Prove that something is a divisor of something else

Question

For every $a > 1$ and $n$ an element of the natural numbers, we have that $a - 1$ is a divisor of $a^n - 1$. Or written with symbols: $$\forall \ a > 1 ∧ n \in \mathbb{N}: (a−1) \ | \ a^n −1.$$

Can someone please give me a proof of this.

Thanks in advance!

Try looking at some polynomial identities. For instance recall that $a^2-1 = (a-1)(a+1)$. — Joel, Nov 06 '13 at 14:59
Look at the roots of each of the two monic polynomials you have: $a-1$ and $a^n -1$ — mv3, Nov 06 '13 at 15:02
Related: http://math.stackexchange.com/questions/497235/how-to-show-that-10n-1-is-divisible-by-9 — lab bhattacharjee, Nov 06 '13 at 16:06

Gautam Shenoy · Answer 1 · 2013-11-06T15:23:43.357

1

You may also use the following:

If $p(x)$ is a polynomial, then the remainder when divided by $x-a$ is $p(a)$.

Proof: $p(x) = q(x)(x-a) + r$. Put $x=a$. $\blacksquare$

edited Nov 06 '13 at 15:23

answered Nov 06 '13 at 15:16

Gautam Shenoy

10,318

score 0 · Answer 2 · answered Nov 06 '13 at 15:06

0

∀a∈R:a>1,n∈N:a-1|a^n-1
a_1-1|a_1^n-1
n,b∈N: b(a_1-1)=a_1^n-1
n→1:b(a_1-1)=a_1^1-1
n→k+1
b(a_1-1)=a_1^(k+1)-1
b(a_1-1)=〖a*a〗_1^k-1
b(a_1-1)=
...

answered Nov 06 '13 at 15:06

Johann Bauer

103

Prove that something is a divisor of something else

2 Answers2