Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm very interested to know how I can factorise $a^{10} +a^5 +1$ in two factors with integer coefficients. I've tried a lot but I don't have any idea how do that.

share|cite|improve this question
You can immediately get the answer using… – Shai Covo Mar 2 '11 at 2:13
Does this look like a geometric series? Base=1, ratio=$a^5$. – Ross Millikan Mar 2 '11 at 6:05

Another hint: can you factor $a^{15}-1$? In more than one way?

share|cite|improve this answer
A big winner. I upvoted all the answers to this, as they are all helpful, but if I could give more votes to one, this would be it. – Ross Millikan Mar 2 '11 at 6:00
Ross, thanks for the compliment! – Michael Lugo Mar 2 '11 at 21:57
Sorry, but I can not realize this hint. Why that? sorry – tom Mar 2 '11 at 23:19
tom: because polynomials have unique factorization, if you can find multiple distinct factorizations of $a^{15}-1$, you're guaranteed to be able to break them down further into a 'common' factorization. Now, your polynomial in $a$ is one factor in an obvious factorization of $a^{15}-1$; there's another obvious factorization, and comparing the two should prove fruitful. – Steven Stadnicki Mar 23 '11 at 18:27
See my answer for the underlying divisibility theory. – Bill Dubuque Mar 24 '11 at 3:24

A nice way to proceed is to exploit the fact that the polynomial sequence $\rm\ f_n = (x^n-1)/(x-1)\ $ is a strong divisibility sequence, i.e. $\rm\: (f_m,f_n)\ =\ f_{\:(m,n)}\:, $ where $\rm\:(u,v)\:$ denotes $\rm\:gcd(u,v)\:.\:$ So, for example, $\rm\ (f_3,f_5) = f_{\:(3,5)} = f_1 = 1\:,\:$ and $\rm\ f_3\:|\:f_{15}\ $ via $\rm\ (f_3,f_{15}) = f_{\:(3,15)} = f_3\:.\:$ Combining this with Euclid's Lemma quickly yields the sought factor, namely

$\rm\quad\quad (f_3,f_5) = 1,\ \ f_3,f_5\:|\:f_{15}\ \ \Rightarrow\ \ f_3\:f_5\:|\:f_{15}\ \ \Rightarrow\ \ f_3\:|\:f_{15}/f_5\:,\ $ i.e. $\rm\ \ x^2+x+1\ |\ x^{10}+x^5+1\ $

Note that the above proof shows $\rm\ (a,b) = 1\ \Rightarrow\ f_a\ |\ f_{a\:b}/f_b\ $ for any strong divisibility sequence $\rm\:f_n\:.\:$ For example, for the fibonacci numbers follows $\rm\ f_5\ |\ f_{20}/f_4\ $ i.e. $\rm\ 5\ |\ 6765/3\: =\: 2255\:.$

More generally, using these properties and a little number theory and combinatorics (inclusion-exclusion) one easily derives the basic factorization properties of cyclotomic polynomials.

The proof of the basic property $\rm\: (f_m,f_n)\: =\: f_{\:(m,n)}\:$ is very simple - essentially the same as the proof of the Bezout identity for integers - see my post here. $\:$ This allows one to view the polynomial Bezout identity as a q-analog of the integer Bezout identity. For example, let's compare the Bezout identity for the gcd $\rm\ 3 = (15,21)\ $ in polynomial and integer form:

$\rm\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\ \: \frac{x^3-1}{x-1}\ =\ (x^{15} + x^9 + 1)\ \frac{x^{15}-1}{x-1}\ -\ (x^9+x^3)\ \frac{x^{21}-1}{x-1}$

for $\rm\ x = 1\ $ specializes to $\ \ 3\ =\ (3)\ 15\ -\ (2)\ 21\:.\ $ It is well-worth mastering these binomial divisibility properties since they occur quite frequently in number theoretical applications and, moreover, they provide excellent motivation for the more general study of divisibility theory. $\quad\ \ $ For an introduction see Borovich and Shafarevich: Number Theory.

share|cite|improve this answer

If you really have no other way: Substitute x for a^5, so you get x^2+x+1.

Then use some quadratic formula action, then use some factoring action on those binomials, then just try all possible products of the factors you get until something nice comes out.

I could see this being not so nice a solution though.

share|cite|improve this answer
Why? Substituting for a^5 is the obvious thing that jumped to mind for me. – user7530 Mar 2 '11 at 8:50

Hint: It clearly does not have a linear factor since $\pm1$ is not a root. Try a quadratic factor $a^2+u a+v$, for some integers $u$ and $v$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.