<s>[INST] For which $k\in \Bbb N$ does $x^2+x+1$ divide $ (x + 1)^{2k}+x^{2k} + 1$? 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.
 [/INST] 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.
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