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I have developed an interest in generalisations of the fibonacci sequence, from tribonacci sequence up to what I'll coin the 'infinibonacci' sequence.

I'm aware that these nth-bonacci sequences require roots of matched nth-degree polynomials in the form $x^n - (x^{n-1} + x^{n-2} + \dots +x^0)=0$ and that 5th-degree polynomials are not guaranteed to be solvable, but I'm wondering if there is a systematic way of showing whether these particular polynomials are solvable by root extraction?

I am prepared to learn anything required to understand the result but am currently starting from smart 2nd year undergraduate level. Thanks.

  • The fundamental theorem of algebra ensures that this polynomial has five roots (of possibly non-unital multiplicity). Are you sure you are interested in factorization by radicals? That's what I believe you mean by "solveable." – parsiad Mar 19 '15 at 05:32
  • That is exactly what I meant by solvable. I meant to be able to generate the 5-nacci and higher, exactly and without any approximating, however it seems that is impossible. – Jeremy Kindler Mar 19 '15 at 11:28
  • It's already challenging to show that all these polynomials are irreducible. The only proof I know uses Rouché's Theorem. – Andrew Dudzik Mar 24 '15 at 01:56
  • Is this out of curiosity, or do you have some particular use in mind for expressing such roots in radicals? I would think that there is a lot you can do, either algebraically or numerically (or, if you're looking at some limiting case, some sort of analysis), that is pretty much agnostic to whether the roots have a particular form. – Milo Brandt Mar 27 '15 at 03:28
  • It was for the purpose of creating specific sequences of natural numbers but I've found a better way since. – Jeremy Kindler Apr 09 '15 at 09:38

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According to Maple's galois function, the polynomial $x^n - (x^{n-1} + \ldots + 1)$ has Galois group $S_n$ for $n$ from $3$ to $9$ (it can't handle polynomials of degree greater than $9$). In particular, for $n = 5$ to $9$ these are not solvable by radicals. I suspect that none of the polynomials for $n \ge 5$ are solvable by radicals.

Robert Israel
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    Dr. Israel, you beat me by a couple of seconds. :) I was testing it with Magma, and it has $S_n$ up to $n=12$. – Tito Piezas III Mar 19 '15 at 06:24
  • Thanks very much for those comments. It seems that I cannot generate 5-nacci and higher sequences without using some approximation. I assume that this conclusion holds for the series sum (which is actually the bit that I'm most interested in)? – Jeremy Kindler Mar 19 '15 at 11:31
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You can test its Galois group using this online Magma calculator. For example, to test the solvable but irreducible quintic $x^5-5x+12=0$, use the command,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-5*x+12; G, R := GaloisGroup(f); G;

Copy and paste. One then finds the order is $10$, hence that quintic is solvable. (All groups with order $<60$ is solvable, but there are solvable groups with order $>60$.)

For the pentanacci,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-x^4-x^3-x^2-x-1; G, R := GaloisGroup(f); G;

It says it is the symmetric group $S_5$. And for higher $n$-nacci (I tried up to $n=12$), we get $S_n$ which are not solvable for $n\geq5$.

P.S. If you are testing other equations, don't forget the asterisk (*) between the numerical coefficient and the variable, like this: 5*x. (I learned that after a while.)

  • Thanks for the great comment and for the link to the Magma calculator. It seems like that only hope for what I'd like to do is that calculating n-th-nacci series, rather than sequences, is solvable, but I somehow doubt it! – Jeremy Kindler Mar 19 '15 at 11:53
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A result of Dedekind says that for any polynomial $p \in \mathbb{Z}[x]$ and any prime $q$ not dividing the discriminant of $p$, then if $p$ factors modulo $q$ into a product of irreducible polynomials with degrees $d_1, \ldots, d_r$, then the Galois group $\text{Gal}(p)$ contains a permutation with cycle structure $(d_1, \ldots, d_r)$.

The discriminant of the Pentanacci polynomial $p_5(x) := x^5 - (x^4 + x^3 + x^2 + x + 1)$ is $9584 = 2^4 \cdot 599$. It is irreducible modulo $5$ and so $\text{Gal}(p_5)$ contains a $5$-cycle. Modulo $3$, we have $$p_5(x) \equiv (x^3 + x^2 + 2 x + 1) (x^2 + x + 2),$$ so $\text{Gal}(p_5)$ contains a product $\sigma$ of a $2$-cycle and a $3$-cycles and thus also the $2$-cycle $\sigma^3$. Now, if $r$ is prime, then a $2$-cycle and an $r$-cycle in $S_r$ together generate all of $S_r$, and in particular the Galois group of $p_5$ is $S_5$. Now, $S_n$ is only solvable iff $n \leq 4$, so the roots of $p_5$ cannot be extracted with radicals.

Tito and Robert's answers used a dedicated CAS command to show that $\text{Gal}(p_n) \cong S_n$ for $n \leq 12$, and we can readily use Dedekind's result to extend this to, say, $n \leq 20$.

For the $13$-nacci polynomial $p_{13}$, applying an argument similar to the $n = 5$ case, and now instead considering the primes $p = 5$ (which gives a $13$-cycle) and $17$ (which gives an element $\sigma$ with $\sigma^{11}$ a transposition), leads to the conclusion that $\text{Gal}(p_{13}) \cong S_{13}$.

For the $14$-nacci polynomial $p_{14}$, factoring modulo $5$ gives a product $\sigma$ of a $4$-cycle and a $9$-cycle, so $\sigma^{18}$ is a transposition, and factoring modulo $19$ gives a $13$-cycle. But $p_{14}$ is irreducible (as it is irreducible modulo $5$), and so its Galois group is transitive, and being a transitive subgroup of $S_n$ that contains a $2$-cycle and an $(n - 1)$-cycle, $\text{Gal}(p_{14})$ is necessarily $S_{14}$ itself.

We can handle the remaining cases similarly. As above, $\sigma$ is an element with cycle structure given by Dedekind's result for the corresponding prime: $$\begin{array}{c|l} n & p \\ \hline 15 & 11 \, (\sigma^5 \text{ a $2$-cycle}), \, 199 \, (\sigma \text{ a $14$-cycle}) \\ 16 & 5 \, (\sigma^6 \text{ a $2$-cycle}), \, 59 \, (\sigma \text{ a $15$-cycle}) \\ 17 & 3 \, (\sigma^{36} \text{ a $2$-cycle}), \, 5 \, (\sigma \text{ a $17$-cycle}) \\ 18 & 5 \, (\sigma^{26} \text{ a $2$-cycle}), \, 17 \, (\sigma \text{ a $17$-cycle}) \\ 19 & 3 \, (\sigma \text{ a $18$-cycle}), \, 13 \, (\sigma^{33} \text{ a $2$-cycle}) \\ 20 & 5 \, (\sigma^{66} \text{ a $2$-cycle}), \, 23 \, (\sigma \text{ a $19$-cycle}) \end{array}$$

Travis Willse
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  • Thanks! That's a great comment. I really need to learn some Galois Theory it would seem. So far the 'furthest' I've gone in maths is some differential equations and vector stuff. It seems like I have a lot left to learn. Where do you think a good place to start would be on learning more about groups? – Jeremy Kindler Mar 19 '15 at 11:51
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    You're welcome, I'm glad you found it useful. Even if you don't know much Galois Theory, you may know that it (and for that matter, Group Theory) was first developed to solve exactly the sort of question you're thinking about, namely, the problem of determining for which quintics one can extract the roots using radicals! I learned some of my group theory from Dummit & Foote's Abstract Algebra. The exposition is generally clear and evocative, but it is very thorough for an introductory text; I enjoy this, though perhaps it's not everyone's style. – Travis Willse Mar 19 '15 at 13:46
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    @jeremykindler Dummit and Foot is a fairly gentle but comprehensive intro to algebra, although my courses used Gallian (gentler still, less comprehensive). Isaacs' Finite Group Theory and Algebra are top notch, both in terms of writing and the amount of work required on your part. – pjs36 Mar 19 '15 at 13:51
  • @JeremyKindler Using a similar argument (that relies on a different set of a cycles that generate the full group $S_n$) I extended the answer to show that $\text{Gal}(p_n) \cong S_n$ for $n \leq 20$. – Travis Willse Mar 24 '15 at 01:43