# I Is quintic only polynomial that needs to be proven imposs?

#### swampwiz

I've been trying to prove the impossibility of the quintic "on the cheap" without having to go through a graduate course in abstract algebra (I haven't even done the undergraduate course, although I've been reading up on it a little bit at a time). I understand Bezout's Lemma, with a practical example being the way that it is possible to tighten all the lug nets on a wheel by incrementing by the same angle, without repeating any, so long as the increment in terms of lugs and the total # of lugs have a Greatest Common Divisor of 1 (there is a better term for this, but it has slipped my mind at the moment).

It seems that if the quintic is impossible, then since a quintic is a special case of a hexic, heptic, etc., then those must be impossible too. Also, there is no point in proving that a quartic is possible since a formula already exists for it. Therefore, the full brunt can be directed at the quintic.

As for the quintic, I was reading this

https://www.quora.com/Why-is-there-no-formula-for-solving-polynomials-with-n-4-explain-the-answer-in-such-a-way-that-a-calculus-1-or-precalculus-student-can-understand-if-possible

which goes into explaining that there can be functions expressed in terms of roots of (A,B,C,D) for degree n and (p,q,r) for degree n-1.

n = 3:

AAB + BBC + CCA = p
ABB + BCC + CAA = q

n = 4:

AB + CD = p
AC + BD = q
AD + BC = r

such that no matter how the roots for (A,B,C,D) are swapped around, the resultant (p,q,r) are the same. And for a formula to exist, every root must be swappable with any other root (although I am not so clear on what this exactly means).

However, it is impossible to come up with any function in (A,B,C,D,E) in which the all the permutations of swapping work, and thus it seems that it could be proven through exhaustion that this is the case, although I am not sure what these functions would look like.

I'd appreciate any insight that will help me understand this "on the cheap".

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#### fresh_42

Mentor
2018 Award
I'd appreciate any insight that will help me understand this "on the cheap".
There is no way to explain it at this level. The answer why it is impossible is: "Because $A_5$ is simple."

Now this already shows the problem with "on the cheap":
• What is $A_5$?
• What is "simple"?
• Why is $A_5$ simple?
• What has $A_5$ to do with it?
Well, at least here is a short relation to what you wrote: It is what makes the roots swap. And as you noted, you have no idea what this means. So we step into the next problem: Field theory "on the cheap".
• How do roots translate into group theory?
• How come fields into play?
• and so on and so on and so on
So sorry, no way.

And the answer to the thread title is: No, all polynomials of degree higher than four cannot be solved in a general way, i.e. by root expressions. Five is only the lowest degree of these. And of course some polynomials can always be solved, just not all.

#### Stephen Tashi

Science Advisor
I'd appreciate any insight that will help me understand this "on the cheap".
It would be profitable to discuss the link you gave, even if we don't reach a full understanding.

Most people I've met who have taken a course in abstract algebra know that properties of groups have something to do with the solvability of polynomial equations in radicals and they know the result that the general equation of the fifth degree cannot be solved in radicals - but few of them can explain the connection! It's probably true that you can't fully understand the unsolvability of the quintic without knowing concepts from abstract algebra - however knowing concepts from abstract algebra, by itself, won't guarantee you understand unsolvability in radicals. It's rare to find a textbook on Galois theory that expounds the connection between Galois theory and the solution of polynomial equations in detail - an exception is "Classical Galois Theory with Examples" by Lisl Gaal.

n = 4:

AB + CD = p
AC + BD = q
AD + BC = r

such that no matter how the roots for (A,B,C,D) are swapped around, the resultant (p,q,r) are the same.
What did you mean by "are the same"? For example $p = AB + CD$ becomes $CB + AD = r$ if $A$ and $C$ are swapped. So $p$ need not keep the same numerical value. Perhaps you mean that swapping $A,B,C,D$ among each other swaps $p,q,r$ among each other.

However, it is impossible to come up with any function in (A,B,C,D,E) in which the all the permutations of swapping work,
What does it mean for a function to "work"?

It is possible to find functions defined by algebraic expressions that remain the same under any way of swapping the variables A,B,C,D,E.

For example
$s_1 = A + B + C + D + E$
$s_2 = AB + AC + AD + AE + BC + BD + BE + CD + CE + DE$
$s_3 = ABC + ABD + ABE + ACD + ACE + ADE + BCD + BCE + BDE + CDE$
$s_4 = ABCD + ABCE + ABDE + ACDE + BCDE$
$s_5 = ABCDE$

These are called the "elementary symmetric functions" in the 5 variables $A,B,C,D,E$. (To that list you could also add a function that had a constant value such as $s_0(A,B,C,D,E) = 1$)

From the elementary symmetric functions, we can build other functions that remain unchanged by swapping.
For example, $f(A,B,C,D,E) = s_1 + 17 (s_1 + s_2)/( s_3 - s_4) + (s_2 + 3s_4)^3 + 27$

Such a complicated expression brings up the issue: What do we mean for the general polynomial equation of degree n to be solvable? We don't mean that the solutions are given by taking $n$-th roots of a polynomial function of the coefficients. If you think of the formula for a quadratic, it involves a ratio of polynomials. So defining "solvable" is an interesting challenge.

(Furthermore, the quintic is "solvable" by some definitions of "solvable" see the section "Beyond Radicals" in https://en.wikipedia.org/wiki/Quintic_function/ )

Marty Green's answer is abbreviated and hard to understand. We have to understand the somewhat cryptic passage:

For the third degree equation, I identified these functions:

AAB + BBC + CCA = p
ABB + BCC + CAA = q

A, B and C are the roots of a cubic, but p and q are the roots of a quadratic. You can see that because if you look at pq and (p+q), the elementary symmetric polynomials in p and q, you will see they are symmetric in A, B and C. So they are easily expressible in terms of the coefficients of our original cubic equation. And that's why p and q are the stepping stone which gets us to the roots of the cubic.

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