Does $p^n$ divide $(p^n-1)!$?












0












$begingroup$


Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09
















0












$begingroup$


Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09














0












0








0





$begingroup$


Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory







share|cite|improve this question











$endgroup$




Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 16:46









Mike

4,171412




4,171412










asked Dec 11 '18 at 16:39









LucaLuca

22719




22719








  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09














  • 3




    $begingroup$
    This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
    $endgroup$
    – Arthur
    Dec 11 '18 at 16:43










  • $begingroup$
    What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:43






  • 1




    $begingroup$
    Not true for $n = 1$
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:44






  • 1




    $begingroup$
    @fleablood Or $n=p=2$.
    $endgroup$
    – Arthur
    Dec 11 '18 at 17:09








3




3




$begingroup$
This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
$endgroup$
– Arthur
Dec 11 '18 at 16:43




$begingroup$
This is almost trivial if you just write out the factorial and count the multiples of $p$, $p^2$, and so on.
$endgroup$
– Arthur
Dec 11 '18 at 16:43












$begingroup$
What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
$endgroup$
– fleablood
Dec 11 '18 at 16:43




$begingroup$
What do you think. All number less than $p^n-1$ divide ($p^n-1)!$ so $p|(p^n-1)!$ and $p^2|(p^n-1)$ and $p^3|(p^n-1)$.....
$endgroup$
– fleablood
Dec 11 '18 at 16:43




1




1




$begingroup$
Not true for $n = 1$
$endgroup$
– fleablood
Dec 11 '18 at 16:44




$begingroup$
Not true for $n = 1$
$endgroup$
– fleablood
Dec 11 '18 at 16:44




1




1




$begingroup$
@fleablood Or $n=p=2$.
$endgroup$
– Arthur
Dec 11 '18 at 17:09




$begingroup$
@fleablood Or $n=p=2$.
$endgroup$
– Arthur
Dec 11 '18 at 17:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

$(p^n - 1)! = 1*2*3......*(p^n-1)$ so



Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



(Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



$(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



So conclusion:



False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
    $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
    Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
    $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
    Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
      $endgroup$
      – Maged Saeed
      Dec 11 '18 at 22:12












    • $begingroup$
      @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
      $endgroup$
      – Frpzzd
      Dec 11 '18 at 23:21










    • $begingroup$
      Oh, I see. Thanks,
      $endgroup$
      – Maged Saeed
      Dec 12 '18 at 1:36











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035479%2fdoes-pn-divide-pn-1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



    Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



    Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



    So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



    In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



    (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



    So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



    Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



    So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



    $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



    So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



    And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



    So conclusion:



    False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



      Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



      Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



      So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



      In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



      (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



      So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



      Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



      So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



      $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



      So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



      And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



      So conclusion:



      False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



        Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



        Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



        So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



        In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



        (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



        So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



        Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



        So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



        $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



        So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



        And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



        So conclusion:



        False for $n =1$. False for $n = 2;p=2$. And true for all other cases.






        share|cite|improve this answer











        $endgroup$



        $(p^n - 1)! = 1*2*3......*(p^n-1)$ so



        Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.



        Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.



        So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.



        In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{frac {(n-1)n}2}$ will divide $(p^n-1)!$.



        (Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)



        So this will be true whenever $frac {n-1}2n ge n$ or if $nge 3$.



        Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.



        So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.



        $(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.



        So if $p-1 ge 2$ or in other words if $p ge 3$ then the will be true for $n=2$.



        And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*color{blue}2*(2^2-1)$ and that's not enough copies of $color{blue}2$.



        So conclusion:



        False for $n =1$. False for $n = 2;p=2$. And true for all other cases.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 17:51

























        answered Dec 11 '18 at 17:08









        fleabloodfleablood

        71.4k22686




        71.4k22686























            3












            $begingroup$

            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36
















            3












            $begingroup$

            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36














            3












            3








            3





            $begingroup$

            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.






            share|cite|improve this answer









            $endgroup$



            It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{n}{p^k}biggrfloor$$
            Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to
            $$sum_{k=1}^infty bigglfloor frac{p^n-1}{p^k}biggrfloor=sum_{k=1}^n (p^{n-k}-1)=frac{p^n-1}{p-1}-n$$
            Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 11 '18 at 16:45









            FrpzzdFrpzzd

            23k841109




            23k841109












            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36


















            • $begingroup$
              how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
              $endgroup$
              – Maged Saeed
              Dec 11 '18 at 22:12












            • $begingroup$
              @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
              $endgroup$
              – Frpzzd
              Dec 11 '18 at 23:21










            • $begingroup$
              Oh, I see. Thanks,
              $endgroup$
              – Maged Saeed
              Dec 12 '18 at 1:36
















            $begingroup$
            how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
            $endgroup$
            – Maged Saeed
            Dec 11 '18 at 22:12






            $begingroup$
            how did you move from $sum_{k=1}^{infty} bigglfloor frac{p^n-1}{p^k} biggrfloor$ to $sum_{k=1}^n (p^{n-k}-1)$? should it be $sum_{k=1}^n (p^{n-k})$ as $bigglfloor frac{1}{p^k}biggrfloor = 0$
            $endgroup$
            – Maged Saeed
            Dec 11 '18 at 22:12














            $begingroup$
            @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
            $endgroup$
            – Frpzzd
            Dec 11 '18 at 23:21




            $begingroup$
            @MagedSaeed No. Notice that $$bigglfloorfrac{p^n-1}{p^k}biggrfloor=lfloor p^{n-k}-1/p^krfloor$$ and that if $xinmathbb Z$ and $ylt 1$, then $lfloor x-yrfloor=x-1$.
            $endgroup$
            – Frpzzd
            Dec 11 '18 at 23:21












            $begingroup$
            Oh, I see. Thanks,
            $endgroup$
            – Maged Saeed
            Dec 12 '18 at 1:36




            $begingroup$
            Oh, I see. Thanks,
            $endgroup$
            – Maged Saeed
            Dec 12 '18 at 1:36


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035479%2fdoes-pn-divide-pn-1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more