3

Prove that: $(x^2+x+1) \mid (x^{6n+2}+x^{3n+1}+1) $ and $(x^2+x+1) \mid (x^{6n+4}+x^{3n+2}+1) $.

I saw proof in book with third roots of unity but i didn't understand it, so i want to see other solution but i dont have idea for that.

Krish
  • 7,102
Sinister
  • 585

3 Answers3

8

Hint $\ {\rm mod}\ x^2\!+\!x\!+\!1\!:\,\ 0\equiv (x^2\!+\!x\!+\!1)(x\!-\!1) \equiv x^3-1\,\Rightarrow\,\color{#c00}{x^3} \equiv 1\,\Rightarrow\, \color{#c00}x^{\color{#c00}3n}\equiv 1\,\Rightarrow\,x^{6n}\equiv 1$

Substituting in both $\ x^{3n},x^{6n}\equiv 1\,$ shows they're $\,\equiv x^2\!+\!x\!+\!1\equiv 0,\ $ using also $\,x^4=x\cdot \color{#c00}{x^3}\equiv x\,$

Remark $ $ Why does integer congruence arithmetic ("modular" arithmetic) extend to polynomials? Simply because the proofs of the basic congruence laws (sum and product rules) work universally, i.e. they use only basic ring laws (distributive,commutative,associative), so the proofs remain valid in any commutative ring. This universality becomes clearer when one studies abstract algebra, which studies more structural versions of these ideas (ideals & quotient rings, a.k.a. residue rings).

Community
  • 1
Bill Dubuque
  • 272,048
  • +1 I have always found polynomial congruences really helpful (like modular arithmetic), but for some reason they don't register so easily. They save a lot of work though. – Mark Bennet Jan 04 '15 at 20:15
  • 1
    @Mark I added a remark which might help it register more intuitively. – Bill Dubuque Jan 04 '15 at 20:56
3

$x^{6n+2}+x^{3n+1}+1 = (x^2+x+1) + x^2(x^{6n}-1) + x(x^{3n}-1)$

We know, $x^3-1\mid x^{3n}-1$ and $x^3-1 \mid x^{6n} - 1$ and $x^2+x+1 \mid x^3-1$ which proves the claim.

As for the second part note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.

r9m
  • 17,938
  • @OP This is essentially the same as the hint in my answer, except it is expressed in terms of divisibility instead of congruences. You may find it helpful to explore the relationship, e.g. my hint in divisibility language is $\ x^2!+!x!+!1\mid x^3!-!1\mid x^{3n}!-1.\ $ – Bill Dubuque Jan 04 '15 at 20:03
2

Although you have asked for another method, but I would still do the method the question wanted you to do to complement the answers already posted. Consider the equation $x^3=1$ So $x^3-1=0$ Which gives us $(x-1)(x^2+x+1)=0$

If $x^2+x+1$ is factorized, we get it's roots as $\frac{-1+\sqrt3i}{2}$ and $\frac{-1-\sqrt3i}{2}$, which are called $\omega$ and ${\omega}^2$. That is why we say 1,$\omega$ and ${\omega}^2$ are the cube roots of unity.

Putting $\omega$ and ${\omega}^2$ in our original equation we know that ${\omega}^3=1$ and $\omega + {\omega}^2 +1=0$.

Put $\omega$ and ${\omega}^2$ in your above question. Makes the question a breeze.

Rohinb97
  • 1,702
  • Remark $\ $ The above implicitly uses the following bivariate extension of the Factor Theorem. $\tag{}$ Bifactor Theorem* $\ $ If $\rm,a\ne b\in \Bbb C\ $ and the polynomial $\rm,f(x)\in \Bbb C[x]\ $ then

    $$\rm f(a) = 0 = f(b)\ \iff\ f(x), =, (x!-!a)(x!-!b)\ h(x)\ \ for\ \ some\ \ h(x)\in \Bbb C[x]$$

    More generally it's true in any commutative ring where $\rm,a!-!b,$ is cancelable, see here. $\ $

     – Bill Dubuque Jan 05 '15 at 00:26