Integer Less One divides Power Less One

Theorem

Let $q, n \in \Z_{>0}$.

Then:

$\paren {q - 1} \divides \paren {q^n - 1}$

where $\divides$ denotes divisibility.

Corollary

Let $m, n, q \in \Z_{>0}$.

Let:

$m \divides n$

where $\divides$ denotes divisibility.

Then:

$\paren {q^m - 1} \divides \paren {q^n - 1}$

Proof

From Sum of Geometric Sequence:

$\ds \frac {q^n - 1} {q - 1} = \sum_{k \mathop = 0}^{n - 1} q^k$

That is:

$q^n - 1 = r \paren {q - 1}$

where $r = 1 + q + q^2 + \cdots + q^{n - 1}$.

As Integer Addition is Closed and Integer Multiplication is Closed, it follows that $r \in \Z$.

Hence the result by definition of divisor of integer.

$\blacksquare$

Sources

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.4$: The rational numbers and some finite fields: Further Exercises $5$