Breaking the barriers at $q=5$ and $q=13$ for $q^k n^2$ an odd perfect number with special prime $q$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)On odd perfects and spoofsOn odd perfect numbers $N$ given in the Eulerian form $N = q^kn^2$, Part IIIf $q^k n^2$ is an odd perfect number with Euler prime $q$, are the following statements known to hold in general?On a conjectured relationship between the least prime factor and the Euler prime of an odd perfect numberIf $q^k n^2$ is an odd perfect number with Euler prime $q$, can $q=17$ hold?Specific evidence for or against the Descartes-Frenicle-Sorli conjecture on odd perfect numbersQuestion about a result on odd perfect numbersCould a Fermat prime divide an odd perfect number?A consequence of assuming the Descartes-Frenicle-Sorli Conjecture on odd perfect numbersIf $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square.
Can a flute soloist sit?
Huge performance difference of the command find with and without using %M option to show permissions
Why are PDP-7-style microprogrammed instructions out of vogue?
how can a perfect fourth interval be considered either consonant or dissonant?
Are spiders unable to hurt humans, especially very small spiders?
Variable with quotation marks "$()"
Did the new image of black hole confirm the general theory of relativity?
Circular reasoning in L'Hopital's rule
Is every episode of "Where are my Pants?" identical?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Can withdrawing asylum be illegal?
"... to apply for a visa" or "... and applied for a visa"?
Do working physicists consider Newtonian mechanics to be "falsified"?
Python - Fishing Simulator
ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?
Identify 80s or 90s comics with ripped creatures (not dwarves)
How did passengers keep warm on sail ships?
US Healthcare consultation for visitors
Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?
Why can't devices on different VLANs, but on the same subnet, communicate?
How to read αἱμύλιος or when to aspirate
Accepted by European university, rejected by all American ones I applied to? Possible reasons?
Are there continuous functions who are the same in an interval but differ in at least one other point?
Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?
Breaking the barriers at $q=5$ and $q=13$ for $q^k n^2$ an odd perfect number with special prime $q$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)On odd perfects and spoofsOn odd perfect numbers $N$ given in the Eulerian form $N = q^kn^2$, Part IIIf $q^k n^2$ is an odd perfect number with Euler prime $q$, are the following statements known to hold in general?On a conjectured relationship between the least prime factor and the Euler prime of an odd perfect numberIf $q^k n^2$ is an odd perfect number with Euler prime $q$, can $q=17$ hold?Specific evidence for or against the Descartes-Frenicle-Sorli conjecture on odd perfect numbersQuestion about a result on odd perfect numbersCould a Fermat prime divide an odd perfect number?A consequence of assuming the Descartes-Frenicle-Sorli Conjecture on odd perfect numbersIf $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square.
$begingroup$
Let $sigma(x)$ be the sum of divisors of the positive integer $x$. If $sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.
Euler proved that an odd perfect number, if one exists, must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.
So now suppose that $sigma(n^2)/q^k$ is a square. This implies that $k=1$, and also that $sigma(n^2) equiv 1 pmod 4$, since $sigma(n^2)/q^k$ is odd and $q equiv k equiv 1 pmod 4$.
The congruence $sigma(n^2) equiv 1 pmod 4$ then implies that $q equiv k pmod 8$. (See this MO post for the details.) Substituting $k=1$, we obtain
$$q equiv 1 pmod 8.$$
This implies that the lowest possible value for the special prime $q$ is $17$. (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $sigma(n^2)/q^k$ is a square.) Note that, if $q=17$, then $(q+1)/2 = 3^2 mid n^2$.
Here is my question:
Can we push the lowest possible value from $q geq 17$, to say, $q geq 41$ or even $q geq 97$, using the ideas in this post, and possibly more?
Reference:
The Abundancy index of divisors of odd perfect numbers – Part III
number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
$endgroup$
add a comment |
$begingroup$
Let $sigma(x)$ be the sum of divisors of the positive integer $x$. If $sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.
Euler proved that an odd perfect number, if one exists, must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.
So now suppose that $sigma(n^2)/q^k$ is a square. This implies that $k=1$, and also that $sigma(n^2) equiv 1 pmod 4$, since $sigma(n^2)/q^k$ is odd and $q equiv k equiv 1 pmod 4$.
The congruence $sigma(n^2) equiv 1 pmod 4$ then implies that $q equiv k pmod 8$. (See this MO post for the details.) Substituting $k=1$, we obtain
$$q equiv 1 pmod 8.$$
This implies that the lowest possible value for the special prime $q$ is $17$. (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $sigma(n^2)/q^k$ is a square.) Note that, if $q=17$, then $(q+1)/2 = 3^2 mid n^2$.
Here is my question:
Can we push the lowest possible value from $q geq 17$, to say, $q geq 41$ or even $q geq 97$, using the ideas in this post, and possibly more?
Reference:
The Abundancy index of divisors of odd perfect numbers – Part III
number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
$endgroup$
$begingroup$
Isn't that more suitable for MO?
$endgroup$
– enedil
Mar 10 at 11:06
$begingroup$
@enedil, I agree. I just wanted to see what the folks here have to say regarding this inquiry before I cross-post it over at MO (after some time, of course).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:09
$begingroup$
@enedil, I have just posted an answer below a few minutes ago. Note the simplicity in the argument. It is for exactly this same reason that I refrain from asking my questions in MO, because they tend to be not so well-received there (as opposed to here at MSE).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:40
add a comment |
$begingroup$
Let $sigma(x)$ be the sum of divisors of the positive integer $x$. If $sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.
Euler proved that an odd perfect number, if one exists, must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.
So now suppose that $sigma(n^2)/q^k$ is a square. This implies that $k=1$, and also that $sigma(n^2) equiv 1 pmod 4$, since $sigma(n^2)/q^k$ is odd and $q equiv k equiv 1 pmod 4$.
The congruence $sigma(n^2) equiv 1 pmod 4$ then implies that $q equiv k pmod 8$. (See this MO post for the details.) Substituting $k=1$, we obtain
$$q equiv 1 pmod 8.$$
This implies that the lowest possible value for the special prime $q$ is $17$. (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $sigma(n^2)/q^k$ is a square.) Note that, if $q=17$, then $(q+1)/2 = 3^2 mid n^2$.
Here is my question:
Can we push the lowest possible value from $q geq 17$, to say, $q geq 41$ or even $q geq 97$, using the ideas in this post, and possibly more?
Reference:
The Abundancy index of divisors of odd perfect numbers – Part III
number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
$endgroup$
Let $sigma(x)$ be the sum of divisors of the positive integer $x$. If $sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.
Euler proved that an odd perfect number, if one exists, must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.
So now suppose that $sigma(n^2)/q^k$ is a square. This implies that $k=1$, and also that $sigma(n^2) equiv 1 pmod 4$, since $sigma(n^2)/q^k$ is odd and $q equiv k equiv 1 pmod 4$.
The congruence $sigma(n^2) equiv 1 pmod 4$ then implies that $q equiv k pmod 8$. (See this MO post for the details.) Substituting $k=1$, we obtain
$$q equiv 1 pmod 8.$$
This implies that the lowest possible value for the special prime $q$ is $17$. (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $sigma(n^2)/q^k$ is a square.) Note that, if $q=17$, then $(q+1)/2 = 3^2 mid n^2$.
Here is my question:
Can we push the lowest possible value from $q geq 17$, to say, $q geq 41$ or even $q geq 97$, using the ideas in this post, and possibly more?
Reference:
The Abundancy index of divisors of odd perfect numbers – Part III
number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
asked Mar 10 at 10:59
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
5,24952045
$begingroup$
Isn't that more suitable for MO?
$endgroup$
– enedil
Mar 10 at 11:06
$begingroup$
@enedil, I agree. I just wanted to see what the folks here have to say regarding this inquiry before I cross-post it over at MO (after some time, of course).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:09
$begingroup$
@enedil, I have just posted an answer below a few minutes ago. Note the simplicity in the argument. It is for exactly this same reason that I refrain from asking my questions in MO, because they tend to be not so well-received there (as opposed to here at MSE).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:40
add a comment |
$begingroup$
Isn't that more suitable for MO?
$endgroup$
– enedil
Mar 10 at 11:06
$begingroup$
@enedil, I agree. I just wanted to see what the folks here have to say regarding this inquiry before I cross-post it over at MO (after some time, of course).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:09
$begingroup$
@enedil, I have just posted an answer below a few minutes ago. Note the simplicity in the argument. It is for exactly this same reason that I refrain from asking my questions in MO, because they tend to be not so well-received there (as opposed to here at MSE).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:40
$begingroup$
Isn't that more suitable for MO?
$endgroup$
– enedil
Mar 10 at 11:06
$begingroup$
Isn't that more suitable for MO?
$endgroup$
– enedil
Mar 10 at 11:06
$begingroup$
@enedil, I agree. I just wanted to see what the folks here have to say regarding this inquiry before I cross-post it over at MO (after some time, of course).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:09
$begingroup$
@enedil, I agree. I just wanted to see what the folks here have to say regarding this inquiry before I cross-post it over at MO (after some time, of course).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:09
$begingroup$
@enedil, I have just posted an answer below a few minutes ago. Note the simplicity in the argument. It is for exactly this same reason that I refrain from asking my questions in MO, because they tend to be not so well-received there (as opposed to here at MSE).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:40
$begingroup$
@enedil, I have just posted an answer below a few minutes ago. Note the simplicity in the argument. It is for exactly this same reason that I refrain from asking my questions in MO, because they tend to be not so well-received there (as opposed to here at MSE).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $(q+1)/2$ is an odd square, then $(q+1)/2 equiv 1 pmod 8$, so that $q equiv 1 pmod 16$. This rules out $41$, $73$, and $89$.
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
$endgroup$
$begingroup$
Welcome to MSE, @PascalOchem! =)
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 17 at 15:20
add a comment |
$begingroup$
Let $N=q^k n^2$ be an odd perfect number with special prime $q$.
Note that if
$$fracsigma(n^2)q^k=fracn^2sigma(q^k)/2$$
is a square, then $k=1$ and $sigma(q^k)/2 = (q+1)/2$ is also a square.
The possible values for the special prime satisfying $q < 100$ and $q equiv 1 pmod 8$ are $17$, $41$, $73$, $89$, and $97$.
For each of these values:
$$fracq_1 + 12 = frac17 + 12 = 9 = 3^2$$
$$fracq_2 + 12 = frac41 + 12 = 21 text which is not a square.$$
$$fracq_3 + 12 = frac73 + 12 = 37 text which is not a square.$$
$$fracq_4 + 12 = frac89 + 12 = 45 text which is not a square.$$
$$fracq_5 + 12 = frac97 + 12 = 49 = 7^2$$
Thus, if $sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q geq 97$.
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
$endgroup$
$begingroup$
Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update.
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:57
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%2f3142243%2fbreaking-the-barriers-at-q-5-and-q-13-for-qk-n2-an-odd-perfect-number-wi%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
$begingroup$
If $(q+1)/2$ is an odd square, then $(q+1)/2 equiv 1 pmod 8$, so that $q equiv 1 pmod 16$. This rules out $41$, $73$, and $89$.
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
$endgroup$
$begingroup$
Welcome to MSE, @PascalOchem! =)
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 17 at 15:20
add a comment |
$begingroup$
If $(q+1)/2$ is an odd square, then $(q+1)/2 equiv 1 pmod 8$, so that $q equiv 1 pmod 16$. This rules out $41$, $73$, and $89$.
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
$endgroup$
$begingroup$
Welcome to MSE, @PascalOchem! =)
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 17 at 15:20
add a comment |
$begingroup$
If $(q+1)/2$ is an odd square, then $(q+1)/2 equiv 1 pmod 8$, so that $q equiv 1 pmod 16$. This rules out $41$, $73$, and $89$.
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
$endgroup$
If $(q+1)/2$ is an odd square, then $(q+1)/2 equiv 1 pmod 8$, so that $q equiv 1 pmod 16$. This rules out $41$, $73$, and $89$.
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
edited Mar 17 at 15:24
Jose Arnaldo Bebita-Dris
5,24952045
5,24952045
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
answered Mar 17 at 11:10
Pascal OchemPascal Ochem
261
261
$begingroup$
Welcome to MSE, @PascalOchem! =)
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 17 at 15:20
add a comment |
$begingroup$
Welcome to MSE, @PascalOchem! =)
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 17 at 15:20
$begingroup$
Welcome to MSE, @PascalOchem! =)
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 17 at 15:20
$begingroup$
Welcome to MSE, @PascalOchem! =)
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 17 at 15:20
add a comment |
$begingroup$
Let $N=q^k n^2$ be an odd perfect number with special prime $q$.
Note that if
$$fracsigma(n^2)q^k=fracn^2sigma(q^k)/2$$
is a square, then $k=1$ and $sigma(q^k)/2 = (q+1)/2$ is also a square.
The possible values for the special prime satisfying $q < 100$ and $q equiv 1 pmod 8$ are $17$, $41$, $73$, $89$, and $97$.
For each of these values:
$$fracq_1 + 12 = frac17 + 12 = 9 = 3^2$$
$$fracq_2 + 12 = frac41 + 12 = 21 text which is not a square.$$
$$fracq_3 + 12 = frac73 + 12 = 37 text which is not a square.$$
$$fracq_4 + 12 = frac89 + 12 = 45 text which is not a square.$$
$$fracq_5 + 12 = frac97 + 12 = 49 = 7^2$$
Thus, if $sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q geq 97$.
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
$endgroup$
$begingroup$
Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update.
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:57
add a comment |
$begingroup$
Let $N=q^k n^2$ be an odd perfect number with special prime $q$.
Note that if
$$fracsigma(n^2)q^k=fracn^2sigma(q^k)/2$$
is a square, then $k=1$ and $sigma(q^k)/2 = (q+1)/2$ is also a square.
The possible values for the special prime satisfying $q < 100$ and $q equiv 1 pmod 8$ are $17$, $41$, $73$, $89$, and $97$.
For each of these values:
$$fracq_1 + 12 = frac17 + 12 = 9 = 3^2$$
$$fracq_2 + 12 = frac41 + 12 = 21 text which is not a square.$$
$$fracq_3 + 12 = frac73 + 12 = 37 text which is not a square.$$
$$fracq_4 + 12 = frac89 + 12 = 45 text which is not a square.$$
$$fracq_5 + 12 = frac97 + 12 = 49 = 7^2$$
Thus, if $sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q geq 97$.
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
$endgroup$
$begingroup$
Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update.
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:57
add a comment |
$begingroup$
Let $N=q^k n^2$ be an odd perfect number with special prime $q$.
Note that if
$$fracsigma(n^2)q^k=fracn^2sigma(q^k)/2$$
is a square, then $k=1$ and $sigma(q^k)/2 = (q+1)/2$ is also a square.
The possible values for the special prime satisfying $q < 100$ and $q equiv 1 pmod 8$ are $17$, $41$, $73$, $89$, and $97$.
For each of these values:
$$fracq_1 + 12 = frac17 + 12 = 9 = 3^2$$
$$fracq_2 + 12 = frac41 + 12 = 21 text which is not a square.$$
$$fracq_3 + 12 = frac73 + 12 = 37 text which is not a square.$$
$$fracq_4 + 12 = frac89 + 12 = 45 text which is not a square.$$
$$fracq_5 + 12 = frac97 + 12 = 49 = 7^2$$
Thus, if $sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q geq 97$.
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
$endgroup$
Let $N=q^k n^2$ be an odd perfect number with special prime $q$.
Note that if
$$fracsigma(n^2)q^k=fracn^2sigma(q^k)/2$$
is a square, then $k=1$ and $sigma(q^k)/2 = (q+1)/2$ is also a square.
The possible values for the special prime satisfying $q < 100$ and $q equiv 1 pmod 8$ are $17$, $41$, $73$, $89$, and $97$.
For each of these values:
$$fracq_1 + 12 = frac17 + 12 = 9 = 3^2$$
$$fracq_2 + 12 = frac41 + 12 = 21 text which is not a square.$$
$$fracq_3 + 12 = frac73 + 12 = 37 text which is not a square.$$
$$fracq_4 + 12 = frac89 + 12 = 45 text which is not a square.$$
$$fracq_5 + 12 = frac97 + 12 = 49 = 7^2$$
Thus, if $sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q geq 97$.
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
edited Mar 31 at 7:37
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
answered Mar 10 at 11:32
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,24952045
5,24952045
$begingroup$
Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update.
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:57
add a comment |
$begingroup$
Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update.
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:57
$begingroup$
Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update.
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:57
$begingroup$
Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update.
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:57
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%2f3142243%2fbreaking-the-barriers-at-q-5-and-q-13-for-qk-n2-an-odd-perfect-number-wi%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
$begingroup$
Isn't that more suitable for MO?
$endgroup$
– enedil
Mar 10 at 11:06
$begingroup$
@enedil, I agree. I just wanted to see what the folks here have to say regarding this inquiry before I cross-post it over at MO (after some time, of course).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:09
$begingroup$
@enedil, I have just posted an answer below a few minutes ago. Note the simplicity in the argument. It is for exactly this same reason that I refrain from asking my questions in MO, because they tend to be not so well-received there (as opposed to here at MSE).
$endgroup$
– Jose Arnaldo Bebita-Dris
Mar 10 at 11:40