How to think of factorising $x^7+x^2+1$ to $(x^2+x+1)(x(x-1)(x^3+1)+1)$ (Thales 2016)

Question

I was just doing the following question:

Find a prime number which divides the number $A=14^7+14^2+1$.

I solved it by finding the result which is $A=105413504+196+1=105413701$ and then trying out all prime numbers till I found that 211 divides it. However, obviously this is extremely tedious. I hence looked at the solution which says that $x^7+x^2+1=(x^2+x+1)(x(x-1)(x^3+1)+1)$ and from here by saying that $x=14$ we get the solution. However I can't seem to think of how to intuitively turn $x^7+x^2+1$ into $(x^2+x+1)(x(x-1)(x^3+1)+1)$. I realize that from the question it is obvious to go looking for factors of A and hence trying to factorize $14^7+14^2+1$, but I can't work out how to go about factorizing it, what are the steps which you need to take in order to factorize a given polynomial. Could you please explain to me how to go about factorizing such an expression and how to intuitively think of each step?

Μιχάλη δε νομίζω πως υπάρχει στάνταρ τρόπος για να παραγοντοποιείς ένα πολυώνυμο. Εφόσον σκέφτηκες ότι πρέπει να παραγοντοποιήσεις το $f(x)=x^7+x^2+1$ τότε μάλλον θα πρέπει να κάνεις κάποιες δοκιμές μέχρι να δεις ότι κάτι δουλεύει. πχ εδώ θα έπρεπε να δεις ότι πρέπει να προσθαφαιρέσεις το $x$ (το οποίο ίσως δεν είναι και τόσο δύσκολο να το φανταστείς). — 1123581321, Sep 27 '20 at 13:37
Αν θες δες αν μπορείς να παραγοντοποιήσεις το $x^{10}+x^5+1$ και το $x^8+x^4+1$ — 1123581321, Sep 27 '20 at 15:22
Does this answer your question? Prime factor of $A=14^7+14^2+1$ — Bill Dubuque, Sep 28 '20 at 08:30
Since that question is closed, I propose that we merge the two questions. — Toby Mak, Sep 28 '20 at 08:31
@Michael Since you ask about intuition, see especially the link I gave in my answer in the proposed dupe to the method of simpler multiples. — Bill Dubuque, Sep 28 '20 at 08:38
@Bill Dubuque I think in the topic https://math.stackexchange.com/questions/2012344 we have no context and attempts. Is this topic should be deleted? — Michael Rozenberg, Sep 30 '20 at 00:19

TheSilverDoe · Answer 1 · 2020-09-27T13:23:42.550

1

You can "see" that $\omega = \exp \left( \frac{2i\pi}{3}\right)$ and $\omega^2= \exp \left( \frac{-2i\pi}{3}\right)$ are roots of $x^7+x^2+1$, therefore $x^7+x^2+1$ factorizes by $x^2+x+1$.

edited Sep 27 '20 at 13:23

answered Sep 27 '20 at 13:22

TheSilverDoe

29,720

1

I'd recommend calling it $\omega$ instead of $j$, as $j$ is a common symbol for $i$. – J.G. Sep 27 '20 at 13:23
hello thanks for your response but what is exp() – Sep 27 '20 at 13:23
@J.G. French notations, my bad :) I edited. – TheSilverDoe Sep 27 '20 at 13:23
@Michael $\exp$ is the exponential... Do you know what are the complex roots of $x^2+x+1$ ? – TheSilverDoe Sep 27 '20 at 13:24
no, I have heard of complex numbers and know how to use them...sort of:). But apart from that no – Sep 27 '20 at 13:25
+1 nice elegance – user2661923 Sep 27 '20 at 13:25
@Michael Don't worry, you'll get to this if you study complex numbers just a little more. – J.G. Sep 27 '20 at 13:26
@Michael Ok, then in a few time studying complex numbers, you will be able to get the roots of any quadratic polynomial. In particular, the two complex roots of $x^2+x+1$ are very useful, and with a little bit of practise, you will see directly that they are also roots of $x^7+x^2+1$... which explains why it factorizes by $x^2+x+1$. – TheSilverDoe Sep 27 '20 at 13:28
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ok thanks a lot @TheSilverDoe – Sep 27 '20 at 13:30

score 1 · Answer 2 · answered Sep 27 '20 at 13:24

By the factor theorem, $x^7+x^2+1$ has no rational roots and no linear factors over $\Bbb Z$, so we look for quadratic factors. To avoid non-integer coefficients, the only such factors are of the form $x^2+ax+1$. The case $a=1$ looks especially promising because, modulo $x^2+x+1$,$$x^3+1\implies x^7=x\implies x^7+x^2+1=0.$$This is essentially the reasoning in @TheSilverDoe's answer, but saves us having to work with explicit roots. As for finding the quotient, note$$\begin{align}x^7+x^2+1&=x(x^6-1)+x^2+x+1\\&=x(x^3-1)(x^3+1)+x^2+x+1\\&=(x^2+x+1)(x(x^3-1)(x+1)+1).\end{align}$$

score 0 · Accepted Answer · answered Sep 27 '20 at 13:51

0

Another way: $$x^7+x^2+1=x^7-x^4+x^4+x^2+1=(x^2+x+1)(x^4(x-1)+x^2-x+1)=$$ $$=(x^2+x+1)(x^5-x^4+x^2-x+1).$$

answered Sep 27 '20 at 13:51

Michael Rozenberg

194,933

Hello Mr. Rozenberg, thank you for responding, however I have one quick question, how did you think of adding $x^4-x^4$ to the polynomial? – Sep 27 '20 at 13:54
@Michael I just saw that $x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$ and $x^7-x^4=x^4(x^3-1)$ gives a factor $x^2+x+1$. – Michael Rozenberg Sep 27 '20 at 13:56
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Understood implicitely, thanks a lot Mr. Rozenberg – Sep 27 '20 at 13:58
@Michael You are welcome! – Michael Rozenberg Sep 27 '20 at 13:59

score -1 · Answer 4 · answered Sep 27 '20 at 17:02

-1

Hint: $x^2+x+1$ always divides $x^a+x^b+x^c$ if $\{a, b, c) \equiv \{0, 1, 2\} \mod 3$. Hence it is easy to note $14^7+14^2+1 \equiv 0 \pmod {14^2+14+1}$, and all that remains is checking primality of $211$.

In case you wish to find the other factor(s), in this case, it may be simpler to consider dividing by $x^3-1$, like so: $$x^7+x^2+1 = x^4(x^3-1) + x(x^3-1)+(x^2+x+1)$$ $$\implies \frac{x^7+x^2+1}{x^2+x+1} = x^4(x-1)+x(x-1)+1=(x-1)(x^4+x)+1$$

answered Sep 27 '20 at 17:02

Macavity

46,381

Please don't repost (other user's) answers from prior duplicates. – Bill Dubuque Sep 28 '20 at 21:32
@BillDubuque I see the resemblance in notation in the first line now that you point out. However yours isn't the first time i have seen that principle, in fact there is a more general version i have mentioned somewhere else in comments in past posts, nor have i posted this one after seeing yours. i hope you agree rest of my answer doesn't have to be from any of your posts either. – Macavity Sep 29 '20 at 03:37

How to think of factorising $x^7+x^2+1$ to $(x^2+x+1)(x(x-1)(x^3+1)+1)$ (Thales 2016)

4 Answers4