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Is $n^2/D(n^2) in mathbbN$, if $q^k n^2$ is an odd perfect number?
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Let $x in mathbbN$, the set of positive integers. The sum of the divisors of $x$ is denoted by $sigma(x)$. Denote the deficiency of $x$ by $D(x):=2x-sigma(x)$, and the sum of the aliquot parts of $x$ by $s(x):=sigma(x)-x$. Finally, denote the abundancy index of $x$ by $I(x):=sigma(x)/x$.
If $m$ is odd and $sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that an odd perfect number (if one exists) must have the form $m=q^k n^2$ where $q$ is the special / Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Here is my question:
Is $n^2/D(n^2) in mathbbN$, if $q^k n^2$ is an odd perfect number?
MY ATTEMPT
From the fundamental equality
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$
one can derive
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)=gcd(n^2,sigma(n^2))$$
so that we ultimately have
$$fracD(n^2)s(q^k)=frac2s(n^2)D(q^k)=gcd(n^2,sigma(n^2)).$$
Thus, we have
$$y:=fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracn^2D(q^k)2s(q^k)s(n^2).$$
In particular, we obtain
$$fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracsigma(q^k)2s(q^k).$$
Using the reasoning in this blog post, we obtain
$$fracq2 + frac12q^k-1 - frac12q^k < y leq fracq2 + frac12q^k-1.$$
Equality holds if and only if $k=1$.
That is, it appears that $y in mathbbN$ if and only if $k=1$ (i.e., the Descartes-Frenicle-Sorli conjecture holds).
So the question "Is $n^2/D(n^2) in mathbbN$?" is equivalent to asking whether $k=1$, for $m=q^k n^2$ an odd perfect number with special / Euler prime $q$.
elementary-number-theory inequality conjectures divisor-sum perfect-numbers
asked Jul 19 at 22:33
Jose Arnaldo Bebita Dris
5,27631941
add a comment |Â
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0
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Let $x in mathbbN$, the set of positive integers. The sum of the divisors of $x$ is denoted by $sigma(x)$. Denote the deficiency of $x$ by $D(x):=2x-sigma(x)$, and the sum of the aliquot parts of $x$ by $s(x):=sigma(x)-x$. Finally, denote the abundancy index of $x$ by $I(x):=sigma(x)/x$.
If $m$ is odd and $sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that an odd perfect number (if one exists) must have the form $m=q^k n^2$ where $q$ is the special / Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Here is my question:
Is $n^2/D(n^2) in mathbbN$, if $q^k n^2$ is an odd perfect number?
MY ATTEMPT
From the fundamental equality
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$
one can derive
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)=gcd(n^2,sigma(n^2))$$
so that we ultimately have
$$fracD(n^2)s(q^k)=frac2s(n^2)D(q^k)=gcd(n^2,sigma(n^2)).$$
Thus, we have
$$y:=fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracn^2D(q^k)2s(q^k)s(n^2).$$
In particular, we obtain
$$fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracsigma(q^k)2s(q^k).$$
Using the reasoning in this blog post, we obtain
$$fracq2 + frac12q^k-1 - frac12q^k < y leq fracq2 + frac12q^k-1.$$
Equality holds if and only if $k=1$.
That is, it appears that $y in mathbbN$ if and only if $k=1$ (i.e., the Descartes-Frenicle-Sorli conjecture holds).
So the question "Is $n^2/D(n^2) in mathbbN$?" is equivalent to asking whether $k=1$, for $m=q^k n^2$ an odd perfect number with special / Euler prime $q$.
elementary-number-theory inequality conjectures divisor-sum perfect-numbers
asked Jul 19 at 22:33
Jose Arnaldo Bebita Dris
5,27631941
Unless n is prime is your first line necessarily true?
– Steve B
Jul 20 at 3:02
@SteveB, the first line $$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$ is necessarily true since $m=q^k n^2$ is perfect implies that $sigma(m)=2m$. Now use the fact that $sigma$ is (weakly) multiplicative and that $gcd(q,n)=1$. You will also need to use the fact that $gcd(q^k, sigma(q^k))=1$.
– Jose Arnaldo Bebita Dris
Jul 20 at 7:00
@SteveB, by the way, it is known (by work of Nielsen, 2015) that $omega(n) geq 9$, where $omega(x)$ is the number of distinct prime factors of $x$. In other words, it is known that $n$ must be composite.
– Jose Arnaldo Bebita Dris
Jul 20 at 13:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
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Let $x in mathbbN$, the set of positive integers. The sum of the divisors of $x$ is denoted by $sigma(x)$. Denote the deficiency of $x$ by $D(x):=2x-sigma(x)$, and the sum of the aliquot parts of $x$ by $s(x):=sigma(x)-x$. Finally, denote the abundancy index of $x$ by $I(x):=sigma(x)/x$.
If $m$ is odd and $sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that an odd perfect number (if one exists) must have the form $m=q^k n^2$ where $q$ is the special / Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Here is my question:
Is $n^2/D(n^2) in mathbbN$, if $q^k n^2$ is an odd perfect number?
MY ATTEMPT
From the fundamental equality
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$
one can derive
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)=gcd(n^2,sigma(n^2))$$
so that we ultimately have
$$fracD(n^2)s(q^k)=frac2s(n^2)D(q^k)=gcd(n^2,sigma(n^2)).$$
Thus, we have
$$y:=fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracn^2D(q^k)2s(q^k)s(n^2).$$
In particular, we obtain
$$fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracsigma(q^k)2s(q^k).$$
Using the reasoning in this blog post, we obtain
$$fracq2 + frac12q^k-1 - frac12q^k < y leq fracq2 + frac12q^k-1.$$
Equality holds if and only if $k=1$.
That is, it appears that $y in mathbbN$ if and only if $k=1$ (i.e., the Descartes-Frenicle-Sorli conjecture holds).
So the question "Is $n^2/D(n^2) in mathbbN$?" is equivalent to asking whether $k=1$, for $m=q^k n^2$ an odd perfect number with special / Euler prime $q$.
elementary-number-theory inequality conjectures divisor-sum perfect-numbers
asked Jul 19 at 22:33
Jose Arnaldo Bebita Dris
5,27631941
Let $x in mathbbN$, the set of positive integers. The sum of the divisors of $x$ is denoted by $sigma(x)$. Denote the deficiency of $x$ by $D(x):=2x-sigma(x)$, and the sum of the aliquot parts of $x$ by $s(x):=sigma(x)-x$. Finally, denote the abundancy index of $x$ by $I(x):=sigma(x)/x$.
If $m$ is odd and $sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that an odd perfect number (if one exists) must have the form $m=q^k n^2$ where $q$ is the special / Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Here is my question:
Is $n^2/D(n^2) in mathbbN$, if $q^k n^2$ is an odd perfect number?
MY ATTEMPT
From the fundamental equality
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$
one can derive
$$fracsigma(n^2)q^k=frac2n^2sigma(q^k)=gcd(n^2,sigma(n^2))$$
so that we ultimately have
$$fracD(n^2)s(q^k)=frac2s(n^2)D(q^k)=gcd(n^2,sigma(n^2)).$$
Thus, we have
$$y:=fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracn^2D(q^k)2s(q^k)s(n^2).$$
In particular, we obtain
$$fracn^2D(n^2)=fracn^2s(q^k)gcd(n^2,sigma(n^2))=fracsigma(q^k)2s(q^k).$$
Using the reasoning in this blog post, we obtain
$$fracq2 + frac12q^k-1 - frac12q^k < y leq fracq2 + frac12q^k-1.$$
Equality holds if and only if $k=1$.
That is, it appears that $y in mathbbN$ if and only if $k=1$ (i.e., the Descartes-Frenicle-Sorli conjecture holds).
So the question "Is $n^2/D(n^2) in mathbbN$?" is equivalent to asking whether $k=1$, for $m=q^k n^2$ an odd perfect number with special / Euler prime $q$.
elementary-number-theory inequality conjectures divisor-sum perfect-numbers
asked Jul 19 at 22:33
Jose Arnaldo Bebita Dris
5,27631941
edited Jul 19 at 22:48
asked Jul 19 at 22:33
Jose Arnaldo Bebita Dris
5,27631941
asked Jul 19 at 22:33
Jose Arnaldo Bebita Dris
5,27631941
asked Jul 19 at 22:33
Jose Arnaldo Bebita Dris
5,27631941
5,27631941
Unless n is prime is your first line necessarily true?
– Steve B
Jul 20 at 3:02
@SteveB, the first line $$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$ is necessarily true since $m=q^k n^2$ is perfect implies that $sigma(m)=2m$. Now use the fact that $sigma$ is (weakly) multiplicative and that $gcd(q,n)=1$. You will also need to use the fact that $gcd(q^k, sigma(q^k))=1$.
– Jose Arnaldo Bebita Dris
Jul 20 at 7:00
@SteveB, by the way, it is known (by work of Nielsen, 2015) that $omega(n) geq 9$, where $omega(x)$ is the number of distinct prime factors of $x$. In other words, it is known that $n$ must be composite.
– Jose Arnaldo Bebita Dris
Jul 20 at 13:57
add a comment |Â
Unless n is prime is your first line necessarily true?
– Steve B
Jul 20 at 3:02
@SteveB, the first line $$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$ is necessarily true since $m=q^k n^2$ is perfect implies that $sigma(m)=2m$. Now use the fact that $sigma$ is (weakly) multiplicative and that $gcd(q,n)=1$. You will also need to use the fact that $gcd(q^k, sigma(q^k))=1$.
– Jose Arnaldo Bebita Dris
Jul 20 at 7:00
@SteveB, by the way, it is known (by work of Nielsen, 2015) that $omega(n) geq 9$, where $omega(x)$ is the number of distinct prime factors of $x$. In other words, it is known that $n$ must be composite.
– Jose Arnaldo Bebita Dris
Jul 20 at 13:57
Unless n is prime is your first line necessarily true?
– Steve B
Jul 20 at 3:02
Unless n is prime is your first line necessarily true?
– Steve B
Jul 20 at 3:02
@SteveB, the first line $$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$ is necessarily true since $m=q^k n^2$ is perfect implies that $sigma(m)=2m$. Now use the fact that $sigma$ is (weakly) multiplicative and that $gcd(q,n)=1$. You will also need to use the fact that $gcd(q^k, sigma(q^k))=1$.
– Jose Arnaldo Bebita Dris
Jul 20 at 7:00
@SteveB, the first line $$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$ is necessarily true since $m=q^k n^2$ is perfect implies that $sigma(m)=2m$. Now use the fact that $sigma$ is (weakly) multiplicative and that $gcd(q,n)=1$. You will also need to use the fact that $gcd(q^k, sigma(q^k))=1$.
– Jose Arnaldo Bebita Dris
Jul 20 at 7:00
@SteveB, by the way, it is known (by work of Nielsen, 2015) that $omega(n) geq 9$, where $omega(x)$ is the number of distinct prime factors of $x$. In other words, it is known that $n$ must be composite.
– Jose Arnaldo Bebita Dris
Jul 20 at 13:57
@SteveB, by the way, it is known (by work of Nielsen, 2015) that $omega(n) geq 9$, where $omega(x)$ is the number of distinct prime factors of $x$. In other words, it is known that $n$ must be composite.
– Jose Arnaldo Bebita Dris
Jul 20 at 13:57
add a comment |Â
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Unless n is prime is your first line necessarily true?
– Steve B
Jul 20 at 3:02
@SteveB, the first line $$fracsigma(n^2)q^k=frac2n^2sigma(q^k)$$ is necessarily true since $m=q^k n^2$ is perfect implies that $sigma(m)=2m$. Now use the fact that $sigma$ is (weakly) multiplicative and that $gcd(q,n)=1$. You will also need to use the fact that $gcd(q^k, sigma(q^k))=1$.
– Jose Arnaldo Bebita Dris
Jul 20 at 7:00
@SteveB, by the way, it is known (by work of Nielsen, 2015) that $omega(n) geq 9$, where $omega(x)$ is the number of distinct prime factors of $x$. In other words, it is known that $n$ must be composite.
– Jose Arnaldo Bebita Dris
Jul 20 at 13:57