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.
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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
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