The polynomial $x^{2k} + 1 + (x+1)^{2k}$ is not divisible by $x^2 + x +1$. Find value of $k$ such that it belongs to natural numbers.

Question

I did some progress by doing this-

I thought if it would be a factor, then

$x^2 +x +1 = 0 $

$x+1=-x^2$

Putting this in the expression, $$x^{4k}+x^{2k}+1$$ Then I tried to solve it further, but I am stuck here...

Between this and your prior question it looks like you are just posting your homework here for us to do for you. — lulu, Feb 20 '20 at 12:20
Lulu , I just registered today here . And I had some doubts in polynomials chapter of my Olympiad book , so I was asking them here ...... — Aayush, Feb 20 '20 at 12:27

score 0 · Accepted Answer · answered Feb 20 '20 at 12:24

0

$$x^{2k}+(x+1)^{2k}+1\equiv x^{2k}+(x^2)^{2k}+1=x^{4k}+x^{2k}+1\equiv0$$ for any $k$, which is not divisible by $3$.

If $k$ is divisible by $3$, so since $x^3\equiv1$, we obtain: $$x^{4k}+x^{2k}+1\equiv3.$$

Can you end it now?

answered Feb 20 '20 at 12:24

Michael Rozenberg

@Aayush If $k$ is divisible by $3$ we see that $x^{4k}+x^{2k}+1=\left(x^3\right)^{\frac{4k}{3}}+\left(x^3\right)^{\frac{2k}{3}}+1\equiv1+1+1=3.$ – Michael Rozenberg Feb 20 '20 at 12:30
Pls explain your step where you told what would happen if it is a multiple of 3 – Aayush Feb 20 '20 at 12:31
@Aayush because $x^3-1=(x+1)(x^2+x+1)\equiv0.$ – Michael Rozenberg Feb 20 '20 at 12:37
Oh sorry I read wrong , thanks for your help – Aayush Feb 20 '20 at 12:39
@Aayush You are welcome! – Michael Rozenberg Feb 20 '20 at 12:39

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