Solution for $x^9 + x^6 + x^4 - x^2 + 1=0$

I learned that $5$ and higher order equations can not all be solved. I read the Abel-Ruffini theorem.

My question:

Let $$x^9+x^6+x^4-x^2+1=0$$

This equation can be expressed in the following way:

$$x^9 + x^6 + x^4 - x^2 + 1=(x^4 + x + 1) (x^5 - x + 1)=0$$

Yes, I know there are only non-radical solution for $x^5-x+1=0$.

But, $x^4+x+1=0$ can be solve.

Finally, I think we can say that, There are $5$ non-radical solution and there are $4$ radical solution for this equation. Is it correct?

My problem is this: For the same equation, I can not understand, Why some of the roots can be expressed in radicals, but the other part can not be expressed? But, the equation is the one/same equation and this polynominal has only one graph..? Is this equation a "semi-solvable" equation?

Note: I do not have math education. I apologize for the flawed question.

• I think the confusion is this: the theorem tells us that the general polynomial equation of degree $>5$ can not be solved, but it says nothing about particular polynomial equations. Clearly we can solve $x^5=0$, say, or $(x-1)(x-2)(x-3)(x-4)(x-5)=0$ say. There is no contradiction in that. – lulu Dec 28 '17 at 14:58
• – Rohan Dec 28 '17 at 15:00
• @lulu Minor typo, it should be $\ge 5$. – Alex Vong Dec 28 '17 at 15:17
• @AlexVong Right, thanks. – lulu Dec 28 '17 at 15:41

Since you have mentioned Abel-Ruffini theorem and user8734617 has mentioned Galois group, I will also mention some Galois theory. As you have noticed, your degree $9$ polynomial can be factored into a product of degree $4$ and degree $5$ polynomial. Indeed, the degree $4$ polynomial is solvable by radicals using Ferrari's method.

However, it is not the case that any degree $5$ polynomial is not solvable by radicals. In fact, Galois's solvability theorem extends Abel-Ruffini theorem by telling us that in a field $F$ with characteristic $0$, any non-constant polynomial $f$ over $F$ is solvable by radicals if and only if its Galois group over $F$, $\operatorname{Gal}(f / F)$, is solvable. Here we are interested in the case $F = \mathbb{Q}$, the rational numbers.

So to prove $x^5 - x + 1$ is not solvable by radicals, we need to show $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ is not solvable. Our strategy is to show $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q}) = S_5$, the symmetric group on $5$ letters. Since $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ is always contained in $S_5$, it suffices to show $S_5$ is contained in $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$.

From this post, $x^p - x + a$ where $a \ne 0$ is irreducible$\bmod{p}$. So we have $x^5 - x + 1$ is irreducible$\bmod{5}$ by letting $p = 5$ and $a = 1$.

Besides, $x^5 - x + 1 = (x^2 + x + 1) (x^3 + x^2 + 1) \bmod{2}$ where both $x^2 + x + 1$ and $x^3 + x^2 + 1$ are irreducible$\bmod{2}$ since they have no roots$\bmod{2}$.

By Dedekind's theorem, $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ contains permutation with cycle type $(a\ b)(c\ d\ e)$ and a $5$-cycle. Now $((a\ b)(c\ d\ e))^3 = (a\ b)^3(c\ d\ e)^3 = (a\ b)$. So $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ contains a transposition and $5$-cycle, which means $S_5$ is contained in $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$.

Finally, since $S_5$ is not solvable, we conclude $x^5 - x + 1$ is not solvable by radicals!

You are correct. Not all roots of the same polynomial are necessarily "equal".

In algebra, we encode this information in so-called Galois group of the field extension $Gal (F:\mathbb Q)$, where $F$ is the smallest field containing all of $\mathbb Q$ and, in addition, all of the roots. It turns out that the elements of that group are some (but not always all!) permutations of the roots.

In your case, the roots of $x^5-x+1$ can only swap with the other roots of $x^5-x+1$, and the roots of $x^4+x+1$ can only swap with the other roots of $x^4+x+1$. In essence, your Galois group is definitely not the whole $S_9$ but is a direct product of some subgroup of $S_5$ and some subgroup of $S_4$.

There are other polynomials of degree 9, however, for which the Galois group is the whole of $S_9$ (all permutations) and for which all the roots are "equal".

• Thank you for the informative and nice answer. – Newstudent Dec 28 '17 at 17:05

That a function has roots of different 'kinds' is not a strange thing. It just expresses the uniqueness of that function. Consider the polynomial function $$(x-\sqrt2)(x-1)(x^2+1)$$ for example. This function has four roots of different kinds, all expressible in terms of radicals, but one is irrational, one is integral and two are complex. This happens all the time. It is not compulsory for an equation to have the same kind of roots.

So when you expect a polynomial equation over the complex field to have either only roots that are radically expressible or otherwise, you create a false and unnecessary distinction. Indeed, what you call non-radical roots are just 'transcendental', so to speak. So there's nothing more strange here than that the diagonal of a square of unit side is an irrational number.

Curious, yes. But not alarming or contradictory in the way you may think.

• Uh...? Because they're roots, then they're algebraic, why transcendental here? – user202729 Dec 28 '17 at 15:55
• @user202729 Strictly speaking, true. But I wanted to talk about that subset of the algebraic numbers that are not radically expressible. What would you call them? – Allawonder Dec 28 '17 at 15:59

What's so special about the fact that some roots can be expressed by radicals whereas some roots can't? Consider the equation $x^3-x^2-2x+2=0$. Its roots are $1$, $\sqrt2$, and $-\sqrt2$. One of them is integer and the other two are irrational. Do you see a problem here? I don't.

I don't see why you claim that “there are only integral solution for $x^5−x+1=0$”. This equation has no integer (or even rational) solution. Or did you mean something else?

• I meant the non-radical solution.. – Newstudent Dec 28 '17 at 15:07