Is it true that $gcd(s(p^k), D(p^k)) = 1$?












5












$begingroup$


Let $sigma(x)$ be the sum of divisors of a positive integer $x$.



Define
$$s(x):=sigma(x)-x$$
to be the sum of the aliquot divisors of $x$, and define
$$D(x):=2x-sigma(x)$$
to be the deficiency of $x$.



Now, let $p$ be a prime number, and let $k$ be a positive integer.



Here is my question:




Is it true that $gcd(s(p^k), D(p^k)) = 1$?




MY ATTEMPT



If $k=1$, then $s(p^k)=s(p)=sigma(p)-p=(p+1)-p=1$, so that clearly
$$gcd(s(p), D(p))=1.$$



Now let $k>1$. We obtain
$$D(p^k) = 2p^k - sigma(p^k) = p^k - bigg(sigma(p^k) - p^kbigg) = p^k - s(p^k)$$
so that we get
$$gcd(s(p^k), D(p^k)) = gcd(s(p^k), p^k - s(p^k)) = gcd(s(p^k), p^k)$$
$$= gcd(sigma(p^k) - p^k, p^k) = gcd(sigma(p^k), p^k) = 1.$$



Note that a proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I will be assuming this assertion without proof.




Is this proof correct?







elementary-number-theory proof-verification prime-numbers greatest-common-divisor divisor-sum







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You only prove that $gcd(s(p^k),p^k) = 1$ for $k=1$ so your proof is missing to show this (or that $gcd(sigma(p^k),p^k) = 1$) for $k>1$
    $endgroup$
    – Winther
    Dec 25 '18 at 10:38










  • $begingroup$
    Thank you for your comment, @Winther! A proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I assumed that assertion without proof.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:44










  • $begingroup$
    Let me add the intermediate steps in my proof so that it is clearer.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:46










  • $begingroup$
    Yes I know it's not hard, but I just pointed it out since you did take time to prove it for $k=1$. Because of this the proof just looks like it's missing something.
    $endgroup$
    – Winther
    Dec 25 '18 at 10:56








  • 2




    $begingroup$
    @JoseArnaldoBebitaDris An easier approach is : If some prime factor $q$ divides both $s(p^k)$ and $D(p^k)$ it also divides $s(p^k)+D(p^k)=p^k$, hence we can conclude $q=p$. The rest should be easy.
    $endgroup$
    – Peter
    Dec 25 '18 at 13:14


















5












$begingroup$


Let $sigma(x)$ be the sum of divisors of a positive integer $x$.



Define
$$s(x):=sigma(x)-x$$
to be the sum of the aliquot divisors of $x$, and define
$$D(x):=2x-sigma(x)$$
to be the deficiency of $x$.



Now, let $p$ be a prime number, and let $k$ be a positive integer.



Here is my question:




Is it true that $gcd(s(p^k), D(p^k)) = 1$?




MY ATTEMPT



If $k=1$, then $s(p^k)=s(p)=sigma(p)-p=(p+1)-p=1$, so that clearly
$$gcd(s(p), D(p))=1.$$



Now let $k>1$. We obtain
$$D(p^k) = 2p^k - sigma(p^k) = p^k - bigg(sigma(p^k) - p^kbigg) = p^k - s(p^k)$$
so that we get
$$gcd(s(p^k), D(p^k)) = gcd(s(p^k), p^k - s(p^k)) = gcd(s(p^k), p^k)$$
$$= gcd(sigma(p^k) - p^k, p^k) = gcd(sigma(p^k), p^k) = 1.$$



Note that a proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I will be assuming this assertion without proof.




Is this proof correct?







elementary-number-theory proof-verification prime-numbers greatest-common-divisor divisor-sum







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You only prove that $gcd(s(p^k),p^k) = 1$ for $k=1$ so your proof is missing to show this (or that $gcd(sigma(p^k),p^k) = 1$) for $k>1$
    $endgroup$
    – Winther
    Dec 25 '18 at 10:38










  • $begingroup$
    Thank you for your comment, @Winther! A proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I assumed that assertion without proof.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:44










  • $begingroup$
    Let me add the intermediate steps in my proof so that it is clearer.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:46










  • $begingroup$
    Yes I know it's not hard, but I just pointed it out since you did take time to prove it for $k=1$. Because of this the proof just looks like it's missing something.
    $endgroup$
    – Winther
    Dec 25 '18 at 10:56








  • 2




    $begingroup$
    @JoseArnaldoBebitaDris An easier approach is : If some prime factor $q$ divides both $s(p^k)$ and $D(p^k)$ it also divides $s(p^k)+D(p^k)=p^k$, hence we can conclude $q=p$. The rest should be easy.
    $endgroup$
    – Peter
    Dec 25 '18 at 13:14
















5












5








5


2



$begingroup$


Let $sigma(x)$ be the sum of divisors of a positive integer $x$.



Define
$$s(x):=sigma(x)-x$$
to be the sum of the aliquot divisors of $x$, and define
$$D(x):=2x-sigma(x)$$
to be the deficiency of $x$.



Now, let $p$ be a prime number, and let $k$ be a positive integer.



Here is my question:




Is it true that $gcd(s(p^k), D(p^k)) = 1$?




MY ATTEMPT



If $k=1$, then $s(p^k)=s(p)=sigma(p)-p=(p+1)-p=1$, so that clearly
$$gcd(s(p), D(p))=1.$$



Now let $k>1$. We obtain
$$D(p^k) = 2p^k - sigma(p^k) = p^k - bigg(sigma(p^k) - p^kbigg) = p^k - s(p^k)$$
so that we get
$$gcd(s(p^k), D(p^k)) = gcd(s(p^k), p^k - s(p^k)) = gcd(s(p^k), p^k)$$
$$= gcd(sigma(p^k) - p^k, p^k) = gcd(sigma(p^k), p^k) = 1.$$



Note that a proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I will be assuming this assertion without proof.




Is this proof correct?







elementary-number-theory proof-verification prime-numbers greatest-common-divisor divisor-sum







share|cite|improve this question











$endgroup$




Let $sigma(x)$ be the sum of divisors of a positive integer $x$.



Define
$$s(x):=sigma(x)-x$$
to be the sum of the aliquot divisors of $x$, and define
$$D(x):=2x-sigma(x)$$
to be the deficiency of $x$.



Now, let $p$ be a prime number, and let $k$ be a positive integer.



Here is my question:




Is it true that $gcd(s(p^k), D(p^k)) = 1$?




MY ATTEMPT



If $k=1$, then $s(p^k)=s(p)=sigma(p)-p=(p+1)-p=1$, so that clearly
$$gcd(s(p), D(p))=1.$$



Now let $k>1$. We obtain
$$D(p^k) = 2p^k - sigma(p^k) = p^k - bigg(sigma(p^k) - p^kbigg) = p^k - s(p^k)$$
so that we get
$$gcd(s(p^k), D(p^k)) = gcd(s(p^k), p^k - s(p^k)) = gcd(s(p^k), p^k)$$
$$= gcd(sigma(p^k) - p^k, p^k) = gcd(sigma(p^k), p^k) = 1.$$



Note that a proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I will be assuming this assertion without proof.




Is this proof correct?







elementary-number-theory proof-verification prime-numbers greatest-common-divisor divisor-sum




elementary-number-theory proof-verification prime-numbers greatest-common-divisor divisor-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 10:49







Jose Arnaldo Bebita-Dris

















asked Dec 25 '18 at 9:40









Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris

5,30452045




5,30452045








  • 2




    $begingroup$
    You only prove that $gcd(s(p^k),p^k) = 1$ for $k=1$ so your proof is missing to show this (or that $gcd(sigma(p^k),p^k) = 1$) for $k>1$
    $endgroup$
    – Winther
    Dec 25 '18 at 10:38










  • $begingroup$
    Thank you for your comment, @Winther! A proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I assumed that assertion without proof.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:44










  • $begingroup$
    Let me add the intermediate steps in my proof so that it is clearer.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:46










  • $begingroup$
    Yes I know it's not hard, but I just pointed it out since you did take time to prove it for $k=1$. Because of this the proof just looks like it's missing something.
    $endgroup$
    – Winther
    Dec 25 '18 at 10:56








  • 2




    $begingroup$
    @JoseArnaldoBebitaDris An easier approach is : If some prime factor $q$ divides both $s(p^k)$ and $D(p^k)$ it also divides $s(p^k)+D(p^k)=p^k$, hence we can conclude $q=p$. The rest should be easy.
    $endgroup$
    – Peter
    Dec 25 '18 at 13:14
















  • 2




    $begingroup$
    You only prove that $gcd(s(p^k),p^k) = 1$ for $k=1$ so your proof is missing to show this (or that $gcd(sigma(p^k),p^k) = 1$) for $k>1$
    $endgroup$
    – Winther
    Dec 25 '18 at 10:38










  • $begingroup$
    Thank you for your comment, @Winther! A proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I assumed that assertion without proof.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:44










  • $begingroup$
    Let me add the intermediate steps in my proof so that it is clearer.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Dec 25 '18 at 10:46










  • $begingroup$
    Yes I know it's not hard, but I just pointed it out since you did take time to prove it for $k=1$. Because of this the proof just looks like it's missing something.
    $endgroup$
    – Winther
    Dec 25 '18 at 10:56








  • 2




    $begingroup$
    @JoseArnaldoBebitaDris An easier approach is : If some prime factor $q$ divides both $s(p^k)$ and $D(p^k)$ it also divides $s(p^k)+D(p^k)=p^k$, hence we can conclude $q=p$. The rest should be easy.
    $endgroup$
    – Peter
    Dec 25 '18 at 13:14










2




2




$begingroup$
You only prove that $gcd(s(p^k),p^k) = 1$ for $k=1$ so your proof is missing to show this (or that $gcd(sigma(p^k),p^k) = 1$) for $k>1$
$endgroup$
– Winther
Dec 25 '18 at 10:38




$begingroup$
You only prove that $gcd(s(p^k),p^k) = 1$ for $k=1$ so your proof is missing to show this (or that $gcd(sigma(p^k),p^k) = 1$) for $k>1$
$endgroup$
– Winther
Dec 25 '18 at 10:38












$begingroup$
Thank you for your comment, @Winther! A proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I assumed that assertion without proof.
$endgroup$
– Jose Arnaldo Bebita-Dris
Dec 25 '18 at 10:44




$begingroup$
Thank you for your comment, @Winther! A proof for the assertion that $gcd(sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I assumed that assertion without proof.
$endgroup$
– Jose Arnaldo Bebita-Dris
Dec 25 '18 at 10:44












$begingroup$
Let me add the intermediate steps in my proof so that it is clearer.
$endgroup$
– Jose Arnaldo Bebita-Dris
Dec 25 '18 at 10:46




$begingroup$
Let me add the intermediate steps in my proof so that it is clearer.
$endgroup$
– Jose Arnaldo Bebita-Dris
Dec 25 '18 at 10:46












$begingroup$
Yes I know it's not hard, but I just pointed it out since you did take time to prove it for $k=1$. Because of this the proof just looks like it's missing something.
$endgroup$
– Winther
Dec 25 '18 at 10:56






$begingroup$
Yes I know it's not hard, but I just pointed it out since you did take time to prove it for $k=1$. Because of this the proof just looks like it's missing something.
$endgroup$
– Winther
Dec 25 '18 at 10:56






2




2




$begingroup$
@JoseArnaldoBebitaDris An easier approach is : If some prime factor $q$ divides both $s(p^k)$ and $D(p^k)$ it also divides $s(p^k)+D(p^k)=p^k$, hence we can conclude $q=p$. The rest should be easy.
$endgroup$
– Peter
Dec 25 '18 at 13:14






$begingroup$
@JoseArnaldoBebitaDris An easier approach is : If some prime factor $q$ divides both $s(p^k)$ and $D(p^k)$ it also divides $s(p^k)+D(p^k)=p^k$, hence we can conclude $q=p$. The rest should be easy.
$endgroup$
– Peter
Dec 25 '18 at 13:14












1 Answer
1






active

oldest

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1












$begingroup$

As already mentioned in the comment, the only prime possibly both dividing $s(p^k)$ and $D(p^k)$ is $p$.



Hence it remains to show that $s(p^k)$ is not divisible by $p$.



$s(p^k)$ is the sum of the divisors of $p^k$ except $p^k$ itself. All the divisors except $1$ are divisible by $p$, hence the sum cannot be divisible by $p$ (The sum consists of $1$ and in the case $k>1$ of other summands divisible by $p$ , in the case $k=1$ , the sum is just $1$). This completes the proof.






share|cite|improve this answer









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    $begingroup$

    As already mentioned in the comment, the only prime possibly both dividing $s(p^k)$ and $D(p^k)$ is $p$.



    Hence it remains to show that $s(p^k)$ is not divisible by $p$.



    $s(p^k)$ is the sum of the divisors of $p^k$ except $p^k$ itself. All the divisors except $1$ are divisible by $p$, hence the sum cannot be divisible by $p$ (The sum consists of $1$ and in the case $k>1$ of other summands divisible by $p$ , in the case $k=1$ , the sum is just $1$). This completes the proof.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As already mentioned in the comment, the only prime possibly both dividing $s(p^k)$ and $D(p^k)$ is $p$.



      Hence it remains to show that $s(p^k)$ is not divisible by $p$.



      $s(p^k)$ is the sum of the divisors of $p^k$ except $p^k$ itself. All the divisors except $1$ are divisible by $p$, hence the sum cannot be divisible by $p$ (The sum consists of $1$ and in the case $k>1$ of other summands divisible by $p$ , in the case $k=1$ , the sum is just $1$). This completes the proof.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As already mentioned in the comment, the only prime possibly both dividing $s(p^k)$ and $D(p^k)$ is $p$.



        Hence it remains to show that $s(p^k)$ is not divisible by $p$.



        $s(p^k)$ is the sum of the divisors of $p^k$ except $p^k$ itself. All the divisors except $1$ are divisible by $p$, hence the sum cannot be divisible by $p$ (The sum consists of $1$ and in the case $k>1$ of other summands divisible by $p$ , in the case $k=1$ , the sum is just $1$). This completes the proof.






        share|cite|improve this answer









        $endgroup$



        As already mentioned in the comment, the only prime possibly both dividing $s(p^k)$ and $D(p^k)$ is $p$.



        Hence it remains to show that $s(p^k)$ is not divisible by $p$.



        $s(p^k)$ is the sum of the divisors of $p^k$ except $p^k$ itself. All the divisors except $1$ are divisible by $p$, hence the sum cannot be divisible by $p$ (The sum consists of $1$ and in the case $k>1$ of other summands divisible by $p$ , in the case $k=1$ , the sum is just $1$). This completes the proof.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 11:12









        PeterPeter

        49.4k1240138




        49.4k1240138






























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