Is it true that $gcd(s(p^k), D(p^k)) = 1$?
$begingroup$
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
asked Dec 25 '18 at 9:40
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,30452045
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Dec 25 '18 at 9:40
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,30452045
$endgroup$
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
|
show 1 more comment
$begingroup$
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
asked Dec 25 '18 at 9:40
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,30452045
$endgroup$
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
asked Dec 25 '18 at 9:40
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,30452045
asked Dec 25 '18 at 9:40
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,30452045
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
asked Dec 25 '18 at 9:40
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,30452045
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
$endgroup$
add a comment |
Your Answer
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051973%2fis-it-true-that-gcdspk-dpk-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
$endgroup$
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.
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
answered Dec 31 '18 at 11:12
PeterPeter
49.4k1240138
49.4k1240138
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051973%2fis-it-true-that-gcdspk-dpk-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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