How to prove that $q+1$ divides $q^k - 1$ for $k \in 2 \mathbb N$

Question

I have to prove that if $k \in 2 \mathbb N $ every $q+1$ divides $q^k-1$ and I really have trouble understanding how to do this.

Answer 1

If you know modular arithmetic, note $q\equiv-1\pmod {q+1}$.

Raise both sides to the $k^{th}$ power, and subtract $1$ from both sides.

Answer 2

I think you will gain insight into this problem by performing synthetic division.

For $k = 2$, you have (showing intermediate steps) \begin{align*} & \begin{array}{c|ccc} & 1 & 0 & -1 \\ -1 & & & \\ \hline & 1 & & \end{array} \\ & \begin{array}{c|ccc} & 1 & 0 & -1 \\ -1 & & -1 & \\ \hline & 1 & -1 & \end{array} \\ & \begin{array}{c|ccc} & 1 & 0 & -1 \\ -1 & & -1 & 1 \\ \hline & 1 & -1 & 0 \end{array} \\ \end{align*} This shows you have remainder zero. Repeating for one or two more $k$ should make the pattern of coefficients in the cofactor clear.

Answer 3

$((q+1)-1)^{2m}-1=$

$\displaystyle{\sum_{k=0}^{2m}}\binom{2m}{k}(q+1)^{2m-k}(-1)^k-1=$

$\displaystyle{\sum_{k=0}^{2m-1}}\binom{2m}{k}(q+1)^{2m-k}(-1)^k.$

All terms contain a factor $(q+1)$.

Answer 4

$$\begin{aligned} q^{2n}-1&=(q^2-1)(1+ q^2 + \dots + q^{2n-2})\\ &=(q-1)(q+1)(1+ q^2 + \dots + q^{2n-2}) \end{aligned}$$

Answer 5

$q^{2k}-1=(q^2-1)\sum_{j=0}^{k-1}q^{2j}$, while $q^2-1=(q-1)(q+1)$.