1
$\begingroup$

Let $p(x)=x^2 +x+1$ divides $f(x)=(x+1)^n + x^n+1$. I need to find values of $n$.

Logically, I can see that $n=2,4$ are the values. But I need to find the rule.

Anyone can help?

$\endgroup$
2
$\begingroup$

HINT: Let $\omega $ be the complex cube root of unity. Then $f(\omega )=0$ and this implies $$f(\omega )=(-1)^n \omega^{2n}+\omega^n+1=0$$

First, suppose $n\equiv 1\pmod{3}$, then we have $(-1)^n\omega^2+\omega +1=0$ i.e. $(-1)^n\omega^2=\omega^2$ and hence $n$ is even. Similarly $\omega^2$ is also a root of $f$ when $n$ is even. Hence $p\mid f$ when $n$ is even and $n\equiv 1\pmod{3}$. Similarly check the cases $n\equiv 0,2\pmod{3} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.