$a^n-a + 1 $ divisible by $n$












19












$begingroup$



Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?







number-theory elementary-number-theory divisibility







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Interesting question. Have you tried anything yet?
    $endgroup$
    – Mohammad Zuhair Khan
    Oct 24 '18 at 8:30






  • 3




    $begingroup$
    The case $a=2$ is popular, see here and the related links.
    $endgroup$
    – Dietrich Burde
    Oct 24 '18 at 8:48






  • 1




    $begingroup$
    $n$ is not a prime number. It also must be odd.
    $endgroup$
    – Oldboy
    Oct 24 '18 at 18:16






  • 2




    $begingroup$
    My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
    $endgroup$
    – Drona
    Oct 26 '18 at 19:54






  • 6




    $begingroup$
    @Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
    $endgroup$
    – Oldboy
    Oct 27 '18 at 8:39


















19












$begingroup$



Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?







number-theory elementary-number-theory divisibility







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Interesting question. Have you tried anything yet?
    $endgroup$
    – Mohammad Zuhair Khan
    Oct 24 '18 at 8:30






  • 3




    $begingroup$
    The case $a=2$ is popular, see here and the related links.
    $endgroup$
    – Dietrich Burde
    Oct 24 '18 at 8:48






  • 1




    $begingroup$
    $n$ is not a prime number. It also must be odd.
    $endgroup$
    – Oldboy
    Oct 24 '18 at 18:16






  • 2




    $begingroup$
    My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
    $endgroup$
    – Drona
    Oct 26 '18 at 19:54






  • 6




    $begingroup$
    @Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
    $endgroup$
    – Oldboy
    Oct 27 '18 at 8:39
















19












19








19


11



$begingroup$



Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?







number-theory elementary-number-theory divisibility







share|cite|improve this question











$endgroup$





Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?







number-theory elementary-number-theory divisibility




number-theory elementary-number-theory divisibility






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 16:57









Klangen

1,73711334




1,73711334










asked Oct 24 '18 at 8:10









DronaDrona

1627




1627








  • 5




    $begingroup$
    Interesting question. Have you tried anything yet?
    $endgroup$
    – Mohammad Zuhair Khan
    Oct 24 '18 at 8:30






  • 3




    $begingroup$
    The case $a=2$ is popular, see here and the related links.
    $endgroup$
    – Dietrich Burde
    Oct 24 '18 at 8:48






  • 1




    $begingroup$
    $n$ is not a prime number. It also must be odd.
    $endgroup$
    – Oldboy
    Oct 24 '18 at 18:16






  • 2




    $begingroup$
    My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
    $endgroup$
    – Drona
    Oct 26 '18 at 19:54






  • 6




    $begingroup$
    @Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
    $endgroup$
    – Oldboy
    Oct 27 '18 at 8:39
















  • 5




    $begingroup$
    Interesting question. Have you tried anything yet?
    $endgroup$
    – Mohammad Zuhair Khan
    Oct 24 '18 at 8:30






  • 3




    $begingroup$
    The case $a=2$ is popular, see here and the related links.
    $endgroup$
    – Dietrich Burde
    Oct 24 '18 at 8:48






  • 1




    $begingroup$
    $n$ is not a prime number. It also must be odd.
    $endgroup$
    – Oldboy
    Oct 24 '18 at 18:16






  • 2




    $begingroup$
    My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
    $endgroup$
    – Drona
    Oct 26 '18 at 19:54






  • 6




    $begingroup$
    @Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
    $endgroup$
    – Oldboy
    Oct 27 '18 at 8:39










5




5




$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30




$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30




3




3




$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48




$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48




1




1




$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16




$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16




2




2




$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54




$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54




6




6




$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39






$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39












1 Answer
1






active

oldest

votes


















1












$begingroup$

$N=a^n-a+1$



$a^{p_1} ≡ a mod p_1$



$a^{p_2 } ≡a mod p_2$



$(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$



$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$



⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$



If $n=p_1p_2 | a^n-a+1$ then we must have:



$p_1p_2 | k_1p_1 + k_2p_2+1$



So we have following linear equation:



$k_1p_1 + k_2 p_2 = m p_1p_2-1$



For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:



with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:



$k_1= 7 t + 11$ and $k_2= -5 t+7$.



Now in first step problem reduces to:



Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.



For example for $a=3$, $p_1=5$ and $p_2=7$ we have:



$3^{35}-3+1=105$ and $(35, 105)=5$



In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.



Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:



$k_1p_1+k_2p_2= M(a)$



Or: $a | k_1p_1+k_2p_2$



We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$



This can help us in choosing a and primes $p_1$ and $p_2$



I see no reason for the lack of more solutions.






share|cite|improve this answer









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    active

    oldest

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    1












    $begingroup$

    $N=a^n-a+1$



    $a^{p_1} ≡ a mod p_1$



    $a^{p_2 } ≡a mod p_2$



    $(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$



    $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$



    ⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$



    If $n=p_1p_2 | a^n-a+1$ then we must have:



    $p_1p_2 | k_1p_1 + k_2p_2+1$



    So we have following linear equation:



    $k_1p_1 + k_2 p_2 = m p_1p_2-1$



    For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:



    with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:



    $k_1= 7 t + 11$ and $k_2= -5 t+7$.



    Now in first step problem reduces to:



    Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.



    For example for $a=3$, $p_1=5$ and $p_2=7$ we have:



    $3^{35}-3+1=105$ and $(35, 105)=5$



    In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.



    Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:



    $k_1p_1+k_2p_2= M(a)$



    Or: $a | k_1p_1+k_2p_2$



    We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$



    This can help us in choosing a and primes $p_1$ and $p_2$



    I see no reason for the lack of more solutions.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $N=a^n-a+1$



      $a^{p_1} ≡ a mod p_1$



      $a^{p_2 } ≡a mod p_2$



      $(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$



      $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$



      ⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$



      If $n=p_1p_2 | a^n-a+1$ then we must have:



      $p_1p_2 | k_1p_1 + k_2p_2+1$



      So we have following linear equation:



      $k_1p_1 + k_2 p_2 = m p_1p_2-1$



      For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:



      with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:



      $k_1= 7 t + 11$ and $k_2= -5 t+7$.



      Now in first step problem reduces to:



      Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.



      For example for $a=3$, $p_1=5$ and $p_2=7$ we have:



      $3^{35}-3+1=105$ and $(35, 105)=5$



      In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.



      Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:



      $k_1p_1+k_2p_2= M(a)$



      Or: $a | k_1p_1+k_2p_2$



      We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$



      This can help us in choosing a and primes $p_1$ and $p_2$



      I see no reason for the lack of more solutions.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $N=a^n-a+1$



        $a^{p_1} ≡ a mod p_1$



        $a^{p_2 } ≡a mod p_2$



        $(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$



        $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$



        ⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$



        If $n=p_1p_2 | a^n-a+1$ then we must have:



        $p_1p_2 | k_1p_1 + k_2p_2+1$



        So we have following linear equation:



        $k_1p_1 + k_2 p_2 = m p_1p_2-1$



        For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:



        with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:



        $k_1= 7 t + 11$ and $k_2= -5 t+7$.



        Now in first step problem reduces to:



        Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.



        For example for $a=3$, $p_1=5$ and $p_2=7$ we have:



        $3^{35}-3+1=105$ and $(35, 105)=5$



        In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.



        Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:



        $k_1p_1+k_2p_2= M(a)$



        Or: $a | k_1p_1+k_2p_2$



        We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$



        This can help us in choosing a and primes $p_1$ and $p_2$



        I see no reason for the lack of more solutions.






        share|cite|improve this answer









        $endgroup$



        $N=a^n-a+1$



        $a^{p_1} ≡ a mod p_1$



        $a^{p_2 } ≡a mod p_2$



        $(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$



        $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$



        ⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$



        If $n=p_1p_2 | a^n-a+1$ then we must have:



        $p_1p_2 | k_1p_1 + k_2p_2+1$



        So we have following linear equation:



        $k_1p_1 + k_2 p_2 = m p_1p_2-1$



        For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:



        with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:



        $k_1= 7 t + 11$ and $k_2= -5 t+7$.



        Now in first step problem reduces to:



        Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.



        For example for $a=3$, $p_1=5$ and $p_2=7$ we have:



        $3^{35}-3+1=105$ and $(35, 105)=5$



        In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.



        Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:



        $k_1p_1+k_2p_2= M(a)$



        Or: $a | k_1p_1+k_2p_2$



        We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$



        This can help us in choosing a and primes $p_1$ and $p_2$



        I see no reason for the lack of more solutions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 13 at 9:19









        siroussirous

        1,6851514




        1,6851514






























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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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