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












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



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










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    $begingroup$
    Another question (math.stackexchange.com/questions/11567/…) was closed as a duplicate of this one where there is a second solution.
    $endgroup$
    – Qiaochu Yuan
    Dec 4 '10 at 14:17






  • 1




    $begingroup$
    Find here: Number Theory for Mathematical Contests, Example#245, Page#36.
    $endgroup$
    – lab bhattacharjee
    Jul 29 '12 at 16:56


















123












$begingroup$


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



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










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  • 6




    $begingroup$
    Another question (math.stackexchange.com/questions/11567/…) was closed as a duplicate of this one where there is a second solution.
    $endgroup$
    – Qiaochu Yuan
    Dec 4 '10 at 14:17






  • 1




    $begingroup$
    Find here: Number Theory for Mathematical Contests, Example#245, Page#36.
    $endgroup$
    – lab bhattacharjee
    Jul 29 '12 at 16:56
















123












123








123


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$begingroup$


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



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










share|cite|improve this question











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



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







elementary-number-theory divisibility faq greatest-common-divisor






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edited Jul 5 '15 at 11:43









Martin Sleziak

44.7k9117272




44.7k9117272










asked Oct 22 '10 at 0:47







Juan Liner















  • 6




    $begingroup$
    Another question (math.stackexchange.com/questions/11567/…) was closed as a duplicate of this one where there is a second solution.
    $endgroup$
    – Qiaochu Yuan
    Dec 4 '10 at 14:17






  • 1




    $begingroup$
    Find here: Number Theory for Mathematical Contests, Example#245, Page#36.
    $endgroup$
    – lab bhattacharjee
    Jul 29 '12 at 16:56
















  • 6




    $begingroup$
    Another question (math.stackexchange.com/questions/11567/…) was closed as a duplicate of this one where there is a second solution.
    $endgroup$
    – Qiaochu Yuan
    Dec 4 '10 at 14:17






  • 1




    $begingroup$
    Find here: Number Theory for Mathematical Contests, Example#245, Page#36.
    $endgroup$
    – lab bhattacharjee
    Jul 29 '12 at 16:56










6




6




$begingroup$
Another question (math.stackexchange.com/questions/11567/…) was closed as a duplicate of this one where there is a second solution.
$endgroup$
– Qiaochu Yuan
Dec 4 '10 at 14:17




$begingroup$
Another question (math.stackexchange.com/questions/11567/…) was closed as a duplicate of this one where there is a second solution.
$endgroup$
– Qiaochu Yuan
Dec 4 '10 at 14:17




1




1




$begingroup$
Find here: Number Theory for Mathematical Contests, Example#245, Page#36.
$endgroup$
– lab bhattacharjee
Jul 29 '12 at 16:56






$begingroup$
Find here: Number Theory for Mathematical Contests, Example#245, Page#36.
$endgroup$
– lab bhattacharjee
Jul 29 '12 at 16:56












7 Answers
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$rm f_{,n}: := a^n!-!1 = a^{n-m} : color{#c00}{(a^m!-!1)} + color{#0a0}{a^{n-m}!-!1}. $ Hence $rm: {f_{,n}! = color{#0a0}{f_{,n-m}}! + k color{#c00}{f_{,m}}},, kinmathbb Z.:$ Apply



Theorem $: $ If $rm f_{, n}: $ is an integer sequence with $rm f_{0} =, 0,: $ $rm :{ f_{,n}!equiv color{#0a0}{f_{,n-m}} (mod color{#c00}{f_{,m})}} $ for $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 = 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 also this post for a conceptual proof exploiting the innate structure - an order ideal.






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  • $begingroup$
    Sort of like the Fibonacci sequence!
    $endgroup$
    – cactus314
    May 23 '15 at 12:03










  • $begingroup$
    @john Yes, they are both strong divisibility sequences, i.e. $,(f_n,f_m) = f_{(n,m)}.,$ See here for the Fibonacci case.
    $endgroup$
    – Bill Dubuque
    May 23 '15 at 13:33





















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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 $rm m,n = 15,21$



$rmdisplaystylequadquadquadquadquadquadquad frac{x^3-1}{x-1} = (x^{15}! +! x^9! +! 1) frac{x^{15}!-!1}{x!-!1} - (x^9!+!x^3) frac{x^{21}!-!1}{x!-!1}$



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



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



Theorem $rmquad (m',n') = ((m,n)') $ 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 $rm:n! +! m.$



$begin{eqnarray}rm &rm x^n! -! 1 &=& rm x^r (x^m! -! 1) + x^r! -! 1 quad rm for r = n! -! m \
quadRightarrowquad &rmqquad n' &=& rm x^r m' + r' quad rm by dividing above by x!-!1 \
quadRightarrow &rm (m', n'), &=& rm (m', r') \
& &=&rm ((m,r)') quad by induction, applicable by: m!+!r = n < n!+!m \
& &=&rm ((m,n)') quad by r equiv n :(mod m)quad bf QED
end{eqnarray}$



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.






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  • $begingroup$
    Is $((rm m,n)')$ supposed to be $((rm m,n))'$ i.e. $rm dfrac{x^{(m,n)}-1}{x-1}$?
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    – Pedro Tamaroff
    Jun 18 '12 at 23:38












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    @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 $rm:f,g,hinmathbb Z[x].:$ 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 $rm:a,bin mathbb Z[x],:$ which implies $rm:f = gcd(g,h).$
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    – Bill Dubuque
    Jun 19 '12 at 0:17





















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Let $mge nge 1$. Apply Euclidean Algorithm.



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



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






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  • $begingroup$
    And this too is a duplicate of an answer in the 5-year-old linked duplicate thread.
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    – Bill Dubuque
    Dec 31 '16 at 2:07





















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More generally, if $gcd(a,b)=1$, $a,b,m,ninmathbb Z^+$, $a> b$, 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.



$$dmid a^m-b^m, a^n-b^niff left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$$



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



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






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  • $begingroup$
    This is precisely the homogenization $(a^n-1to a^n-b^n)$ 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!
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    – Bill Dubuque
    Dec 31 '16 at 2:03












  • $begingroup$
    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.
    $endgroup$
    – Bill Dubuque
    Jul 13 '17 at 20:44












  • $begingroup$
    I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$?
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    – Vmimi
    Dec 2 '18 at 23:48



















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More generally, if $a,b,m,ninmathbb Z_{ge 1}$, $a>b$ 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 $nmid a,biff nmid (a,b)$ to prove:



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



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



$a^mequiv b^m,, a^nequiv b^n$ mod $d$ by definition of $d$.



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



$(a,b)=1iff (a,d)=(b,d)=1$, 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 $d$.



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



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






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  • $begingroup$
    What is $d$? I don't understand why gcd$(b, d)=1 $ and why do you need that to be true?
    $endgroup$
    – Vmimi
    Nov 30 '18 at 0:22





















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Let
$$gcd(a^n - 1, a^m - 1) = t$$
then
$$a^n equiv 1 ,big(text{ mod } tbig),quadtext{and}quad,a^m equiv 1 ,big(text{ mod } tbig)$$
And thus
$$a^{nx + my} equiv 1, big(text{ mod } tbig)$$
$forall,x,,yin mathbb{Z}$



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



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






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    3












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





    For $xge0$,
    $$
    left(p^a-1right)sum_{k=0}^{x-1}p^{ak}=p^{ax}-1
    $$
    Therefore, we have that
    $$
    left.G,middle|,p^{ax}-1right.
    $$
    If $left.G,middle|,p^{ax-(b-1)y}-1right.$, then
    $$
    left(p^{ax-(b-1)y}-1right)-p^{ax-by}left(p^b-1right)=p^{ax-by}-1
    $$
    Therefore, for any $x,yge0$ so that $ax-byge0$,
    $$
    left.G,middle|,p^{ax-by}-1right.
    $$
    which means that
    $$
    left.G,middle|,p^g-1right.
    $$





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






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      7 Answers
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      7 Answers
      7






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      63












      $begingroup$

      $rm f_{,n}: := a^n!-!1 = a^{n-m} : color{#c00}{(a^m!-!1)} + color{#0a0}{a^{n-m}!-!1}. $ Hence $rm: {f_{,n}! = color{#0a0}{f_{,n-m}}! + k color{#c00}{f_{,m}}},, kinmathbb Z.:$ Apply



      Theorem $: $ If $rm f_{, n}: $ is an integer sequence with $rm f_{0} =, 0,: $ $rm :{ f_{,n}!equiv color{#0a0}{f_{,n-m}} (mod color{#c00}{f_{,m})}} $ for $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 = 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 also this post for a conceptual proof exploiting the innate structure - an order ideal.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Sort of like the Fibonacci sequence!
        $endgroup$
        – cactus314
        May 23 '15 at 12:03










      • $begingroup$
        @john Yes, they are both strong divisibility sequences, i.e. $,(f_n,f_m) = f_{(n,m)}.,$ See here for the Fibonacci case.
        $endgroup$
        – Bill Dubuque
        May 23 '15 at 13:33


















      63












      $begingroup$

      $rm f_{,n}: := a^n!-!1 = a^{n-m} : color{#c00}{(a^m!-!1)} + color{#0a0}{a^{n-m}!-!1}. $ Hence $rm: {f_{,n}! = color{#0a0}{f_{,n-m}}! + k color{#c00}{f_{,m}}},, kinmathbb Z.:$ Apply



      Theorem $: $ If $rm f_{, n}: $ is an integer sequence with $rm f_{0} =, 0,: $ $rm :{ f_{,n}!equiv color{#0a0}{f_{,n-m}} (mod color{#c00}{f_{,m})}} $ for $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 = 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 also this post for a conceptual proof exploiting the innate structure - an order ideal.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Sort of like the Fibonacci sequence!
        $endgroup$
        – cactus314
        May 23 '15 at 12:03










      • $begingroup$
        @john Yes, they are both strong divisibility sequences, i.e. $,(f_n,f_m) = f_{(n,m)}.,$ See here for the Fibonacci case.
        $endgroup$
        – Bill Dubuque
        May 23 '15 at 13:33
















      63












      63








      63





      $begingroup$

      $rm f_{,n}: := a^n!-!1 = a^{n-m} : color{#c00}{(a^m!-!1)} + color{#0a0}{a^{n-m}!-!1}. $ Hence $rm: {f_{,n}! = color{#0a0}{f_{,n-m}}! + k color{#c00}{f_{,m}}},, kinmathbb Z.:$ Apply



      Theorem $: $ If $rm f_{, n}: $ is an integer sequence with $rm f_{0} =, 0,: $ $rm :{ f_{,n}!equiv color{#0a0}{f_{,n-m}} (mod color{#c00}{f_{,m})}} $ for $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 = 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 also this post for a conceptual proof exploiting the innate structure - an order ideal.






      share|cite|improve this answer











      $endgroup$



      $rm f_{,n}: := a^n!-!1 = a^{n-m} : color{#c00}{(a^m!-!1)} + color{#0a0}{a^{n-m}!-!1}. $ Hence $rm: {f_{,n}! = color{#0a0}{f_{,n-m}}! + k color{#c00}{f_{,m}}},, kinmathbb Z.:$ Apply



      Theorem $: $ If $rm f_{, n}: $ is an integer sequence with $rm f_{0} =, 0,: $ $rm :{ f_{,n}!equiv color{#0a0}{f_{,n-m}} (mod color{#c00}{f_{,m})}} $ for $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 = 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 also this post for a conceptual proof exploiting the innate structure - an order ideal.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Apr 13 '17 at 12:19









      Community

      1




      1










      answered Oct 23 '10 at 2:21









      Bill DubuqueBill Dubuque

      210k29191639




      210k29191639












      • $begingroup$
        Sort of like the Fibonacci sequence!
        $endgroup$
        – cactus314
        May 23 '15 at 12:03










      • $begingroup$
        @john Yes, they are both strong divisibility sequences, i.e. $,(f_n,f_m) = f_{(n,m)}.,$ See here for the Fibonacci case.
        $endgroup$
        – Bill Dubuque
        May 23 '15 at 13:33




















      • $begingroup$
        Sort of like the Fibonacci sequence!
        $endgroup$
        – cactus314
        May 23 '15 at 12:03










      • $begingroup$
        @john Yes, they are both strong divisibility sequences, i.e. $,(f_n,f_m) = f_{(n,m)}.,$ See here for the Fibonacci case.
        $endgroup$
        – Bill Dubuque
        May 23 '15 at 13:33


















      $begingroup$
      Sort of like the Fibonacci sequence!
      $endgroup$
      – cactus314
      May 23 '15 at 12:03




      $begingroup$
      Sort of like the Fibonacci sequence!
      $endgroup$
      – cactus314
      May 23 '15 at 12:03












      $begingroup$
      @john Yes, they are both strong divisibility sequences, i.e. $,(f_n,f_m) = f_{(n,m)}.,$ See here for the Fibonacci case.
      $endgroup$
      – Bill Dubuque
      May 23 '15 at 13:33






      $begingroup$
      @john Yes, they are both strong divisibility sequences, i.e. $,(f_n,f_m) = f_{(n,m)}.,$ See here for the Fibonacci case.
      $endgroup$
      – Bill Dubuque
      May 23 '15 at 13:33













      33












      $begingroup$

      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 $rm m,n = 15,21$



      $rmdisplaystylequadquadquadquadquadquadquad frac{x^3-1}{x-1} = (x^{15}! +! x^9! +! 1) frac{x^{15}!-!1}{x!-!1} - (x^9!+!x^3) frac{x^{21}!-!1}{x!-!1}$



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



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



      Theorem $rmquad (m',n') = ((m,n)') $ 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 $rm:n! +! m.$



      $begin{eqnarray}rm &rm x^n! -! 1 &=& rm x^r (x^m! -! 1) + x^r! -! 1 quad rm for r = n! -! m \
      quadRightarrowquad &rmqquad n' &=& rm x^r m' + r' quad rm by dividing above by x!-!1 \
      quadRightarrow &rm (m', n'), &=& rm (m', r') \
      & &=&rm ((m,r)') quad by induction, applicable by: m!+!r = n < n!+!m \
      & &=&rm ((m,n)') quad by r equiv n :(mod m)quad bf QED
      end{eqnarray}$



      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.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Is $((rm m,n)')$ supposed to be $((rm m,n))'$ i.e. $rm dfrac{x^{(m,n)}-1}{x-1}$?
        $endgroup$
        – Pedro Tamaroff
        Jun 18 '12 at 23:38












      • $begingroup$
        @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 $rm:f,g,hinmathbb Z[x].:$ 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 $rm:a,bin mathbb Z[x],:$ which implies $rm:f = gcd(g,h).$
        $endgroup$
        – Bill Dubuque
        Jun 19 '12 at 0:17


















      33












      $begingroup$

      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 $rm m,n = 15,21$



      $rmdisplaystylequadquadquadquadquadquadquad frac{x^3-1}{x-1} = (x^{15}! +! x^9! +! 1) frac{x^{15}!-!1}{x!-!1} - (x^9!+!x^3) frac{x^{21}!-!1}{x!-!1}$



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



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



      Theorem $rmquad (m',n') = ((m,n)') $ 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 $rm:n! +! m.$



      $begin{eqnarray}rm &rm x^n! -! 1 &=& rm x^r (x^m! -! 1) + x^r! -! 1 quad rm for r = n! -! m \
      quadRightarrowquad &rmqquad n' &=& rm x^r m' + r' quad rm by dividing above by x!-!1 \
      quadRightarrow &rm (m', n'), &=& rm (m', r') \
      & &=&rm ((m,r)') quad by induction, applicable by: m!+!r = n < n!+!m \
      & &=&rm ((m,n)') quad by r equiv n :(mod m)quad bf QED
      end{eqnarray}$



      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.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Is $((rm m,n)')$ supposed to be $((rm m,n))'$ i.e. $rm dfrac{x^{(m,n)}-1}{x-1}$?
        $endgroup$
        – Pedro Tamaroff
        Jun 18 '12 at 23:38












      • $begingroup$
        @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 $rm:f,g,hinmathbb Z[x].:$ 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 $rm:a,bin mathbb Z[x],:$ which implies $rm:f = gcd(g,h).$
        $endgroup$
        – Bill Dubuque
        Jun 19 '12 at 0:17
















      33












      33








      33





      $begingroup$

      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 $rm m,n = 15,21$



      $rmdisplaystylequadquadquadquadquadquadquad frac{x^3-1}{x-1} = (x^{15}! +! x^9! +! 1) frac{x^{15}!-!1}{x!-!1} - (x^9!+!x^3) frac{x^{21}!-!1}{x!-!1}$



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



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



      Theorem $rmquad (m',n') = ((m,n)') $ 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 $rm:n! +! m.$



      $begin{eqnarray}rm &rm x^n! -! 1 &=& rm x^r (x^m! -! 1) + x^r! -! 1 quad rm for r = n! -! m \
      quadRightarrowquad &rmqquad n' &=& rm x^r m' + r' quad rm by dividing above by x!-!1 \
      quadRightarrow &rm (m', n'), &=& rm (m', r') \
      & &=&rm ((m,r)') quad by induction, applicable by: m!+!r = n < n!+!m \
      & &=&rm ((m,n)') quad by r equiv n :(mod m)quad bf QED
      end{eqnarray}$



      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.






      share|cite|improve this answer











      $endgroup$



      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 $rm m,n = 15,21$



      $rmdisplaystylequadquadquadquadquadquadquad frac{x^3-1}{x-1} = (x^{15}! +! x^9! +! 1) frac{x^{15}!-!1}{x!-!1} - (x^9!+!x^3) frac{x^{21}!-!1}{x!-!1}$



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



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



      Theorem $rmquad (m',n') = ((m,n)') $ 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 $rm:n! +! m.$



      $begin{eqnarray}rm &rm x^n! -! 1 &=& rm x^r (x^m! -! 1) + x^r! -! 1 quad rm for r = n! -! m \
      quadRightarrowquad &rmqquad n' &=& rm x^r m' + r' quad rm by dividing above by x!-!1 \
      quadRightarrow &rm (m', n'), &=& rm (m', r') \
      & &=&rm ((m,r)') quad by induction, applicable by: m!+!r = n < n!+!m \
      & &=&rm ((m,n)') quad by r equiv n :(mod m)quad bf QED
      end{eqnarray}$



      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.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jul 4 '15 at 19:25

























      answered Oct 22 '10 at 1:51









      Bill DubuqueBill Dubuque

      210k29191639




      210k29191639












      • $begingroup$
        Is $((rm m,n)')$ supposed to be $((rm m,n))'$ i.e. $rm dfrac{x^{(m,n)}-1}{x-1}$?
        $endgroup$
        – Pedro Tamaroff
        Jun 18 '12 at 23:38












      • $begingroup$
        @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 $rm:f,g,hinmathbb Z[x].:$ 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 $rm:a,bin mathbb Z[x],:$ which implies $rm:f = gcd(g,h).$
        $endgroup$
        – Bill Dubuque
        Jun 19 '12 at 0:17




















      • $begingroup$
        Is $((rm m,n)')$ supposed to be $((rm m,n))'$ i.e. $rm dfrac{x^{(m,n)}-1}{x-1}$?
        $endgroup$
        – Pedro Tamaroff
        Jun 18 '12 at 23:38












      • $begingroup$
        @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 $rm:f,g,hinmathbb Z[x].:$ 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 $rm:a,bin mathbb Z[x],:$ which implies $rm:f = gcd(g,h).$
        $endgroup$
        – Bill Dubuque
        Jun 19 '12 at 0:17


















      $begingroup$
      Is $((rm m,n)')$ supposed to be $((rm m,n))'$ i.e. $rm dfrac{x^{(m,n)}-1}{x-1}$?
      $endgroup$
      – Pedro Tamaroff
      Jun 18 '12 at 23:38






      $begingroup$
      Is $((rm m,n)')$ supposed to be $((rm m,n))'$ i.e. $rm dfrac{x^{(m,n)}-1}{x-1}$?
      $endgroup$
      – Pedro Tamaroff
      Jun 18 '12 at 23:38














      $begingroup$
      @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 $rm:f,g,hinmathbb Z[x].:$ 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 $rm:a,bin mathbb Z[x],:$ which implies $rm:f = gcd(g,h).$
      $endgroup$
      – Bill Dubuque
      Jun 19 '12 at 0:17






      $begingroup$
      @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 $rm:f,g,hinmathbb Z[x].:$ 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 $rm:a,bin mathbb Z[x],:$ which implies $rm:f = gcd(g,h).$
      $endgroup$
      – Bill Dubuque
      Jun 19 '12 at 0:17













      13












      $begingroup$

      Let $mge nge 1$. Apply Euclidean Algorithm.



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



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






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        And this too is a duplicate of an answer in the 5-year-old linked duplicate thread.
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:07


















      13












      $begingroup$

      Let $mge nge 1$. Apply Euclidean Algorithm.



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



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






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        And this too is a duplicate of an answer in the 5-year-old linked duplicate thread.
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:07
















      13












      13








      13





      $begingroup$

      Let $mge nge 1$. Apply Euclidean Algorithm.



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



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






      share|cite|improve this answer











      $endgroup$



      Let $mge nge 1$. Apply Euclidean Algorithm.



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



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







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 8 '16 at 18:55

























      answered Oct 25 '15 at 10:23









      user236182user236182

      11.9k11233




      11.9k11233












      • $begingroup$
        And this too is a duplicate of an answer in the 5-year-old linked duplicate thread.
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:07




















      • $begingroup$
        And this too is a duplicate of an answer in the 5-year-old linked duplicate thread.
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:07


















      $begingroup$
      And this too is a duplicate of an answer in the 5-year-old linked duplicate thread.
      $endgroup$
      – Bill Dubuque
      Dec 31 '16 at 2:07






      $begingroup$
      And this too is a duplicate of an answer in the 5-year-old linked duplicate thread.
      $endgroup$
      – Bill Dubuque
      Dec 31 '16 at 2:07













      8












      $begingroup$

      More generally, if $gcd(a,b)=1$, $a,b,m,ninmathbb Z^+$, $a> b$, 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.



      $$dmid a^m-b^m, a^n-b^niff left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$$



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



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






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        This is precisely the homogenization $(a^n-1to a^n-b^n)$ 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!
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:03












      • $begingroup$
        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.
        $endgroup$
        – Bill Dubuque
        Jul 13 '17 at 20:44












      • $begingroup$
        I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$?
        $endgroup$
        – Vmimi
        Dec 2 '18 at 23:48
















      8












      $begingroup$

      More generally, if $gcd(a,b)=1$, $a,b,m,ninmathbb Z^+$, $a> b$, 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.



      $$dmid a^m-b^m, a^n-b^niff left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$$



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



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






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        This is precisely the homogenization $(a^n-1to a^n-b^n)$ 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!
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:03












      • $begingroup$
        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.
        $endgroup$
        – Bill Dubuque
        Jul 13 '17 at 20:44












      • $begingroup$
        I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$?
        $endgroup$
        – Vmimi
        Dec 2 '18 at 23:48














      8












      8








      8





      $begingroup$

      More generally, if $gcd(a,b)=1$, $a,b,m,ninmathbb Z^+$, $a> b$, 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.



      $$dmid a^m-b^m, a^n-b^niff left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$$



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



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






      share|cite|improve this answer











      $endgroup$



      More generally, if $gcd(a,b)=1$, $a,b,m,ninmathbb Z^+$, $a> b$, 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.



      $$dmid a^m-b^m, a^n-b^niff left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$$



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



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







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 1 '17 at 10:31

























      answered Oct 4 '15 at 17:03









      user236182user236182

      11.9k11233




      11.9k11233












      • $begingroup$
        This is precisely the homogenization $(a^n-1to a^n-b^n)$ 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!
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:03












      • $begingroup$
        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.
        $endgroup$
        – Bill Dubuque
        Jul 13 '17 at 20:44












      • $begingroup$
        I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$?
        $endgroup$
        – Vmimi
        Dec 2 '18 at 23:48


















      • $begingroup$
        This is precisely the homogenization $(a^n-1to a^n-b^n)$ 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!
        $endgroup$
        – Bill Dubuque
        Dec 31 '16 at 2:03












      • $begingroup$
        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.
        $endgroup$
        – Bill Dubuque
        Jul 13 '17 at 20:44












      • $begingroup$
        I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$?
        $endgroup$
        – Vmimi
        Dec 2 '18 at 23:48
















      $begingroup$
      This is precisely the homogenization $(a^n-1to a^n-b^n)$ 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!
      $endgroup$
      – Bill Dubuque
      Dec 31 '16 at 2:03






      $begingroup$
      This is precisely the homogenization $(a^n-1to a^n-b^n)$ 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!
      $endgroup$
      – Bill Dubuque
      Dec 31 '16 at 2:03














      $begingroup$
      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.
      $endgroup$
      – Bill Dubuque
      Jul 13 '17 at 20:44






      $begingroup$
      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.
      $endgroup$
      – Bill Dubuque
      Jul 13 '17 at 20:44














      $begingroup$
      I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$?
      $endgroup$
      – Vmimi
      Dec 2 '18 at 23:48




      $begingroup$
      I didn't understand why if gcd$(a,b) =1$ then gcd$(b, d) =1$? and why $left(ab^{-1}right)^mequiv left(ab^{-1}right)^nequiv 1pmod{d}$?
      $endgroup$
      – Vmimi
      Dec 2 '18 at 23:48











      8












      $begingroup$

      More generally, if $a,b,m,ninmathbb Z_{ge 1}$, $a>b$ 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 $nmid a,biff nmid (a,b)$ to prove:



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



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



      $a^mequiv b^m,, a^nequiv b^n$ mod $d$ by definition of $d$.



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



      $(a,b)=1iff (a,d)=(b,d)=1$, 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 $d$.



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



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






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        What is $d$? I don't understand why gcd$(b, d)=1 $ and why do you need that to be true?
        $endgroup$
        – Vmimi
        Nov 30 '18 at 0:22


















      8












      $begingroup$

      More generally, if $a,b,m,ninmathbb Z_{ge 1}$, $a>b$ 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 $nmid a,biff nmid (a,b)$ to prove:



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



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



      $a^mequiv b^m,, a^nequiv b^n$ mod $d$ by definition of $d$.



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



      $(a,b)=1iff (a,d)=(b,d)=1$, 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 $d$.



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



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






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        What is $d$? I don't understand why gcd$(b, d)=1 $ and why do you need that to be true?
        $endgroup$
        – Vmimi
        Nov 30 '18 at 0:22
















      8












      8








      8





      $begingroup$

      More generally, if $a,b,m,ninmathbb Z_{ge 1}$, $a>b$ 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 $nmid a,biff nmid (a,b)$ to prove:



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



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



      $a^mequiv b^m,, a^nequiv b^n$ mod $d$ by definition of $d$.



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



      $(a,b)=1iff (a,d)=(b,d)=1$, 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 $d$.



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



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






      share|cite|improve this answer











      $endgroup$



      More generally, if $a,b,m,ninmathbb Z_{ge 1}$, $a>b$ 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 $nmid a,biff nmid (a,b)$ to prove:



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



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



      $a^mequiv b^m,, a^nequiv b^n$ mod $d$ by definition of $d$.



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



      $(a,b)=1iff (a,d)=(b,d)=1$, 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 $d$.



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



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







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 5 '18 at 23:28


























      community wiki





      8 revs
      user26486













      • $begingroup$
        What is $d$? I don't understand why gcd$(b, d)=1 $ and why do you need that to be true?
        $endgroup$
        – Vmimi
        Nov 30 '18 at 0:22




















      • $begingroup$
        What is $d$? I don't understand why gcd$(b, d)=1 $ and why do you need that to be true?
        $endgroup$
        – Vmimi
        Nov 30 '18 at 0:22


















      $begingroup$
      What is $d$? I don't understand why gcd$(b, d)=1 $ and why do you need that to be true?
      $endgroup$
      – Vmimi
      Nov 30 '18 at 0:22






      $begingroup$
      What is $d$? I don't understand why gcd$(b, d)=1 $ and why do you need that to be true?
      $endgroup$
      – Vmimi
      Nov 30 '18 at 0:22













      5












      $begingroup$

      Let
      $$gcd(a^n - 1, a^m - 1) = t$$
      then
      $$a^n equiv 1 ,big(text{ mod } tbig),quadtext{and}quad,a^m equiv 1 ,big(text{ mod } tbig)$$
      And thus
      $$a^{nx + my} equiv 1, big(text{ mod } tbig)$$
      $forall,x,,yin mathbb{Z}$



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



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






      share|cite|improve this answer











      $endgroup$


















        5












        $begingroup$

        Let
        $$gcd(a^n - 1, a^m - 1) = t$$
        then
        $$a^n equiv 1 ,big(text{ mod } tbig),quadtext{and}quad,a^m equiv 1 ,big(text{ mod } tbig)$$
        And thus
        $$a^{nx + my} equiv 1, big(text{ mod } tbig)$$
        $forall,x,,yin mathbb{Z}$



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



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






        share|cite|improve this answer











        $endgroup$
















          5












          5








          5





          $begingroup$

          Let
          $$gcd(a^n - 1, a^m - 1) = t$$
          then
          $$a^n equiv 1 ,big(text{ mod } tbig),quadtext{and}quad,a^m equiv 1 ,big(text{ mod } tbig)$$
          And thus
          $$a^{nx + my} equiv 1, big(text{ mod } tbig)$$
          $forall,x,,yin mathbb{Z}$



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



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






          share|cite|improve this answer











          $endgroup$



          Let
          $$gcd(a^n - 1, a^m - 1) = t$$
          then
          $$a^n equiv 1 ,big(text{ mod } tbig),quadtext{and}quad,a^m equiv 1 ,big(text{ mod } tbig)$$
          And thus
          $$a^{nx + my} equiv 1, big(text{ mod } tbig)$$
          $forall,x,,yin mathbb{Z}$



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



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







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '17 at 14:40

























          answered Dec 1 '17 at 14:34









          Darío A. GutiérrezDarío A. Gutiérrez

          2,59941530




          2,59941530























              3












              $begingroup$

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





              For $xge0$,
              $$
              left(p^a-1right)sum_{k=0}^{x-1}p^{ak}=p^{ax}-1
              $$
              Therefore, we have that
              $$
              left.G,middle|,p^{ax}-1right.
              $$
              If $left.G,middle|,p^{ax-(b-1)y}-1right.$, then
              $$
              left(p^{ax-(b-1)y}-1right)-p^{ax-by}left(p^b-1right)=p^{ax-by}-1
              $$
              Therefore, for any $x,yge0$ so that $ax-byge0$,
              $$
              left.G,middle|,p^{ax-by}-1right.
              $$
              which means that
              $$
              left.G,middle|,p^g-1right.
              $$





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






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

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





                For $xge0$,
                $$
                left(p^a-1right)sum_{k=0}^{x-1}p^{ak}=p^{ax}-1
                $$
                Therefore, we have that
                $$
                left.G,middle|,p^{ax}-1right.
                $$
                If $left.G,middle|,p^{ax-(b-1)y}-1right.$, then
                $$
                left(p^{ax-(b-1)y}-1right)-p^{ax-by}left(p^b-1right)=p^{ax-by}-1
                $$
                Therefore, for any $x,yge0$ so that $ax-byge0$,
                $$
                left.G,middle|,p^{ax-by}-1right.
                $$
                which means that
                $$
                left.G,middle|,p^g-1right.
                $$





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






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

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





                  For $xge0$,
                  $$
                  left(p^a-1right)sum_{k=0}^{x-1}p^{ak}=p^{ax}-1
                  $$
                  Therefore, we have that
                  $$
                  left.G,middle|,p^{ax}-1right.
                  $$
                  If $left.G,middle|,p^{ax-(b-1)y}-1right.$, then
                  $$
                  left(p^{ax-(b-1)y}-1right)-p^{ax-by}left(p^b-1right)=p^{ax-by}-1
                  $$
                  Therefore, for any $x,yge0$ so that $ax-byge0$,
                  $$
                  left.G,middle|,p^{ax-by}-1right.
                  $$
                  which means that
                  $$
                  left.G,middle|,p^g-1right.
                  $$





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






                  share|cite|improve this answer











                  $endgroup$



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





                  For $xge0$,
                  $$
                  left(p^a-1right)sum_{k=0}^{x-1}p^{ak}=p^{ax}-1
                  $$
                  Therefore, we have that
                  $$
                  left.G,middle|,p^{ax}-1right.
                  $$
                  If $left.G,middle|,p^{ax-(b-1)y}-1right.$, then
                  $$
                  left(p^{ax-(b-1)y}-1right)-p^{ax-by}left(p^b-1right)=p^{ax-by}-1
                  $$
                  Therefore, for any $x,yge0$ so that $ax-byge0$,
                  $$
                  left.G,middle|,p^{ax-by}-1right.
                  $$
                  which means that
                  $$
                  left.G,middle|,p^g-1right.
                  $$





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







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 15 '18 at 14:32

























                  answered Mar 15 '18 at 14:11









                  robjohnrobjohn

                  266k27306630




                  266k27306630






























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