# Unsolvability of Quintic + equations from a logical perspective

From what I understand, equations higher than the fourth degree cannot have a general solution. I am curious if there are logical reasons for this. By extending ZFC in some non-trivial way, could we possibly attain a solution? More generally, I am trying to understand what it means for a solution to 'not exist'. Do they truly not exist in any way, or do they not exist 'given our current mathematical systems'?

• Just to clarify your title and text, a quartic formula exists (although not very fun to use), but the general quintic is unsolvable by radicals. – Ethan Hunt Jun 1 '16 at 1:45
• A solution to the general polynomial equation of a degree $n \geq 5$ always exists for some domains (it exists sometimes over $\mathbb{R}$, and always over $\mathbb{C}$, for example). What does not exist is a formula, using only elementary operations ($+, -, \times, \div, \sqrt{~}$), for the solutions of the general equation. So if a formula for the solution exists, it must use in someway a non-elementary function or operation. – Henrique Augusto Souza Jun 1 '16 at 1:57

## 2 Answers

It really is a case of "solutions truly don't exist" - at least, solutions of a specific form.

Here's the question, formulated roughly:

If $Q$ is a (non-degenerate) quinitic with rational coefficients, must $Q$ have a solution which can be expressed in terms of the arithmetic operations (plus, times, minus, divides) and $n$th roots, applied to rational numbers?

For a more precise statement of the question, look up "solvable by radicals."

Then Abel's Theorem is that the answer is no: there are quintics, all of whose solutions are not expressible in this form. In particular, they aren't weird quintics at all, on the face of things - I don't know why wikipedia doesn't give an explicit example, but one quintic with no solution which can be expressed by radicals is $$x^5-4x+2$$ (see page 27 of http://vc.bridgew.edu/cgi/viewcontent.cgi?article=1182&context=undergrad_rev).

EDIT: One take-away from all this is that we really shouldn't say "Quintics are unsolvable" - we need to specify what kind of "solution" we're asking for!

Think about it this way: clearly not every (nondegenerate) quadratic equation (with rational coefficients) has a rational solution - e.g. $x^2-2=0$. The unsolvability of the quintic is a statement of a similar form: there are quintics whose solutions lie outside a given "nice" class of numbers, specifically those which can be expressed in a certain way.

Also, complementing my comment, the nonexistence of a radical formula (that's how we call such a formula) follows from Galois Theory, applied to extensions of fields (the domains we're talking about) by roots of polynomials. We see what happens when we put those solutions inside the field, in terms of ''symmetry''. That symmetry is given by an algebraic structure called a Galois group.

A fundamental theorem of Galois Theory states that a polynomial has a radical solution formula if, and only if, the symmetry associated with it is nice enough (it's Galois group must be a solvable group). What is shown is that, for polynomials with degree $> 4$, the Galois group associated is never solvable (this has to do with something called the alternating group, in case you want to look at it).

Now, on a logical point of view. Well, once our current logical system implies the unsolvability by radicals of the higher degree polynomials, trying to force it would not result on a consistent system. We would need to relax certain axioms to the minimum point where we could postulate solvability by radicals (that minimum point must still have the necessary axioms to formulate what solvability by radicals is!) and get something consistent. My guess is that such point does not exist, or at least in a way that leaves the other parts of mathematics sufficiently untouched.

Also, as a philosophical note, our logic system, our fundamentals, are actually more ruled by mathematicians than rulers of mathematics. Reasons for it being this way is that it is sufficiently good to imply the validity of everything we ever proved to day and composed of axioms simple enough to not generate polemic over it. Finding an alternative logic system that fixed solvability of the quintics while breaking everything else on mathematics would be very undesired and certainly uninteresting for a great deal of the mathematical community.

Think of it as a choice: we must start mathematics from somewhere, so let's start from what gets us the most bang for the buck.

• To comment on your third paragraph: you would need to go well below even Peano Arithmetic, and probably below the much much weaker theory $I\Sigma_1$, in order for "quintics are solvable" to be consistent. So yes, the vast majority of mathematics would have to go! – Noah Schweber Jun 1 '16 at 2:20
• Not sure about etiquette, still learning to use this site, but both of these responses answered my question. Thank you – nobody Jun 1 '16 at 2:44