore what nested roots do to our argument
6. Prove that there is no equation for a quintic with finitely many layers of nested roots
Note that that final statement is the strongest thing we can actually prove here–not that no equation can exist, but that any equation that does exist will need something stronger than nth roots. Things like infinite sums are not disproven here and in practice iterative methods like that are how people solve quintic equations most of the time.
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##Polar Complex Numbers
The most popular way to write a complex number is in the form a + bi, where a is the real part and b is the imaginary part. This is great when you want to decompose it into its real and imaginary parts and if you want to multiply two numbers together then you can just foil it out, but it isn’t obvious how you’d go about taking the nth root of such a number. To make that easier we can convert the number to polar form, r*e^(iθ). Here r is the magnitude of the number–just do some Pythagoras on a and b–and θ is the angle that the point makes with the real axis. Euler’s Identity is helpful in understanding this format.
What’s really useful about this way of writing a complex number is that multiplying is trivial: if you have r1e^(iθ1) * r2e^(iθ2) then the product is r1*r2*e^(i(θ1+θ2)). You multiply the magnitudes and *add* the angles. This makes operations like squaring or cube rooting a lot easier: you apply the square or cube root to the magnitude and then the angle gets doubled or divided by 3, etc.
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##Ordering Roots
We know from the Fundamental Theorem of Algebra that an N-degree polynomial will have n roots. That is to say, we can write the polynomial in the form a(x-x1)(x-x2)(x-x3) … (x -xn) where x1, x2, x3, … , xn are complex numbers that constitute the roots (and a is just a scaling constant, trivially found as the coefficient of x^n in the original polynomial). We then ask: which of the roots is x1, which one is x2, and so on? What order do they come in?
To show that we can’t answer that question we first assume that we can–we’re going to prove by contradiction. We take a simple polynomial: x^2 = 4. This trivially has the roots x = +2 and x = -2. We assume that our ordering rule selects that x1 = +2 and x2 = -2.
From here we go on a journey: we are going to slowly and smoothly vary the coefficients of the polynomial. In response the roots will also vary smoothly. If we can get the coefficients back to where they started then they should give the same roots.
The journey we take is through the complex plane. We use polar representations to make the square root easier. We write 4 as 4*e^(θi) which means our roots are x1 = 2*e^(.5θi) and x2 = -2e^(.5θi). As we increment θ up from 0 we watch the roots move in a circle. When the coefficient gets to 4*e^(πθi) (i.e. just -4) the roots are x1 = 2e^(.5πi) (i.e. 2i) and x2 = -2e^(.5πi) (i.e. -2i), as expected. We keep on going until we get the coefficient back to 4e^(2πθ). This is just 4, right where we started. Here the roots are x1 = 2e^(πθ) and x2 = -2e^(πθ), or x1 = -2 and x2 = +2. As expected we have returned to the same roots, but they have changed order!
What this shows is that there is no way to put the roots in a specific ordering if that ordering is to be continuous
The quadratic formula is a few short operations. The [cubic](https://math.vanderbilt.edu/schectex/courses/cubic/) formula is a few lines. The [quartic](https://en.wikipedia.org/wiki/Quartic_equation#The_general_case) formula takes a couple of pages to describe. The quintic equation is unknown, but presumably much much worse.
At which point the question is, “who needs it?” Any scientist or engineer faced with a quintic equation can use an iterative numerical method, which repeats a short easy bit of math over and over to find an approximate answer that can be as accurate as you need. A mathematician cares about proving the *existence* and *properties* of the solution, but doesn’t care what the solution actually *is*.
The [properties of the solutions](https://en.wikipedia.org/wiki/Geometrical_properties_of_polynomial_roots) of *all* polynomial equations of any degree are well understood, so the mathematician’s work is done, and there are plenty of great [root-finding algorithms](https://en.wikipedia.org/wiki/Root-finding_algorithms) for the scientists and engineers to use, so the quintic formula, if it existed, would be just a useless curiosity.
First, when they say “no formula”, they mean “no formula with this limitations”:
* only contains 4 arithmetic operations and roots (of any degree)
* only finite amount of operations (no infinite sums or such)
It is absolutely possible to create a quintic formula by violating those rules. Quintics can be, for example, solved with Newton’s method (which involves an infinitely repeating steps, that converge to the answer).
Now, let’s examine the relationship between the roots and coefficients. We have to use *complex numbers* here – unfortunately, many nice polynomial properties just doesn’t work when observed only on real numbers. Just pretend, that coefficients and roots are 2D points on the plane, instead of points on the line.
1. The relationship is **continuous** – it has no sudden jumps. Slightly “wiggling” coefficients will only slightly “wiggle” the roots, and the reverse is also true.
2. The coefficients are **ordered** – swapping two coefficients changes the roots.
3. The roots are **unordered** – swapping roots does not change the coefficients
Now, what properties should a root formula have?
1. If the formula is continuous (has no sudden jumps), then it must be **multi-valued** – it must produce all roots at once. Otherwise, it would be impossible to gradually swap the roots – continuity demands the formula to track the root as it moves, but single-valueness demands it to return to the original value.
2. If we run formula “in reverse”, it must take any possible rearrangement of roots, and “consume” it – reorder back into a fixed order of coefficients.
3. The n-th root in complex numbers has exactly “n” answers. However, it has only one “coefficient” to play with (the number under the root), so the answers of the root cannot go all possible rearrangements – they can only go **cyclic rearrangements**.
4. The 4 arithmetic operations are single-valued and cannot rearrange roots.
So all boils down to one question – can rearrangements of N elements be “consumed” by only cyclic rearrangements?
* For N=2, 3, 4 the answer is “yes”.
* However, starting with N=5 there is a problem: a cycle of 3 elements can be broken down into two other cycles of 3 elements. That means, each time we “consume” a 3-cycle, we break 2 other ones. So it is impossible to consume all 3-cycles in 5 element group – therefore a 5-degree solution formula with roots cannot exist.
There are formulae, but what there aren’t are formulae that only involve addition, subtraction, multiplication, division, and nth roots (eg, square roots or cube roots).
As for why, that’s a question that vexed almost 300 years of mathematicians, until it was finally proven by a guy called Abel, and later another guy called Galois, who provided a general framework for this kind of question before being killed in a duel at the age of 19.
Lesson 1: if you like mathematics, don’t get involved in 19th century French politics.
Here’s roughly how Galois’ proof works.
**The first big insight** is that if you have a system of numbers, (eg, the real numbers, you can take a polynomial that doesn’t have a solution (eg, x^2 + 1 = 0), then pretend it does have solutions, and get a bigger number system. Eg, if we say “actually, x^2 + 1 = 0 has a solution, and we call it i”, then the real numbers turn into the complex numbers.
Galois came up with a general system for “extending” number systems with roots of polynomials, and used the system to shoot down a whole lot of questions that had been plaguing mathematicians since the Ancient Greeks, eg “how do you trisect an angle with ruler and compass”? (Galois: you can’t, and here’s why). or “how do you use ruler and compass to make a cube whose volume is twice that of a given one?” (Galois: you can’t, and here’s why).
Eg, we can extend the rational numbers with a solution to x^3 – 2 = 0, getting numbers of the form a + b 2^(1/3) + c 2^(2/3).
**The second big insight** is that sometimes, the roots of a polynomial are all pretty much indistinguishable. Not the *same*, but indistinguishable. Eg, x^2 + 1 has two roots, i and -i. But these two roots have all the exact same properties. The engineers use j instead of i to represent “the” root of x^2 + 1. But what if they were actually using j for -i all along? There’s no way anyone could know. If you take any chunk of maths, and replace all the i with -i, and all the -i with i, then it comes out exactly the same.
We could call this a “symmetry” of the roots of x^2 + 1. And there are, it turns out, two symmetries: {i -> -i and -i -> i}, and {i -> i and -i -> -i}.
If we collect all the symmetries of the roots of a polynomial together into a collection (mathematicians call this a “group” of symmetries), then there are all kinds of possibilities for what that “group” of symmetries might look like.
**The third big insight** is some interesting things about what kinds of “groups” of symmetries are possible – in general, and of roots of polynomials.
One important type of group is when you just rotate a number of things: For example {A -> B, B -> C, C -> D and D -> A}. The letters A, B, C and D just get rotated.
There are other, more complicated groups of symmetries: for example, if we look at all possible permutations of three things, there are 6 of them: first, there are three cycles:
* {A->A, B->B, C->C}, (kind of a trivial cycle, I know).
* {A->B, B->C, C->A},
* {A->C, B->A, C->B},
And there are also three swaps:
* {A->A, B->C, C->B},
* {A->C, B->B, C->A},
* {A->B, B->A, C->C}.
If A, B and C these were roots of polynomials, it would represent a situation (like the complex numbers i and -i), where the roots could be swapped or cycled any old way, and nobody would be able to tell the difference.
Not all polynomials are like this. For example,
x^3 + 2 = 0 over rational numbers has three roots, but if you decide to shuffle them, then as soon as you decide what to replace the first one with, then your decisions about the others are forced. It turns out that all the shuffles of the roots are cycles.
This is typical of polynomials of the form x^n – a = 0.
And it turns out this cuts both ways: if your group of roots has only cycles (or powers of cycles), then your polynomial has to be x^n – a = 0 in disguise.
**The fourth big Insight:**
Not every group of root shuffles has just cycles, but many of them can be built out of things that are just cycles.
For example: We could build up this group:
* {A->A, B->B, C->C},
* {A->B, B->C, C->A},
* {A->C, B->A, C->B},
* {A->A, B->C, C->B},
* {A->C, B->B, C->A},
* {A->B, B->A, C->C}
by saying “let’s just have all the cycles of A, B and C”:
* {A->A, B->B, C->C},
* {A->B, B->C, C->A},
* {A->C, B->A, C->B},
(so this would correspond to taking a cube root) and then saying “let’s also have the cycles that swap A and B” (so that would correspond to a square root).
So if you had a polynomial whose 3 roots could be shuffled in every possible way, it would be possible to solve it by doing a cube root, and then a square root, with possibly some normal addition, multiplication etc between, before, and after.
**The fifth big Insight:**
So, what polynomials can be solved with roots? Galois showed this is the same question as “which groups of shuffles can be built up out of cycles?”
And people who studied group theory already knew enough about that second question to give the famous answer about quintics.
It turns out that any group of shuffles of 1, 2, 3 or 4 roots can be built out of cycles, but as soon as you add that 5th root, there are some groups of shuffles that can’t; and so, the 5th degree polynomials corresponding to that group can’t be solved with just roots.
As you can see, there’s a lot .. a lot of somewhat heavy maths behind it, which I’ve tried (and probably failed) to ELI5.
Galois was a genius, and spent the night before his duel frantically writing letters to friends, and writing down pages and pages of advanced maths that was still in his head and nowhere else. This was at the age of 19. If he had stayed out of French politics, or carefully avoided the girl his political enemies sent to entrap him, he might be a household name – who knows what else he would have discovered, given a chance at a long and productive life.
The (most common) proof is way, WAY beyond ELI5 level. It occupied about half a semester of graduate-level abstract algebra at my university, and the progression of the proof is not at all obvious, or at least wasn’t at all obvious to me (and I am pretty good at this sort of thing). It’s a pretty extraordinary piece of mathematics, all the more so for having been done by one guy at age 18.
But I’ll sketch it out, broadly speaking.
——
So. We need to have some way of describing a mathematical object that tells us whether a polynomial has a solution. It’s not even obvious where we would *start* with this. But here’s an idea: what if we think about what would happen if we had the rational numbers and we “attached” the roots of the polynomial to it.
So for example, if we have the polynomial x^2 + 1 = 0, we can “attach” +i and -i (the two roots of this polynomial) to the rational numbers. This turns out to get us all numbers of the form a + bi where a and b are both *rational* numbers (note that this is a subset of the complex numbers, not all of them). We call this operation a **field extension**, because both objects are fields, a kind of mathematical object that “acts like” the rational numbers in some sense. (Specifically, a field is a set on which you can add, subtract, multiply, and divide by non-zero values, and it turns out that “take the rationals and stick the roots of a polynomial on them” always results in such an object.)
More abstractly, we take the field Q of rational numbers (Q is the usual symbol for them), and for any polynomial P with roots x1, x2, x3, …, xn we can create a new field which we write Q[x1, x2, x3…, xn] of the rational numbers with these extra roots. We call this new, bigger, field the splitting field of our polynomial P.
This is helpful because it takes us out of the realm of our polynomial, and into the realm of talking about abstract algebraic objects (which is usually where mathematicians like to live). But how the hell does this help us?
—–
Well, it turns out – and again this is not at all obvious and takes a lot of work to prove – that the relationship between Q and the splitting field Q[x1, x2, …, xn] encodes the information we want.
Return to our earlier example: we took the rationals and added +i and -i to them. But the properties of these “rational complex” numbers wouldn’t change if we swapped +i and -i, and we wouldn’t be messing with the rationals that lack any imaginary part by doing so. In other words, we have an operation that:
* Preserves the properties of the bigger field (the splitting field) *and*
* Does not change the smaller field, in our case Q, at all.
If we take all the operations that do this – and there may be quite a few of them – they form another kind of mathematical structure called a *group*. Groups are a very common type of mathematical object, because they in some sense describe symmetries and transformations in a very general way, and studying the symmetries of an object is often a way to understand its properties.
This particular group, which we call the *Galois group*, encodes information about how our polynomial’s roots extend the rational numbers. In some cases, when the polynomial’s roots are all rational themselves, it doesn’t extend the rationals at all (because you could already “get to” those numbers). In other cases, it extends the rationals in various ways.
—–
Okay, but how does *that* help us?
Well, it turns out that the properties of the Galois group tell us something about the original polynomial’s roots – namely, if they can be described using just arithmetic and nth root operations.
It turns out that roots that can be described this way extend the rationals in a very specific kind of way. The resulting extensions – and their corresponding Galois groups from the previous section – can only take on a particular kind of structure.
Since we can build up our full extension by *all* the roots by extending by each one one at a time, we get a sequence of Galois groups for each of those extensions in turn. And it turns out that if that sequence has particular properties – which turn out to be equivalent to the full Galois group being something called a [solvable group](https://en.wikipedia.org/wiki/Solvable_group), the original polynomial had a solution that could be written using only arithmetic and radicals.
—–
Finally, we show that there exists at least one polynomial of degree 5 – it turns out that x^5 – x – 1 works – whose Galois group is *not* solvable. Then we work backward:
* This polynomial has a non-solvable Galois group.
* Therefore, its chain of extensions does not have the property that it would have if every root could be written with only arithmetic and radials.
* Therefore, **this** polynomial has no solution with only arithmetic and radicals
* Therefore, there is no general formula that can solve every polynomial that way.
It turns out that the smallest non-solvable Galois group requires at least five roots, and therefore a polynomial of degree at least 5, which explains why degree five can’t be solved with a formula using only arithmetic and roots (and 2, 3, and 4 can be).
First, let’s clarify some terms. When we talk about “solving” a polynomial, we’re talking about finding its roots. The roots of a polynomial are the values of x that make the polynomial equal to zero. For example, if we have a quadratic polynomial like (x^2 – 3x + 2), the roots are the values of x that make (x^2 – 3x + 2 = 0). In this case, the roots are (x = 1) and (x = 2).
Now, for polynomials of degree 2 (quadratics), 3 (cubics), and 4 (quartics), we have formulas that can find the roots. These are the quadratic formula, Cardano’s formula, and Ferrari’s formula, respectively. These formulas are great because they give us a systematic way to find the roots of any polynomial of degree 2, 3, or 4.
However, for polynomials of degree 5 and higher, things get more complicated. In the 19th century, a mathematician named Évariste Galois proved that there is no general formula, using only the usual algebraic operations (addition, subtraction, multiplication, division, and root extraction), that can find the roots of a fifth-degree polynomial or higher. This is known as the Abel–Ruffini theorem.
The reason for this has to do with the nature of the symmetries of the roots of polynomials. For polynomials of degree 4 and lower, these symmetries form what’s called a “solvable group,” which means that there’s a systematic way to break down the problem of finding the roots into simpler problems. But for degree 5 and higher, the symmetries form a more complex type of group, called a “non-solvable group,” and there’s no way to break down the problem in the same way.
This doesn’t mean that we can’t find the roots of a fifth-degree polynomial at all. It just means that there’s no one-size-fits-all formula that works for all fifth-degree polynomials. We have to use other methods, like numerical approximation, to find the roots in general.
The real answer involves [Galois theory](https://math.stackexchange.com/questions/1733072/the-quintic-equation-why-is-there-no-closed-formula), but I’m not sure any 5-year old would understand that. It was apparently too difficult for even the famous mathematician Poisson (of Poisson distribution and Poisson equation fame), who said “”[Galois’s] argument is neither sufficiently clear nor sufficiently developed to allow us to judge its rigor”.
I’ll add that in practice, nobody usually cares, because for a fifth-order polynomial, you are guaranteed at least one real root. You can divide by that root to get a fourth-order polynomial that you could then solve analytically. Or, just keep solving numerically…
There is a formula. You can use long division to factor any length polynomial into 2 and 3 term factors. From there the roots can be determined using the quadratic formula. This is a trial and error method tho, so generally using a computer or plotting a graph is easier.
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