$x^a - 1$ divides $x^b - 1$ if and only if $a$ divides $b$

Let $x > 1$ and $a$, $b$ be positive integers. I know that $a$ divides $b$ implies that $x^a - 1$ divides $x^b - 1$. If $b = qa$, then

$x^b - 1 = (x^a)^q - 1^q = (x^a - 1)((x^a)^{q-1} + \ldots + x^a + 1)$

I'm interested in the converse of the statement. If $x^a - 1$ divides $x^b - 1$, does this imply that $a$ divides $b$?

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By the division algorithm, you can write $b=qa+r$ with $0 \leq r < a$. When you do long division of $x^{b} - 1$ by $x^{a} - 1$, you will eventuall encounter the remainder $x^{r} - 1$, and the process stops. If $x^{a} - 1$ divides $x^{b} - 1$, the remainder must be $0$, whence $r=0$. – Chris Leary Apr 4 '12 at 14:57
Generalization : math.stackexchange.com/questions/7473/… – lab bhattacharjee Mar 9 '14 at 7:13

Let $b=a \cdot q+r$.

Then

$$x^b-1=x^b-x^r+x^r-1=x^r(x^{aq}-1)+x^r-1 \,.$$

Use that $x^a-1$ divides $x^{aq}-1$.

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Thanks, very simple. – spin Apr 8 '12 at 20:02
@N.S. Aren't you using recursion there? The last sentence. – Dhruv Mar 5 at 14:47
@DhruvSomani The last step is just the formula $$y^q-1^q=(y-1)(y^{q-1}+..+y+1)$$ – N. S. Mar 5 at 19:52

Hint $\:$ By Theorem below $\rm\:(x^a\!-1,x^b\!-1) = x^{(a,b)}\!-1.\:$ This $\rm\: =\: x^a\!-1\!\iff\! (a,b)=a\!\iff\! a\ |\ b$

Apply Theorem to $\rm\ f_n =\ x^n\!-1\ =\ x^{n-m} \: (x^m\!-1) + x^{n-m}\!-1 =\: g\ f_m + f_{n-m}\equiv f_{n-m}\pmod{f_m}$

Theorem $\:$ If $\rm\:f_n\:$ is a sequence in a GCD domain (domain where gcds exist) satisfying $\rm\ f_{\:0} =\: 0\:\$ and $\rm\ \: f_n\equiv f_{n-m}\ (mod\ f_m)\$ for $\rm\: n > m\$ then $\rm\:\ (f_n,f_m)\ =\ f_{(n,\:m)}\ \:$ where $\rm\ (i,\:j) = gcd(i,\:j)\:$.

Proof $\$ By induction on $\rm\:n + m\:$. The theorem is trivially true if $\rm\ n = m\$ or $\rm\ n = 0\$ or $\rm\: m = 0\:$.
So we may assume $\rm\:n > m > 0\:$.$\$ Note $\rm\ (f_n,f_m)\: =\ (f_{n-m},f_m)\$ follows from the hypothesis.
Since $\rm\:\ (n-m)+m \ <\ n+m\: ,\$ induction yields $\rm\ \ (f_{n-m},f_m)\ =\ f_{(n-m,\:m)}\ =\ f_{(n,\:m)}\$

See this post and its links for more on such divisibility sequences.

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Thanks for the answer, it's always interesting a more general approach – spin Apr 8 '12 at 20:05