Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...

Question

How can we prove that

$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...

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I attempted to solve this with Mathematical Induction as follows:

Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..

Basic Step

Let n = 1 ⇒ $x^2 - x + 1$ | $^8 − ^7 + 1$

I then proved that the remainder is 0 using polynomial long division.

$\frac{^{6n+2} − ^{6n+1} + 1}{x^2 - x + 1}$ = $x^6 - x^4 - x^3 + x + 1$ R 0

∴ s(1) is true

Assumption Step

Assume that s(m) is true

⇒ $\frac{^{6m+2} − ^{6m+1} + 1}{x^2 - x + 1}$ = Q(x) where Q(x) is a polynomial

Inductive Step

To prove that s(m+1) is true

⇒ $\frac{^{6(m+1)+2} − ^{6(m+1)+1} + 1}{x^2 - x + 1}$ = T(x) where T(x) is a polynomial(x) is a polynomial

⇒ $\frac{^{6m+8} − ^{6m+7} + 1}{x^2 - x + 1}$ = T(x)

⇒ $\frac{x^6(^{6m+2} − ^{6m+1}) + 1}{x^2 - x + 1}$ = T(x)

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However, I'm unsure of how to proceed from here. I would appreciate it if anyone could help me with this. Thanks!

Answer 1

Let $$x^{6m+2}-x^{6m+1}+1=\lambda(x^2-x+1)$$ $$x^{6m+8}-x^{6m+7}+1=S$$ $$S=x^6(x^{6m+2}-x^{6m+1})+1$$ $$S=x^6(\lambda(x^2-x+1)-1)+1$$ $$S=x^6\lambda(x^2-x+1)-x^6+1$$ Now dividing $S$ by $x^2-x+1$ yields $$\frac{x^6\lambda(x^2-x+1)}{x^2-x+1}+\frac{1-x^6}{x^2-x+1}$$ $$\frac{x^6\lambda(x^2-x+1)}{x^2-x+1}+\frac{(1+x-x^3-x^4)(x^2-x+1)}{x^2-x+1}$$

Answer 2

1. Finds the roots of $x^2-x+1$; they are complex
2. Pick one of them and prove that is a root of $x^{6n+2}-x^{6n+1}+1$; automatically the other one is also a root therefore the two polynomials have common roots, hence the one with lesser number of roots is a factor of the other polynomial.

Answer 3

Define the polynomial

$$P_n(x):=x^{6n+2}−x^{6n+1} + 1$$

Observe that, $x=-1$ is not a root of $x^2-x+1=0$, then multiplying both sides of the equation by $(x+1)$, yields:

$$(x+1)(x^2-x+1)=0$$

This implies that, $x^3= -1,\;x≠-1$.

Therefore, making $x^3\equiv -1$, by $\mod x^2-x+1$, we have:

$$ \begin{align}P_n(x)&\equiv x^2\cdot \left(x^3\right)^{2n}-x\cdot \left(x^3\right)^{2n}+1\\ &\equiv x^2-x+1.\end{align} $$

This completes the proof.

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More explicit explanation:

We can rewrite the polynomial $P_n(x)$ as follows:

$$P_n(x):=(x^2-x+1)Q(x)+R(x)$$

where, $Q(x)$ and $R(x)=ax+b$ are polynomials with some real coefficients.

Let $z\in\mathbb C\setminus \mathbb R$ be a root of $x^2-x+1=0$. Since $z≠-1$, multiplication both sides by $(z+1)$ yields $z^3+1=0$. This implies that, $z^3=-1$ and $z\not\in \mathbb R.$

Putting $x=z$ in the original polynomial indentity, we get:

$$ \begin{align}&P_n(z):=\frac {z^3+1}{z+1}Q(z)+R(z)\\ \implies &P_n(z)=0+R(z)\\ \implies &P_n(z)=R(z)\end{align} $$

Then, using $z^3=-1$, we obtain:

$$ \begin{align}P_n(z):&=z^2\cdot \left(z^3\right)^{2n}-z\cdot \left(z^3\right)^{2n}+1\\ &=z^2-z+1\\ &=0\end{align} $$

This leads to:

$$R(z)=az+b=0.$$

This means, $a=b=0$. Therefore, $R(x)\equiv 0$.

Because, if $a=0$, then $b=0$. Thus $R(x)\equiv 0$.

Otherwise, if $a≠0$, then $z=-\frac ba \in \mathbb R$ which gives a contradiction.

Answer 4

you could use the relationship between S(m) and S(m+1) such that:

$ x^{6m+2} − x^{6m+1} + 1 $ and $ x^6 * (x^{6m+2} − x^{6m+1} )+1 $ if you use the famous trick of +1 -1 in () we get :

S(m+1) = $ x^6 * (x^{6m+2} − x^{6m+1} +1 )/(x^2 −x +1) +(1-x^6) / (x^2 −x +1) $ = $ (x^6 )* S(m) + (1-x^6) / (x^2 −x +1) $ =$ (x^6 )* Q(x) + (-x^4 -x^3 +x +1 ) $ for some Q(x) polynomial = K(x) polynomial as sum of 2 polynomials

so we get if S(m) is true so is S(m+1) , from the induction it always holds (you already calculated the case S(1) ).