For which $k\in \Bbb N$ does $x^2+x+1$ divide $ (x + 1)^{2k}+x^{2k} + 1$?

Question

I tried many attempt to solve this problem but failed every time.

The polynomial $x^{2k} + 1 + (x + 1)^{2k}$ is not divisible by $x^2 + x + 1$. Find the value of $k \in \mathbb{N}$

My question is: can this be solved with using remainder/factor theorem only?

I haven't learned Number Theory yet so maybe I don't know sufficient number of lemmas.

Answer 1

The roots of $x^2 + x + 1$ are cube roots of $1,$ the one in the second quadrant is $$ \omega = \frac{-1 + i \sqrt 3}{2}\; , $$ the other root is $\omega^2 \; . \;$ We automatically have $$ \omega + 1 = - \omega^2 $$

Since $$ \omega^3 = 1 \; , \; $$ any integer power of $\omega$ is a cube root of $1$ or $1$ itself.

If $k \equiv 0 \pmod 3,$ then both $\omega^{2k}$ and $(\omega + 1)^{2k}$ are equal to $1,$ and that polynomial gets value $3$

If $k \not\equiv 0 \pmod 3,$ then $$(\omega + 1)^{2k} = (- \omega^2)^{2k} = \left( \omega^{2k}\right)^2$$ So this time we have a nontrivial cube root of $1$ added to its square and then add $1,$ giving zero.

Answer 2

Hint: recall Lemma: $\ {x^{2}\!+\!x\!+\!1\mid x^A\! +\! x^B\! +\! x^C\ \ {\rm if} \ \ \{A,B,C\}\equiv \color{#0af}{\{2,1,0\}}\pmod{\!3}}$

$\!\!\bmod \color{#c00}{x^2}\!+\!\color{#0a0}{x\!+\!1}\!:\,\ \color{#0a0}{x\!+\!1}\equiv -\color{#c00}{x^2}\,\Rightarrow\, f\equiv (-\color{#c00}{x^{\large 2}})^{\large 2k}\! + x^{\large 2k}+ 1\equiv x^{\color{darkorange}{\large 4k}}+x^{\color{darkorange}{\large 2k}}+x^{\large \color{darkorange}0}$

$\!\!\bmod 3\!:\ \{A,B,C\} = \{\color{darkorange}{4k,2k,0}\}\equiv \{k,-k,0\},\:$ which is $\,\equiv \color{#0af}{\{-1,1,0\}}\! \iff \color{#90f}{k\not \equiv 0}$

Therefore by applying the above Lemma we conclude that $\,x^2\!+\!x\!+\!1\mid f \iff \color{#90f}{3\nmid k}$

Answer 3

You said you haven't learned number theory yet.

I don't know if you have learned about complex roots of unity.

Here is a proof where you just need to know the binomial expansion and mathematical induction.

Let $f_n(x)=(x+1)^n+x^n+1.$

I claim $f_{6s}(x)=q_1(x)(x^2+x+1)+3,$

$ f_{6s+2}(x)=q_2(x)(x^2+x+1),$ and

$f_{6s+4}(x)=q_3(x)(x^2+x+1).$

Base case $(s=0)$:

$f_0(x)=(x+1)^0+x^0+1=1+1+1=3.$

$f_2(x)=(x+1)^2+x^2+1=x^2+2x+1+x^2+1=2x^2+2x+2=2(x^2+x+1)$

$f_4(x)=(x+1)^4+x^4+1=x^4+4x^3+6x^2+4x+1+x^4+1=2x^4+4x^3+6x^2+4x+2$

$=(2x^2+2x+2)(x^2+x+1).$

Induction step:

$f_{n+6}(x)-f_n(x)=(x+1)^{n+6}-(x+1)^n+x^{n+6}-x^n$

$=(x+1)^n((x+1)^6-1)+x^n(x^6-1).$

Note that $x^6-1=(x^3+1)(x^3-1)=(x^3+1)(x-1)(x^2+x+1)$ is a multiple of $x^2+x+1$.

Therefore also $(x+1)^6-1=((x+1)^3+1)((x+1)-1)((x+1)^2+(x+1)+1)$ is

a multiple of $x^2+x+1$ [since $(x+1)^3+1=x^3+3x^2+3x+2=(x+2)(x^2+x+1)],$

and we are basically done.