If $q equiv k equiv 1 pmod 4$, is it necessarily true that $gcdbigg(sigma(q^k),sigma(q^{(k-1)/2})bigg)=1$?












1












$begingroup$


Let $sigma$ denote the classical sum-of-divisors function. In what follows, we let $q$ be a prime number.



Here is my question:




If $q equiv k equiv 1 pmod 4$, is it necessarily true that $gcdbigg(sigma(q^k),sigma(q^{(k-1)/2})bigg)=1$?




MY ATTEMPT



I pattern my approach after the answer to this closely related MSE question.



$$sigma(q^{frac{k-1}{2}})=frac{q^{frac{k+1}{2}} - 1}{q - 1}$$
$$sigma(q^k)=frac{q^{k+1} - 1}{q - 1}$$



Therefore,
$$gcdbigg(sigma(q^{frac{k-1}{2}}),sigma(q^k)bigg)=bigg(frac{1}{q-1}bigg)gcdbigg(q^{frac{k+1}{2}} - 1, q^{k+1} - 1bigg)=frac{q^{gcdbigg(frac{k+1}{2}, hspace{0.05in} k+1bigg)} - 1}{q - 1}$$
$$=frac{q^{frac{k+1}{2}} - 1}{q - 1} = sigma(q^{frac{k-1}{2}}).$$



But $sigma(q^{(k-1)/2}) = 1$ is true if and only if
$$q^{(k-1)/2} = 1 iff (k-1)/2 = 0 iff k = 1.$$




Is this proof correct?







elementary-number-theory proof-verification greatest-common-divisor divisor-sum arithmetic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For what it's worth, keeping in mind I'm not a number theory expert, what you show in your proof attempt all looks fine to me.
    $endgroup$
    – John Omielan
    Jan 7 at 5:53












  • $begingroup$
    @JohnOmielan, thank you for your feedback! Could you write down your last comment as an actual answer, so that I would be able to accept it?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 10:18






  • 1




    $begingroup$
    You are welcome. As you requested, I have added an answer below.
    $endgroup$
    – John Omielan
    Jan 7 at 11:04
















1












$begingroup$


Let $sigma$ denote the classical sum-of-divisors function. In what follows, we let $q$ be a prime number.



Here is my question:




If $q equiv k equiv 1 pmod 4$, is it necessarily true that $gcdbigg(sigma(q^k),sigma(q^{(k-1)/2})bigg)=1$?




MY ATTEMPT



I pattern my approach after the answer to this closely related MSE question.



$$sigma(q^{frac{k-1}{2}})=frac{q^{frac{k+1}{2}} - 1}{q - 1}$$
$$sigma(q^k)=frac{q^{k+1} - 1}{q - 1}$$



Therefore,
$$gcdbigg(sigma(q^{frac{k-1}{2}}),sigma(q^k)bigg)=bigg(frac{1}{q-1}bigg)gcdbigg(q^{frac{k+1}{2}} - 1, q^{k+1} - 1bigg)=frac{q^{gcdbigg(frac{k+1}{2}, hspace{0.05in} k+1bigg)} - 1}{q - 1}$$
$$=frac{q^{frac{k+1}{2}} - 1}{q - 1} = sigma(q^{frac{k-1}{2}}).$$



But $sigma(q^{(k-1)/2}) = 1$ is true if and only if
$$q^{(k-1)/2} = 1 iff (k-1)/2 = 0 iff k = 1.$$




Is this proof correct?







elementary-number-theory proof-verification greatest-common-divisor divisor-sum arithmetic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For what it's worth, keeping in mind I'm not a number theory expert, what you show in your proof attempt all looks fine to me.
    $endgroup$
    – John Omielan
    Jan 7 at 5:53












  • $begingroup$
    @JohnOmielan, thank you for your feedback! Could you write down your last comment as an actual answer, so that I would be able to accept it?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 10:18






  • 1




    $begingroup$
    You are welcome. As you requested, I have added an answer below.
    $endgroup$
    – John Omielan
    Jan 7 at 11:04














1












1








1


1



$begingroup$


Let $sigma$ denote the classical sum-of-divisors function. In what follows, we let $q$ be a prime number.



Here is my question:




If $q equiv k equiv 1 pmod 4$, is it necessarily true that $gcdbigg(sigma(q^k),sigma(q^{(k-1)/2})bigg)=1$?




MY ATTEMPT



I pattern my approach after the answer to this closely related MSE question.



$$sigma(q^{frac{k-1}{2}})=frac{q^{frac{k+1}{2}} - 1}{q - 1}$$
$$sigma(q^k)=frac{q^{k+1} - 1}{q - 1}$$



Therefore,
$$gcdbigg(sigma(q^{frac{k-1}{2}}),sigma(q^k)bigg)=bigg(frac{1}{q-1}bigg)gcdbigg(q^{frac{k+1}{2}} - 1, q^{k+1} - 1bigg)=frac{q^{gcdbigg(frac{k+1}{2}, hspace{0.05in} k+1bigg)} - 1}{q - 1}$$
$$=frac{q^{frac{k+1}{2}} - 1}{q - 1} = sigma(q^{frac{k-1}{2}}).$$



But $sigma(q^{(k-1)/2}) = 1$ is true if and only if
$$q^{(k-1)/2} = 1 iff (k-1)/2 = 0 iff k = 1.$$




Is this proof correct?







elementary-number-theory proof-verification greatest-common-divisor divisor-sum arithmetic-functions







share|cite|improve this question











$endgroup$




Let $sigma$ denote the classical sum-of-divisors function. In what follows, we let $q$ be a prime number.



Here is my question:




If $q equiv k equiv 1 pmod 4$, is it necessarily true that $gcdbigg(sigma(q^k),sigma(q^{(k-1)/2})bigg)=1$?




MY ATTEMPT



I pattern my approach after the answer to this closely related MSE question.



$$sigma(q^{frac{k-1}{2}})=frac{q^{frac{k+1}{2}} - 1}{q - 1}$$
$$sigma(q^k)=frac{q^{k+1} - 1}{q - 1}$$



Therefore,
$$gcdbigg(sigma(q^{frac{k-1}{2}}),sigma(q^k)bigg)=bigg(frac{1}{q-1}bigg)gcdbigg(q^{frac{k+1}{2}} - 1, q^{k+1} - 1bigg)=frac{q^{gcdbigg(frac{k+1}{2}, hspace{0.05in} k+1bigg)} - 1}{q - 1}$$
$$=frac{q^{frac{k+1}{2}} - 1}{q - 1} = sigma(q^{frac{k-1}{2}}).$$



But $sigma(q^{(k-1)/2}) = 1$ is true if and only if
$$q^{(k-1)/2} = 1 iff (k-1)/2 = 0 iff k = 1.$$




Is this proof correct?







elementary-number-theory proof-verification greatest-common-divisor divisor-sum arithmetic-functions




elementary-number-theory proof-verification greatest-common-divisor divisor-sum arithmetic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 7:31







Jose Arnaldo Bebita Dris

















asked Jan 6 at 1:59









Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris

5,38141944




5,38141944








  • 1




    $begingroup$
    For what it's worth, keeping in mind I'm not a number theory expert, what you show in your proof attempt all looks fine to me.
    $endgroup$
    – John Omielan
    Jan 7 at 5:53












  • $begingroup$
    @JohnOmielan, thank you for your feedback! Could you write down your last comment as an actual answer, so that I would be able to accept it?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 10:18






  • 1




    $begingroup$
    You are welcome. As you requested, I have added an answer below.
    $endgroup$
    – John Omielan
    Jan 7 at 11:04














  • 1




    $begingroup$
    For what it's worth, keeping in mind I'm not a number theory expert, what you show in your proof attempt all looks fine to me.
    $endgroup$
    – John Omielan
    Jan 7 at 5:53












  • $begingroup$
    @JohnOmielan, thank you for your feedback! Could you write down your last comment as an actual answer, so that I would be able to accept it?
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 10:18






  • 1




    $begingroup$
    You are welcome. As you requested, I have added an answer below.
    $endgroup$
    – John Omielan
    Jan 7 at 11:04








1




1




$begingroup$
For what it's worth, keeping in mind I'm not a number theory expert, what you show in your proof attempt all looks fine to me.
$endgroup$
– John Omielan
Jan 7 at 5:53






$begingroup$
For what it's worth, keeping in mind I'm not a number theory expert, what you show in your proof attempt all looks fine to me.
$endgroup$
– John Omielan
Jan 7 at 5:53














$begingroup$
@JohnOmielan, thank you for your feedback! Could you write down your last comment as an actual answer, so that I would be able to accept it?
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 7 at 10:18




$begingroup$
@JohnOmielan, thank you for your feedback! Could you write down your last comment as an actual answer, so that I would be able to accept it?
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 7 at 10:18




1




1




$begingroup$
You are welcome. As you requested, I have added an answer below.
$endgroup$
– John Omielan
Jan 7 at 11:04




$begingroup$
You are welcome. As you requested, I have added an answer below.
$endgroup$
– John Omielan
Jan 7 at 11:04










1 Answer
1






active

oldest

votes


















1












$begingroup$

The work shown here follows the same pattern as that of the closely related MSE question that was referenced. Based on that, plus my own checking (but keep in mind though that I'm not a number theory expert), what you've shown in proof attempt all looks fine to me.



One small thing to note is that since $q^{k + 1} - 1 = left(q^{frac{k + 1}{2}} - 1right)left(q^{frac{k + 1}{2}} + 1right)$, this is another way to see that $gcdleft(q^{k + 1} - 1, q^{frac{k + 1}{2}} - 1right) = q^{frac{k + 1}{2}} - 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer, @JohnOmielan! Gladly accepting it now.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 11:07











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

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1












$begingroup$

The work shown here follows the same pattern as that of the closely related MSE question that was referenced. Based on that, plus my own checking (but keep in mind though that I'm not a number theory expert), what you've shown in proof attempt all looks fine to me.



One small thing to note is that since $q^{k + 1} - 1 = left(q^{frac{k + 1}{2}} - 1right)left(q^{frac{k + 1}{2}} + 1right)$, this is another way to see that $gcdleft(q^{k + 1} - 1, q^{frac{k + 1}{2}} - 1right) = q^{frac{k + 1}{2}} - 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer, @JohnOmielan! Gladly accepting it now.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 11:07
















1












$begingroup$

The work shown here follows the same pattern as that of the closely related MSE question that was referenced. Based on that, plus my own checking (but keep in mind though that I'm not a number theory expert), what you've shown in proof attempt all looks fine to me.



One small thing to note is that since $q^{k + 1} - 1 = left(q^{frac{k + 1}{2}} - 1right)left(q^{frac{k + 1}{2}} + 1right)$, this is another way to see that $gcdleft(q^{k + 1} - 1, q^{frac{k + 1}{2}} - 1right) = q^{frac{k + 1}{2}} - 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer, @JohnOmielan! Gladly accepting it now.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 11:07














1












1








1





$begingroup$

The work shown here follows the same pattern as that of the closely related MSE question that was referenced. Based on that, plus my own checking (but keep in mind though that I'm not a number theory expert), what you've shown in proof attempt all looks fine to me.



One small thing to note is that since $q^{k + 1} - 1 = left(q^{frac{k + 1}{2}} - 1right)left(q^{frac{k + 1}{2}} + 1right)$, this is another way to see that $gcdleft(q^{k + 1} - 1, q^{frac{k + 1}{2}} - 1right) = q^{frac{k + 1}{2}} - 1$.






share|cite|improve this answer











$endgroup$



The work shown here follows the same pattern as that of the closely related MSE question that was referenced. Based on that, plus my own checking (but keep in mind though that I'm not a number theory expert), what you've shown in proof attempt all looks fine to me.



One small thing to note is that since $q^{k + 1} - 1 = left(q^{frac{k + 1}{2}} - 1right)left(q^{frac{k + 1}{2}} + 1right)$, this is another way to see that $gcdleft(q^{k + 1} - 1, q^{frac{k + 1}{2}} - 1right) = q^{frac{k + 1}{2}} - 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 7 at 11:01

























answered Jan 7 at 10:53









John OmielanJohn Omielan

1,73629




1,73629












  • $begingroup$
    Thank you for your answer, @JohnOmielan! Gladly accepting it now.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 11:07


















  • $begingroup$
    Thank you for your answer, @JohnOmielan! Gladly accepting it now.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 7 at 11:07
















$begingroup$
Thank you for your answer, @JohnOmielan! Gladly accepting it now.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 7 at 11:07




$begingroup$
Thank you for your answer, @JohnOmielan! Gladly accepting it now.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 7 at 11:07


















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