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

Question

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

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

Answer 1

Let $\,\rm f_{\,n} :=\, a^n-\!1.\,$ Mimic in expts a subtractive Euclidean algorithm $\rm\,(n,m) = (\color{#0a0}{n\!-\!m},m)$

$$\begin{align} \rm{e.g.}\ \ &\rm (f_5,f_2) = (f_3,f_2) = (f_1,f_2) = (f_1,f_1) = (f_1,\color{darkorange}{f_0})= f_1 = f_{\,(5,\,2)}\\[.4em] {\rm like}\ \ \ &(5,\ 2)\, =\:\! (3,\ 2)\, =\:\! (1,\ 2)\:\! =\:\! (1,\ 1)\:\! =\:\! (1,\ \color{darkorange}0)\:\! = 1,\ \ {\rm true\ by}\end{align}\qquad$$

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

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 $\rm\, \ (f_{\,n-m},f_{\,m})\, =\, f_{\,(n-m,\:m)} =\, f_{\,(n,\:m)}.$

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

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

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Alternatively it follows simply by the Order Theorem $\bmod d\!:\ a^k\equiv 1\iff {\rm ord}\,(a)\mid k,\,$ viz.

$$\begin{eqnarray}\ {\rm mod}\:\ d\!:\ a^M\!\equiv 1\equiv a^N&\!\iff\!& {\rm ord}(a)\ |\ M,N\!\color{#c00}\iff\! {\rm ord}(a)\ |\ (M,N)\!\iff\! \color{#0a0}{a^{(M,N)}\!\equiv 1}\\[.5em] {\rm i.e.}\ \ \ d\ |\ a^M\!-\!1,\:a^N\!-\!1\! &\!\iff\!\!&\ d\ |\ \color{#0a0}{a^{(M,N)}\!-\!1},\qquad\,\ {\rm where} \quad\! (M,N)\, :=\, \gcd(M,N) \end{eqnarray}\ \ \ \ \ $$

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

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 to $\ a<b,c \!\iff\! a< \min(b,c),\, $ 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 - just like above.

The conceptual structure at the core 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.

Answer 2

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

Theorem $\rm\,\ (m',n') = ((m,n)')\, $ as ideals in $\rm R\!=\!\Bbb Z[x],\,$ i.e. $\,\rm m'R\!+\!n'R = (m,n)'R, $ $\rm\,m,n\in\Bbb N$

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 $\rm\:n\! +\! m.$

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

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.

Answer 3

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 Bezout's identity , we have $$nx + my =\gcd(n,m)$$ for some $x,y\in\mathbb{Z}$. This follows $$a^{nx + my} \equiv 1\pmod t \implies a^{\gcd(n,m)} \equiv 1 \pmod t\implies t\big|\big( a^{\gcd(n,m)} - 1\big) $$ Assume that $n=\gcd(n,m)n_0$ where $n_0 \in \mathbb{Z}^*$and let $m=\gcd(n,m)m_0$ where $m_0 \in \mathbb{Z}^*$ Hence we have that $$ a^n=(a^{gcd(n,m)})^{n_0}-1=k_n(a^{gcd(n,m)}-1) \implies (a^{gcd(n,m)}-1) \big|\big (a^n-1). $$ Likewise we can prove that

$$ (a^{gcd(n,m)}-1) \big|\big (a^m-1). $$ Hence we get that $$ (a^{gcd(n,m)}-1) \big|\ t $$ Therefore

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

since, as proved above, each side divides the other.

Answer 4

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

Answer 5

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.

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

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

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

Answer 6

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

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 $d$.

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

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

Answer 7

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

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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 $$ \left.p^g-1\,\middle|\,G\right.\tag3 $$

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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 $$ \left.G\,\middle|\,p^g-1\right.\tag8 $$

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

Answer 8

I thought I'd contribute a rather nonstandard answer using ideas from category theory. I wouldn't say that any undergraduate number theory student should be expected to follow this reasoning nor am I claiming this is a particularly great approach to number theory! It's merely a bit of fun for those who like this kind of thing.

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We consider $\mathbb{N}$ as a poset where $a\leq b\iff a \mid b$ and thus a category. Posetal categories are particularly pleasant types of categories as all squares, triangles etc. commute automatically. This will mean in a number of places we have little to check (whereas usually we'd have to check various "naturality" conditions).

Step 1:

The product $A\times B$ of two objects $A,B$ in a category, if it exists, is characterised by:

• it coming with two projection morphisms $\pi_1:A\times B \to A,\pi_2:A\times B\to B$
• given maps $f:C\to A,g:C\to B$ we get a unique map $\langle f,g\rangle :C\to A\times B$ such that $\pi_1\circ\langle f,g\rangle = f, \pi_2\circ\langle f,g\rangle = g$.

Here the objects in question are natural numbers $m,n$ and if $k$ is their categorical product (supposing it exists) then:

• the first bullet point just tells us that $k \mid m$ and $k \mid n$
• the second bullet point says that if $l \mid m$ and $l \mid n$ then $l \mid k$

In other words, $k$ is precisely $\text{gcd}(m,n)$ and always exists.

Step 2:

We now claim that the mapping $n \mapsto a^n-1$ is in fact a functor $\mathbb{N}\to \mathbb{N}$. Clearly it gives an assignment of objects of $\mathbb{N}$ to other objects of that category but to be a functor it must also send morphisms to morphisms, i.e. if $m \mid n$ then it must be the case $a^m-1 \mid a^n-1$. But this is clearly the case just by thinking about the abstract polynomials $x^m-1 \mid x^n-1$ dividing each other as $$x^n-1 = (x^m-1)(x^{n-m} + x^{n-2m} + \ldots + 1) $$ The rest of the conditions of a functor (normally called "functoriality") immediately follow due to the fact that all diagrams commute in $\mathbb{N}$.

Step 3:

Now the statement of the theorem $$\text{gcd}(a^n-1,a^m-1) = a^{\text{gcd}(n,m)}-1 $$ then just says that the functor $a^{(-)}-1$ preserves products. To prove this, we could of course just immediately do some number theory (which will eventually be necessary). However a natural categorical approach here would be to check if it has a left adjoint as all functors with left adjoints preserve all small limits and binary categorical products are one of the simplest examples of a nontrivial limit.

A left adjoint would be a functor $F:\mathbb{N}\to \mathbb{N}$ such that for all pairs of integers $m,n$ we have $$F(m) \mid n \iff m \mid a^n-1.$$ The naturality condition normally part of an adjunction will automatically follow once such an $F$ is followed as, again, all diagrams commute in this category $\mathbb{N}$.

Well $m \mid a^n-1$ is equivalent to saying that $a^n \equiv 1 \pmod{m}$ which is itself (rather obviously) equivalent to the statement that $n$ is a multiple of the multiplicative order of $a$ modulo $m$, which we'll write as $\text{ord}_m(a)$. (We also learn that $a$ must be coprime to $m$ for this situation to ever arise but that's not important here.)

We will thus be done if the assignment $m\mapsto \text{ord}_m(a)$ is functorial because then we can take this to be $F$. In particular, we require that if $m \mid n$ then $\text{ord}_a(m) \mid \text{ord}_a(n)$.

But note that $$a^{\text{ord}_a(n)}\equiv 1\pmod{n}\implies a^{\text{ord}_a(n)}\equiv 1\pmod{m}$$ since $m \mid n$. Thus we must have $\text{ord}_a(m) \mid \text{ord}_a(n)$ and we are done.

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Perhaps this was fun to a certain audience; it definitely makes me smile. I think that in fact there are a lot of adjoints hiding away in various elementary number theory statements (even in the proofs of Bill Dubuque across MSE) and perhaps at some point I'll have to make a mini-survey of such ideas.

Answer 9

Apology for adding Answer(Already lot of answers)

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.

Answer 10

Let more generally $R$ be a commutative ring, $a \in R$ and $n,m \geq 0$ with $d := \mathrm{gcd}(n,m)$. Then we have an equation of ideals $$\langle a^n - 1 ,\, a^m - 1 \rangle = \langle a^d - 1 \rangle.$$ To prove this, one may simply recycle this answer. Here is an alternative proof which uses some ideas of category theory.

Two ideals $I,J \subseteq R$ are equal iff their quotients $R/I$, $R/J$ represent the same subfunctor of $\mathrm{Hom}(R,-)$. This follows from the Yoneda Lemma. (You can also prove it directly, but essentially, as always, you just repeat the proof of the Yoneda Lemma). If you are not familar with category theory, you can also rephrase this as follows:

We have $I = J$ iff for all homomorphisms $\varphi : R \to S$, where $S$ is any commutative ring, we have $\varphi(I)=0 \iff \varphi(J)=0$.

For $\langle a^n - 1 , a^m - 1 \rangle$ the subfunctor consists of those homomorphisms $\varphi : R \to S$ such that $\varphi(a^n - 1) = \varphi(a^m - 1)=0$, which is equivalent to $u^n = 1$, $u^m = 1$ for $u := \varphi(a)$. For $\langle a^d - 1 \rangle$ the subfunctor consists of those homomorphisms $\varphi : R \to S$ such that $\varphi(a^d - 1)=0$, which is equivalent to $u^d = 1$.

This reduces the whole problem to showing $$(u^n = 1 \land u^m = 1) \iff u^d = 1$$ for $u \in S$. The implication $\Leftarrow$ is straight forward from $d \mid m,n$, the implication $\Rightarrow$ follows from Bézout's theorem, and we are done.

Yes, this is overkill, and yes, these arguments have already appeared in other answers here. But I think it still gives a different perspective. As a matter of fact, showing that two ideals are equal by comparing their quotient rings and their universal properties, and likewise with normal subgroups in case of a group, is a general technique that is useful in many situations.

Answer 11

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.

Answer 12

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 $R_{m,n} = 1$.