# Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

For all $a, m, n \in \mathbb{Z}^+$,

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

• Another question (math.stackexchange.com/questions/11567/…) was closed as a duplicate of this one where there is a second solution. Dec 4, 2010 at 14:17
• Find here: Number Theory for Mathematical Contests, Example#245, Page#36. Jul 29, 2012 at 16:56
• And what if we consider GCD over $\mathbb{C} [X]$? Dec 10, 2019 at 6:11
• @labbhattacharjee hey i know it's been 8 years but I just stumbled upon this post and I found that link quite interesting. Do you have more of that kind of material (mathematical contests)? Nov 3, 2020 at 4:04
• @NotAMathematician, See pdfdrive.com/… and fmf.uni-lj.si/~lavric Nov 5, 2020 at 5:07

Mimic in expts a subtractive Euclidean algorithm $$\rm\,(n,m) = (\color{#0a0}{n\!-\!m},m)$$

\begin{align} \rm{e.g.}\ \ &\rm (f_5,f_2) = (f_3,f_2) = (f_1,f_2) = (f_1,f_1) = (f_1,\color{darkorange}{f_0})= f_1 = f_{\:\!(5,\,2)}\\[.3em] {\rm like}\ \ \ &(5,\ 2)\, =\:\! (3,\ 2)\, =\:\! (1,\ 2)\:\! =\:\! (1,\ 1)\:\! =\:\! (1,\ \color{darkorange}0)\:\! = 1,\ \ {\rm since}\end{align}\qquad

$$\rm\ f_{\,n}\: :=\ a^n\!-\!1\ =\ a^{n-m} \: \color{#c00}{(a^m\!-\!1)} + \color{#0a0}{a^{n-m}\!-\!1},\,\$$ hence $$\rm\:\ {f_{\,n}\! = \color{#0a0}{f_{\,n-m}}\! + k\ \color{#c00}{f_{\,m}}},\,\ k\in\mathbb Z,\:$$ thus

Theorem $$\:$$ If $$\rm\ f_{\, n}\:$$ is an integer sequence with $$\rm\ \color{darkorange}{f_{0} =\, 0},\:$$ $$\rm \:{ f_{\,n}\!\equiv \color{#0a0}{f_{\,n-m}}\ (mod\ \color{#c00}{f_{\,m})}}\$$ for all $$\rm\: n > m,\$$ then $$\rm\: (f_{\,n},f_{\,m})\ =\ f_{\,(n,\:m)}, \:$$ where $$\rm\ (i,\:j)\$$ denotes $$\rm\ gcd(i,\:j).\:$$

Proof $$\$$ By induction on $$\rm\:n + m\:$$. The theorem is trivially true if $$\rm\ n = m\$$ or $$\rm\ n = \color{darkorange}0\$$ or $$\rm\: m = \color{darkorange}0.\:$$
So we may assume $$\rm\:n > m > 0\:$$.$$\$$ Note $$\rm\ (f_{\,n},f_{\,m}) = (\color{#0a0}{f_{\,n-m}},\color{#c00}{f_{\,m}})\$$ follows by $$\rm\color{#90f}{Euclid}$$ & hypothesis.
Since $$\rm\ (n-m)+m \ <\ n+m,\$$ induction yields $$\rm\, \ (f_{\,n-m},f_{\,m})\, =\, f_{\,(n-m,\:m)} =\, f_{\,(n,\:m)}.$$

$$\rm\color{#90f}{Euclid}\!:\ A\equiv a\pmod{\! m}\,\Rightarrow\ (A,m) = (a,m)\,$$ is the reduction (descent) step used both above and in the Euclidean algorithm $$\rm\: (A,m) = (A\bmod m,\,m),\,$$ the special case $$\,\rm f_{\:\!n} = n\,$$ above.

This is a prototypical strong divisibility sequence. Same for Fibonacci numbers.

Alternatively it has a natural proof via the Order Theorem $$\ a^k\equiv 1\iff {\rm ord}(a)\mid k,\,$$ viz.

$$\begin{eqnarray}\ {\rm mod}\:\ d\!:\ a^M\!\equiv 1\equiv a^N&\!\iff\!& {\rm ord}(a)\ |\ M,N\!\color{#c00}\iff\! {\rm ord}(a)\ |\ (M,N)\!\iff\! \color{#0a0}{a^{(M,N)}\!\equiv 1}\\[.2em] {\rm i.e.}\ \ \ d\ |\ a^M\!-\!1,\:a^N\!-\!1\! &\!\iff\!\!&\ d\ |\ \color{#0a0}{a^{(M,N)}\!-\!1},\qquad\,\ {\rm where} \quad\! (M,N)\, :=\, \gcd(M,N) \end{eqnarray}\ \ \ \ \$$

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

Note  We used the GCD universal property $$\ a\mid b,c \color{#c00}\iff a\mid (b,c)\$$ [which is the definition of a gcd in more general rings].  Compare that with $$\ a and, analogously, $$\,\ 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 that lies 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.

• Sort of like the Fibonacci sequence! May 23, 2015 at 12:03
• @john Yes, they are both strong divisibility sequences, i.e. $\,(f_n,f_m) = f_{(n,m)}.\,$ See here for the Fibonacci case. May 23, 2015 at 13:33

Below is a proof which has the neat feature that it immediately specializes to a proof of the integer Bezout identity for $$\rm\:x = 1,\:$$ allowing us to view it as a q-analog of the integer case.

E.g. for $$\rm\ m,n\ =\ 15,21$$

$$\rm\displaystyle\quad\quad\quad\quad\quad\quad\quad \frac{x^3-1}{x-1}\ =\ (x^{15}\! +\! x^9\! +\! 1)\ \frac{x^{15}\!-\!1}{x\!-\!1} - (x^9\!+\!x^3)\ \frac{x^{21}\!-\!1}{x\!-\!1}$$

for $$\rm\ x = 1\$$ specializes to $$\ 3\ \ =\ \ 3\ (15)\ \ -\ \ 2\ (21)\:,\$$ i.e. $$\rm\ (3)\ =\ (15,21) := gcd(15,21)$$

Definition $$\rm\displaystyle \quad n' \: :=\ \frac{x^n - 1}{x-1}\:$$. $$\quad$$ Note $$\rm\quad n' = n\$$ for $$\rm\ x = 1$$.

Theorem $$\rm\quad (m',n')\ =\ ((m,n)')\$$ as ideals in $$\rm\,\Bbb Z[x],\,$$for naturals $$\rm\:m,n.$$

Proof $$\$$ It is trivially true if $$\rm\ m = n\$$ or if $$\rm\ m = 0\$$ or $$\rm\ n = 0.\:$$

W.l.o.g. suppose $$\rm\:n > m > 0.\:$$ We proceed by induction on $$\rm\:n\! +\! m.$$

$$\begin{eqnarray}\rm &\rm x^n\! -\! 1 &=&\ \rm x^r\ (x^m\! -\! 1)\ +\ x^r\! -\! 1 \quad\ \ \rm for\ \ r = n\! -\! m \\[.1em] \quad\Rightarrow\quad\! &\rm\qquad n' &=&\ \rm x^r\ m'\ +\ r' \quad\ \ \rm by\ dividing\ above\ by\ \ x\!-\!1 \\ \quad\Rightarrow\ \ &\rm (m', n')\, &=&\ \rm (m', r') \\ & &=&\rm ((m,r)') \quad\ \ by\ induction, applicable\ by\:\ m\!+\!r = n < n\!+\!m \\[.1em] & &=&\rm ((m,n)') \quad\ \:\! by\ \ r \equiv n\ \:(mod\ m)\quad\ \ \bf\small QED \end{eqnarray}$$

Corollary $$\$$ Integer Bezout Theorem $$\$$ Proof:  set $$\rm\ x = 1\$$ above, i.e. erase primes.

A deeper understanding comes when one studies Divisibility Sequences and Divisor Theory.

• Is $((\rm m,n)')$ supposed to be $((\rm m,n))'$ i.e. $\rm \dfrac{x^{(m,n)}-1}{x-1}$?
– Pedro
Jun 18, 2012 at 23:38
• @Peter  Let $\rm\:(m,n)' = \dfrac{x^{\,(m,n)}\!-\!1}{x\!-\!1} =: f.\:$ Then $\rm\:((m,n)') = (f) = f\:\mathbb Z[x]\:$ is a principal ideal, thus the equality $\rm\:(m',n') = ((m,n)')\:$ denotes the ideal equality $\rm\:(g,h) = (f)\:$ for polynomials $\rm\:f,g,h\in\mathbb Z[x].\:$ If you have no knowledge of ideals you can instead simply interpret it as saying that $\rm\:f\:|\:g,h\:$ and $\rm\:f = a\,g+b\,h\:$ for some $\rm\:a,b\in \mathbb Z[x],\:$ which implies $\rm\:f = gcd(g,h).$ Jun 19, 2012 at 0:17

Let $m\ge n\ge 1$. Apply Euclidean Algorithm.

$\gcd\left(a^m-1,a^n-1\right)=\gcd\left(a^{n}\left(a^{m-n}-1\right),a^n-1\right)$. Since $\gcd(a^n,a^n-1)=1$, we get

$\gcd\left(a^{m-n}-1,a^n-1\right)$. Iterate this until it becomes $$\gcd\left(a^{\gcd(m,n)}-1,a^{\gcd(m,n)}-1\right)=a^{\gcd(m,n)}-1$$

• And this too is a duplicate of an answer in the 5-year-old linked duplicate thread. Dec 31, 2016 at 2:07

Let $$\gcd(a^n - 1, a^m - 1) = t$$ then $$a^n \equiv 1 \pmod t\,\quad\text{and}\quad\,a^m \equiv 1 \,\pmod t$$ And thus $$a^{nx + my} \equiv 1\, \pmod t$$ $$\forall\,x,\,y\in \mathbb{Z}$$

According to the Extended Euclidean algorithm, we have $$nx + my =\gcd(n,m)$$ This follows $$a^{nx + my} \equiv 1\pmod t \implies a^{\gcd(n,m)} \equiv 1 \pmod t\implies t\big|\big( a^{\gcd(n,m)} - 1\big)$$

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

Since it is easy to show $$(a^{\gcd(n,m)}-1)\big|t$$.

• I don't understand why you can conclude that $(a^{\gcd(n,m)}-1)\mid t$ from $a^{\gcd(n,m)}\equiv 1\pmod t$. The latter will give you $t\mid (a^{\gcd(n,m)}-1)\$.
– Bach
Dec 27, 2020 at 20:28
• How did you manage to conclude, which would require that $k=1$: $$a^{\gcd(m,n)}-1\, =\gcd(a^m-1, a^n-1)$$ May 31, 2021 at 4:25
• To prove $(a^{\gcd(n,m)}-1)\big|t$, I would try to show that $(a^{\gcd(n,m)}-1)$ divides both $(a^{m}-1)$ and $(a^{n}-1)$. As it is a common divisor, it must divide their greatest common divisor, $t$. Oct 3, 2021 at 17:42

More generally, if $\gcd(a,b)=1$, $a,b,m,n\in\mathbb Z^+$, $a> b$, then $$\gcd(a^m-b^m,a^n-b^n)=a^{\gcd(m,n)}-b^{\gcd(m,n)}$$

Proof: Since $\gcd(a,b)=1$, we get $\gcd(b,d)=1$, so $b^{-1}\bmod d$ exists.

$$d\mid a^m-b^m, a^n-b^n\iff \left(ab^{-1}\right)^m\equiv \left(ab^{-1}\right)^n\equiv 1\pmod{d}$$

$$\iff \text{ord}_{d}\left(ab^{-1}\right)\mid m,n\iff \text{ord}_{d}\left(ab^{-1}\right)\mid \gcd(m,n)$$

$$\iff \left(ab^{-1}\right)^{\gcd(m,n)}\equiv 1\pmod{d}\iff a^{\gcd(m,n)}\equiv b^{\gcd(m,n)}\pmod{d}$$

• This is precisely the homogenization $(a^n-1\to a^n-b^n)$ of a proof in the 5-year-old duplicate thread linked in Yuan's comment on the question. To avoid posting such duplicate answers it's a good ides to first peruse duplicate links before posting an answer to a five year old question! Dec 31, 2016 at 2:03
• Update: actually this homegenized version was posted 5 months prior in this answer.. There are probably older dupes too since this is a FAQ. Posting the link in case anyone decides to organize. Jul 13, 2017 at 20:44
• I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $\left(ab^{-1}\right)^m\equiv \left(ab^{-1}\right)^n\equiv 1\pmod{d}$? Dec 2, 2018 at 23:48

More generally, if $a,b,m,n\in\mathbb Z_{\ge 1}$, $a>b$ and $(a,b)=1$ (as usual, $(a,b)$ denotes $\gcd(a,b)$), then $$(a^m-b^m,a^n-b^n)=a^{(m,n)}-b^{(m,n)}$$

Proof: Use $\,x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+\cdots+xy^{k-2}+x^{k-1})\,$

and use $n\mid a,b\iff n\mid (a,b)$ to prove:

$a^{(m,n)}-b^{(m,n)}\mid a^m-b^m,\, a^n-b^n\iff$

$a^{(m,n)}-b^{(m,n)}\mid (a^m-b^m,a^n-b^n)=: d\ \ \ (1)$

$a^m\equiv b^m,\, a^n\equiv b^n$ mod $d$ by definition of $d$.

Bezout's lemma gives $\,mx+ny=(m,n)\,$ for some $x,y\in\Bbb Z$.

$(a,b)=1\iff (a,d)=(b,d)=1$, so $a^{mx},b^{ny}$ mod $d$ exist (notice $x,y$ can be negative).

$a^{mx}\equiv b^{mx}$, $a^{ny}\equiv b^{ny}$ mod $d$.

$a^{(m,n)}\equiv a^{mx}a^{ny}\equiv b^{mx}b^{ny}\equiv b^{(m,n)}\pmod{\! d}\ \ \ (2)$

$(1)(2)\,\Rightarrow\, a^{(m,n)}-b^{(m,n)}=d$

Written for a duplicate question, this may be a bit more elementary than the other answers here, so I will post it:

If $$g=(a,b)$$ and $$G=\left(p^a-1,p^b-1\right)$$, then $$\left(p^g-1\right)\sum_{k=0}^{\frac ag-1}p^{kg}=p^a-1\tag1$$ and $$\left(p^g-1\right)\sum_{k=0}^{\frac bg-1}p^{kg}=p^b-1\tag2$$ Thus, we have that $$\left.p^g-1\,\middle|\,G\right.\tag3$$

For $$x\ge0$$, $$\left(p^a-1\right)\sum_{k=0}^{x-1}p^{ak}=p^{ax}-1\tag4$$ Therefore, we have that $$\left.G\,\middle|\,p^{ax}-1\right.\tag5$$ If $$\left.G\,\middle|\,p^{ax-b(y-1)}-1\right.$$, then $$\left.G\,\middle|\,\left(p^{ax-b(y-1)}-1\right)-p^{ax-by}\left(p^b-1\right)\right.=p^{ax-by}-1\tag6$$ Therefore, by induction on $$y$$ (with $$(5)$$ as the base case and $$(6)$$ as the inductive step), for any $$x,y\ge0$$ so that $$ax-by\ge0$$, $$\left.G\,\middle|\,p^{ax-by}-1\right.\tag7$$ which means that $$\left.G\,\middle|\,p^g-1\right.\tag8$$

Putting all this together gives $$G=p^g-1\tag9$$

• Can you elaborate on the step $if G \mid p^{ax-(b-1)y}$. How you arrived on this term. Oct 2, 2019 at 9:26
• That was a typo. Thanks for noticing it. Does it make more sense now?
– robjohn
Oct 2, 2019 at 14:07
• Why you are using "if....". What if, "if..." doesn't hold true Oct 2, 2019 at 14:14
• It does in the case of $y=1$ due to the preceding line. The subsequent lines finish the induction.
– robjohn
Oct 2, 2019 at 14:18
• I feel you should mention the induction argument more clearly. After you edited solution, it appears inductive in nature. Still some more clarity will be helpful. Oct 2, 2019 at 14:21

Besides excellent answers above, you can use a property that

$$\gcd((y+1)x, y)= \gcd(x,y)$$

where $$x=a^m - 1, y = a^n - 1$$ to find the proof.

It's a beautiful question

In fact, I tried to check on computer.(When I didn't know Bezout's Identity)

I tried to prove as:

Let d = gcd($$a^m-1, a^n-1$$)

implies: $$a^m ≡ 1$$ $$mod(d)$$ and $$a^n ≡ 1$$ $$mod(d)$$

Now, $$gcd(m,n) = mx+ny$$ .........#Bezout's Identity

$$a^{gcd(m, n)} ≡ a^{(mx+ny)} ≡ a^{mx}a^{ny} ≡ 1$$ $$mod(d)$$

Therefore, $$d |a^{gcd(m,n)} −1.$$ We now show that $$a^{gcd(m,n)} −1 |d.$$ Since gcd(m,n) |m, we have

$$a^{gcd(m,n)} −1 |a^m −1$$ .....#1

Similarly, $$a^{gcd(m,n)} −1 |a^n −1$$ .....#2

Since, $$a^{gcd(m, n)}-1$$ divides both $$a^m-1$$ and $$a^n-1$$ so it must also divide their GCD :

$$a^{gcd(m, n)}-1| gcd(a^m-1, a^n-1)$$$$mod(d)$$

Since, $$d |a^{gcd(m,n)}−1$$ and $$a^{gcd(m,n)}−1 |d$$, we must have $$d = gcd(a^m−1,a^n−1)$$ = $$a^{gcd(m,n)} −1$$

So, Bezout's Identity makes the proof simpler.

It is enough to check that if $$(m,n) = 1$$ then the resultant

$$R_{m,n} \colon =\operatorname{res}(\frac{x^m-1}{x-1}, \frac{x^n-1}{x-1}) = 1$$

Now we have

$$R_{m,n} = \prod_{\xi^n=1, \xi \ne 1} \frac{\xi^m-1}{\xi-1}$$

Now, since $$(m,n) = 1$$, the map $$\xi \mapsto \xi^m$$ is a bijection of $$\{\xi^n=1, \xi \ne 1\}$$. We conclude that $$R_{m,n} = 1$$.