# $\gcd(b^x - 1, b^y - 1, b^ z- 1,…) = b^{\gcd(x, y, z,…)} -1$ [duplicate]

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Number theory proving question?

Dear friends,

Since $b$, $x$, $y$, $z$, $\ldots$ are integers greater than 1, how can we prove that $$\gcd (b ^ x - 1, b ^ y - 1, b ^ z - 1 ,\ldots)= b ^ {\gcd (x, y, z, .. .)} - 1$$ ?

Thank you!

Paulo Argolo

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## marked as duplicate by Qiaochu Yuan, Aryabhata, Robin Chapman, kennytmNov 25 '10 at 20:07

This is essentially a duplicate of this previous question coupled with induction math.stackexchange.com/questions/7473/… –  Timothy Wagner Nov 23 '10 at 20:09
@Timothy: ah, you're right. I should've checked for duplicates. –  Qiaochu Yuan Nov 23 '10 at 20:20

It suffices to prove it for two terms, that is, $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n,m)} - 1$. The basic idea is that we can use the Euclidean algorithm on the exponents, as follows: if $n > m$, then

$$\gcd(a^n - 1, a^m - 1) = \gcd(a^n - 1, a^n - a^{n-m}) = \gcd(a^{n-m} - 1, a^m - 1).$$

So we can keep subtracting one exponent from the other until we get $\gcd(n, m)$ as desired. Another way to look at this computation is to write $d = \gcd(a^n - 1, a^m - 1)$ and note that

$$a^n \equiv 1 \bmod d, a^m \equiv 1 \bmod d \Rightarrow a^{nx+my} \equiv 1 \bmod d$$

from which it readily follows, as before, that $a^{\gcd(n,m)} \equiv 1 \bmod d$, so $d$ dividess $a^{\gcd(n,m)} - 1$. On the other hand, $a^{\gcd(n, m)} - 1$ also divides $d$. To see this, denote $e \cdot \gcd(n,m) = n$ and $f \cdot \gcd(n,m) = m$ (verify for yourrself that this makes sense). We then have \begin{align*}a^n-1 = (a^{\gcd(n,m)})^e -1 \equiv & 1 \pmod{a^{\gcd(n,m)}-1} \\ a^m-1 = (a^{\gcd(n,m)})^f-1 \equiv & 1 \pmod{a^{\gcd(n,m)}-1}. \end{align*} Hence we have $$a^{\gcd(n,m)}-1 \; |\; d .$$

What's really nice about this result is that it holds both for particular values of $a$ and also for $a$ as a variable, e.g. in a polynomial ring with indeterminate $a$.

You can readily deduce several seemingly nontrivial results from this; for example, the sequence defined by $a_0 = 2, a_n = 2^{a_{n-1}} - 1$ is a sequence of pairwise relatively prime integers, from which it follows that there are infinitely many primes. By working only slightly harder you can deduce that in fact there are infinitely many primes congruent to $1 \bmod p$ for any prime $p$.

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Hint $\,$ By below $\, a^M\!-\!1,\:a^N\!-\!1\$ and $\, a^{(M,N)}\!-\!1\$ have the same set $\,S$ of common divisors $\,d,\,$ therefore they have the same greatest common divisor $\ (= \max\ S).$

$$\begin{eqnarray}\ {\rm mod}\,\ d\!:\ \ a^M,\:a^N\equiv 1&\!\iff\!\!& {\rm ord}(a)\ |\ M,N\color{#0a0}\iff {\rm ord}(a)\ |\ (M,N)\iff \color{#c00}{a^{\,(M,N)}\!\equiv 1}\\ {\rm i.e.}\ \ \ d\ |\ a^M\!-\!1,\:a^N\!-\!1 &\!\iff\!\!&\ d\ |\ \color{#c00}{a^{\,(M,N)}\!-\!1},\qquad\ \ \, {\rm where} \quad\! (M,N)\, :=\, \gcd(M,N) \end{eqnarray}$$

Remark  We e used $\ a\mid b,c \color{#0a0}\iff a\mid (b,c),\,$ the fundamental universal property of the gcd. Compare that with $\,\ a<b,c \!\iff\! a< \min(b,c),\$ and $\,\ a\subset b,c\iff a\subset b\cap c.\$ Such universal "iff" characterizations enable quick and easy simultaneous proof of both directions.

The conceptual structure at the heart of this simple proof is the ubiquitous order ideal. $\$ See this answer for more on this and the more familiar additive form of a denominator ideal.

Generally: $\ \gcd(f(m), f(n))\, =\, f(\gcd(m,n))\ \,$ if $\, \ f(n) \equiv f(n\!-\!m)\ \ ({\rm mod}\ f(m))\$ and $\ f(0) = 0.\,$ See this answer for a simple inductive proof.

In fact there is a q-analog: the result also holds true for polynomials $\ f(n) = (x^n\!-\!1)/(x\!-\!1),\,$ and $\ x\to 1\$ yields the integer case (Bezout identity) - see this answer for a simple proof.

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