2

I want to prove $\forall (p,k) \in\mathbb{N}$$^{2}$ with k not divisible by $3$ : $1+p+p^2\mid 1+p^{2k}+(1+p)^{2k}$

An attempt.

$1+p+p²=(p-j)(p-\bar{j})$ with $j=e^{i\frac{2\pi}{3}}$.

Then I prove that j and $\bar{j}$ are roots of the polynomial $1+x^{2k}+(1+x)^{2k}$.So $1+x+x^2\mid1+x^{2k}+(1+x)^{2k}$.

Can I conclude $1+p+p^2\mid 1+p^{2k}+(1+p)^{2k}$ ?

Thank you in advance

Thomas Andrews
  • 177,126
yoda
  • 315

1 Answers1

1

Consider the polynomials $1+x+x^2$ and $P(x)=1+x^{2k}+(1+x)^{2k}$.

The polynomial $x^2+x+1$ is monic (lead coefficient $1$). Therefore there exist polynomials $Q(x)$, $R(x)$, with integer coefficients, and with $R(x)$ of degree less than $2$ such that $$P(x)=(1+x+x^2)Q(x)+R(x).$$ From the facts you have etablished, you can conclude that $R(j)=0$ and $R(\bar{j})=0$, meaning that $R$ is identically $0$.

André Nicolas
  • 507,029