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Are there any natural numbers $n$ that satisfy the condition $7921sigma(n) = 15840n$?
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Are there any natural numbers $n$ that satisfy the condition $7921sigma(n) = 15840n$, where $sigma(n)$ denotes the sum of divisors of $n$?
This question arises from the theory of immaculate groups (or, equivalently, Leinster groups). An immaculate group is a group, such that its order is equal to the sum of all orders of its proper normal subgroups.
It is easy to see, that if $A$ is a non-abelian simple group then $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist such $n$, that $M_11timesmathbbZ_n$ is immaculate?", where $M_11$ stands for Mathieu simple group of order $7920$.
Currently I know only two facts about such $n$-s: if they exist, then $7921|n$, and that such $n$-s, if they exist, are too large to be found by exhaustive search.
Any help will be appreciated.
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked Jul 20 at 7:57
Yanior Weg
1,0061629
 |Â
show 12 more comments
up vote
29
down vote
favorite
Are there any natural numbers $n$ that satisfy the condition $7921sigma(n) = 15840n$, where $sigma(n)$ denotes the sum of divisors of $n$?
This question arises from the theory of immaculate groups (or, equivalently, Leinster groups). An immaculate group is a group, such that its order is equal to the sum of all orders of its proper normal subgroups.
It is easy to see, that if $A$ is a non-abelian simple group then $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist such $n$, that $M_11timesmathbbZ_n$ is immaculate?", where $M_11$ stands for Mathieu simple group of order $7920$.
Currently I know only two facts about such $n$-s: if they exist, then $7921|n$, and that such $n$-s, if they exist, are too large to be found by exhaustive search.
Any help will be appreciated.
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked Jul 20 at 7:57
Yanior Weg
1,0061629
10
This seems like such a random question to ask until you see the entirely reasonable motivation.
– Arthur
Jul 20 at 9:13
@Arthur Tbh I saw a motivation at the start, this being essentially a question of how close $sigma(n)$ can be to $2n$.
– fretty
Jul 20 at 9:54
2
An explantion why it is hopeless to search such a number with brute force, would be nice.
– Peter
Jul 20 at 11:25
6
I figured out a positive integer $n$ with $$7921sigma(n)=2cdot 15840cdot n$$ $$n=563432025678613816032$$ Not sure whether this helps
– Peter
Jul 20 at 12:17
1
I have tried hard to find a solution, but I failed. My conjecture is that there is no solution, but a proof will be very difficult considering that it is , for example , an open question whether there is an odd perfect number.
– Peter
Jul 21 at 16:46
 |Â
show 12 more comments
up vote
29
down vote
favorite
up vote
29
down vote
favorite
Are there any natural numbers $n$ that satisfy the condition $7921sigma(n) = 15840n$, where $sigma(n)$ denotes the sum of divisors of $n$?
This question arises from the theory of immaculate groups (or, equivalently, Leinster groups). An immaculate group is a group, such that its order is equal to the sum of all orders of its proper normal subgroups.
It is easy to see, that if $A$ is a non-abelian simple group then $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist such $n$, that $M_11timesmathbbZ_n$ is immaculate?", where $M_11$ stands for Mathieu simple group of order $7920$.
Currently I know only two facts about such $n$-s: if they exist, then $7921|n$, and that such $n$-s, if they exist, are too large to be found by exhaustive search.
Any help will be appreciated.
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked Jul 20 at 7:57
Yanior Weg
1,0061629
Are there any natural numbers $n$ that satisfy the condition $7921sigma(n) = 15840n$, where $sigma(n)$ denotes the sum of divisors of $n$?
This question arises from the theory of immaculate groups (or, equivalently, Leinster groups). An immaculate group is a group, such that its order is equal to the sum of all orders of its proper normal subgroups.
It is easy to see, that if $A$ is a non-abelian simple group then $AtimesmathbbZ_n$ is immaculate iff $(|A|+1)sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5timesmathbbZ_15128$ and $A_6timesmathbbZ_366776$. In terms of immaculate groups this question thus can be reworded as:
"Does there exist such $n$, that $M_11timesmathbbZ_n$ is immaculate?", where $M_11$ stands for Mathieu simple group of order $7920$.
Currently I know only two facts about such $n$-s: if they exist, then $7921|n$, and that such $n$-s, if they exist, are too large to be found by exhaustive search.
Any help will be appreciated.
group-theory elementary-number-theory finite-groups normal-subgroups divisor-sum
asked Jul 20 at 7:57
Yanior Weg
1,0061629
asked Jul 20 at 7:57
Yanior Weg
1,0061629
asked Jul 20 at 7:57
Yanior Weg
1,0061629
asked Jul 20 at 7:57
Yanior Weg
1,0061629
1,0061629
10
This seems like such a random question to ask until you see the entirely reasonable motivation.
– Arthur
Jul 20 at 9:13
@Arthur Tbh I saw a motivation at the start, this being essentially a question of how close $sigma(n)$ can be to $2n$.
– fretty
Jul 20 at 9:54
2
An explantion why it is hopeless to search such a number with brute force, would be nice.
– Peter
Jul 20 at 11:25
6
I figured out a positive integer $n$ with $$7921sigma(n)=2cdot 15840cdot n$$ $$n=563432025678613816032$$ Not sure whether this helps
– Peter
Jul 20 at 12:17
1
I have tried hard to find a solution, but I failed. My conjecture is that there is no solution, but a proof will be very difficult considering that it is , for example , an open question whether there is an odd perfect number.
– Peter
Jul 21 at 16:46
 |Â
show 12 more comments
10
This seems like such a random question to ask until you see the entirely reasonable motivation.
– Arthur
Jul 20 at 9:13
@Arthur Tbh I saw a motivation at the start, this being essentially a question of how close $sigma(n)$ can be to $2n$.
– fretty
Jul 20 at 9:54
2
An explantion why it is hopeless to search such a number with brute force, would be nice.
– Peter
Jul 20 at 11:25
6
I figured out a positive integer $n$ with $$7921sigma(n)=2cdot 15840cdot n$$ $$n=563432025678613816032$$ Not sure whether this helps
– Peter
Jul 20 at 12:17
1
I have tried hard to find a solution, but I failed. My conjecture is that there is no solution, but a proof will be very difficult considering that it is , for example , an open question whether there is an odd perfect number.
– Peter
Jul 21 at 16:46
10
10
This seems like such a random question to ask until you see the entirely reasonable motivation.
– Arthur
Jul 20 at 9:13
This seems like such a random question to ask until you see the entirely reasonable motivation.
– Arthur
Jul 20 at 9:13
@Arthur Tbh I saw a motivation at the start, this being essentially a question of how close $sigma(n)$ can be to $2n$.
– fretty
Jul 20 at 9:54
@Arthur Tbh I saw a motivation at the start, this being essentially a question of how close $sigma(n)$ can be to $2n$.
– fretty
Jul 20 at 9:54
2
2
An explantion why it is hopeless to search such a number with brute force, would be nice.
– Peter
Jul 20 at 11:25
An explantion why it is hopeless to search such a number with brute force, would be nice.
– Peter
Jul 20 at 11:25
6
6
I figured out a positive integer $n$ with $$7921sigma(n)=2cdot 15840cdot n$$ $$n=563432025678613816032$$ Not sure whether this helps
– Peter
Jul 20 at 12:17
I figured out a positive integer $n$ with $$7921sigma(n)=2cdot 15840cdot n$$ $$n=563432025678613816032$$ Not sure whether this helps
– Peter
Jul 20 at 12:17
1
1
I have tried hard to find a solution, but I failed. My conjecture is that there is no solution, but a proof will be very difficult considering that it is , for example , an open question whether there is an odd perfect number.
– Peter
Jul 21 at 16:46
I have tried hard to find a solution, but I failed. My conjecture is that there is no solution, but a proof will be very difficult considering that it is , for example , an open question whether there is an odd perfect number.
– Peter
Jul 21 at 16:46
 |Â
show 12 more comments
3 Answers
3
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oldest
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4
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accepted
$$mathbfcolorgreenFixed version$$
$$mathbfcolorbrownConstraints of the task$$
The issue equation is
$$89^2sigma(n)=2^53^25^111^1n.tag1$$
Let WLOG
$$n=2^A3^B5^C11^D89^U+2m,quad (A,B,C,D,U)inmathbb Ncup 0 ,quad gcd(m, 2cdot3cdot5cdot11cdot89)=1,tag2$$
then $(1)$ takes the form of
$$89^2cdotdfrac2^A+1-11 cdotdfrac3^B+1-12 cdotdfrac5^C+1-14cdotdfrac11^D+1-110cdotdfrac89^U+3-188cdotsigma(m) =$$
$$2^A+53^B+25^C+111^D+189^U+2m,tag3$$
or
$$dfrac2^A+1-12^A+1cdotdfrac3^B+1-13^B+1cdotdfrac5^C+1-15^C+1cdotdfrac11^D+1-111^D+1cdotdfrac89^U+3-188cdot89^Ucdotsigma(m) = 2^8cdot3cdot5cdotm,tag4$$
$$left(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac89^3-89^-U88cdotsigma(m) = 3840m.tag5$$
Easy to see that
$$dfrac89^3-89^-U88 ge dfrac89^3-188 = 8011.$$
At the same time, $sigma(m)$ is the sum of divisors, so
$$sigma(m)ge 1+m.$$
Thus,
$$8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 3840.tag6$$
Similarly,
begincases
8011cdotleft(1-dfrac2^-A2right)cdotdfrac23cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right)cdotdfrac23cdotdfrac1011 le3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)cdotdfrac23cdotdfrac45 le 3840\
8011cdotleft(1-dfrac2^-12right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac23 le 3840,tag7\
endcases
or
begincases
8011cdotleft(1-dfrac2^-A2right) le 7920\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right) le 5280\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right) le 6336\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)le 7200\
8011cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 7680,quad A=1, B=0.tag8\
endcases
The system $(8)$ has the solution
$$left[
beginaligned
&A=0\
&A=1, B=0, C=D=1\
&A=1, B=0, C>2\
&A=2, B=C=0\
&5ge Age3, B=C=D=0,
endalignedright.tag9\$$
or
$$left[
beginalign
&2 not| nhspace40pt\
&((n=2m)vee(n=8m)vee(n=16m)vee(n=32m)vee(n=110m))\ &quadwedge(gcd(2cdot3cdot5cdot11,m)=1)\
&(n=50m)wedge(gcd(2cdot3, m)=1)\
&(n=4m)wedge(gcd(2cdot3cdot5,m)=1)\
endalignright..tag10
$$
$$mathbfcolorbrownApplying of the constraints$$
Obtained constraints $(10)$ allow to filter the possible solutions.
At first, becames impossible the case
$$n=2^1089^3k.$$
Secondly, if
$$n=89^2cdot8011cdot2003cdot167k,quad gcd(89cdot167cdot2003cdot8011,k) = 1,$$
with the equation
$$28sigma(k)=55k,$$
then
begincases
49sigma(m)=55m,text if k=4m, gcd(2cdot3cdot5,m)=1 \
21sigma(m)=22m,text if k=8m, gcd(2cdot3cdot5cdot11,m)=1 \
31cdot7sigma(m)=4cdot55mtext if k=16m, gcd(2cdot3cdot5cdot11,m)=1 \
21^2sigma(m)=8cdot55mtext if k=32m, gcd(2cdot3cdot5cdot11,m)=1 .tag11
endcases
Taking in account that
$$sigma(7)=2^3,quadsigma(49)= 2^2cdot5^2,$$
the system $(11)$ really can not be satisfied.
So I think that the issue equation $mathbfcolorbrownhas not solutions in the positive integer numbers.$
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
1
$m$ not divisible by 2, 3, 5, 11, and 89?
– qwr
Jul 29 at 4:50
1
How did you get from $89^2cdotldotscdotfrac89^U+1-188$ to $frac89^U+3-188$?
– jpvee
Jul 29 at 6:50
1
@YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^11cdot3cdot5cdot11cdot m$.
– jpvee
Jul 30 at 11:35
1
@YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :)
– jpvee
Jul 30 at 12:19
2
@YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^11$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3).
– jpvee
Jul 30 at 13:14
 |Â
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7
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This is not an answer, but an extended comment, that should help anyone interested in using brute-force numerical search to find the solution.
The problem is to find $n in mathbbN$ for which
$$7921 sigma(n) = 15840 n tag1labelNA1$$
where $sigma(n)$ is the sum of all divisors of $n$, as defined in the Wikipedia divisor function article, and as a sequence in OEIS A000203.
Because $sigma(n) = n + 1$ if $n$ is a prime, and
$$7921 (n + 1) ne 15840 n, quad n in mathbbN$$
we already know there is no prime solution to $eqrefNA1$.
Consider the prime factorization of $n$. Let $p_i in mathbbN$ be nonrepeating primes ($p_i = p_j$ if and only if $i = j$), and $1 le k_i in mathbbN$. Then,
$$n = prod_i=0^N-1 p_i^k_i$$
and
$$sigma(n) = prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1$$
because $sigma(p^k) = sum_j=0^k p^j = (p^k+1-1)/(p-1)$ when $p$ is a prime.
We can now rewrite the problem $eqrefNA1$ as
$$7921 prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1 = 15840 prod_i=0^N-1 p_i^k_i tag2labelNA2$$
Rearranging the terms yields
$$prod_i=0^N-1 frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = frac792115840 = fracnsigma(n) = frac89^22^5 cdot 3^2 cdot 5 cdot 11
tag3labelNA3$$
Note the term
$$f_i = frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = fracp_i^k_isum_j=0^k_i p^j, quad frac12 lt f_i lt 1
tag4labelNA4$$
i.e.,
$$beginarrayll
f_i = fracp_ip_i + 1, & k_i = 1 \
f_i = fracp_i^2p_i^2 + p_i + 1, & k_i = 2 \
f_i = fracp_i^3p_i^3 + p_i^2 + p_i + 1 , & k_i = 3 \
f_i = fracp_i^k_ip_i^k_i + p_i^k_i-1 + dots + p_i + 1 & \
endarray$$
Thus, the numerical search problem is now reduced to find the set of terms $f_i$ based on primes $p_i$ and their positive powers $k_i$, so that the product $$prod_i=0^N-1 f_i = frac792115840$$
In particular, because $f_i lt 1$, a particular set can be rejected immediately if the product falls below the target ratio.
For example, if $p_0 = 89$, $k_0 = 2$, to eliminate the prime factor in the numerator. Repeating, that leads to $p_1 = 8011$, $k_1 = 1$; $p_2 = 2003$, $k_2 = 1$; and $p_3 = 167$, $k_3 = 1$, to get us to a result with a composite numerator and a denominator:
$$beginarrayl
n & fracnsigma(n) & frac15840 n7921 sigma(n) \
hline
89^2 cdot 8011 cdot 2003 cdot 167
& frac79218064 = frac79212^7 cdot 3^2 cdot 7
& frac5528 = frac5 cdot 112^2 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 7
& frac79219216 = frac79212^10 cdot 3^2
& frac5532 = frac5 cdot 112^5
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2
& frac792112096 = frac79212^6 cdot 3^3 cdot 7
& frac5542 = frac5 cdot 112 cdot 3 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2 cdot 7
& frac792113824 = frac79212^9 cdot 3^3
& frac5548 = frac5 cdot 112^4 cdot 3
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2^2
& frac792114112 = frac79212^5 cdot 3^2 cdot 7^2
& frac5549 = frac5 cdot 117^2
\ endarray$$
If you append $p_5 = 7$, $k_5 = 1$ or $k_5 = 2$ to $n$ in the final row above, the rightmost field drops below 1 (to $55/56$ for $k_5 = 1$, and to $55/57$ for $k_5 = 2$), leading nowhere. Similarly, appending $p_6 = 3$, $k_6 = 1$ or $k_6 = 2$ to $n$ in the second-to-last row (to $55/64$ for $k_6 = 1$, and to $165/208$ for $k_6 = 2$) leads nowhere.
It looks to (very non-mathematician) me that an exhaustive search over primes $p$ is possible, due to terms $f_i$ having a power of a prime in the numerator, as specified in $eqrefNA4$. Whether an exhaustive search is possible or not is an open question (and is important for those looking for proof), but the efficient numerical brute force search strategies are straightforward; especially if one is looking for some other ratios than $fracnsigma(n) = frac792115840$.
To continue the search above, I'd need a prime $p$ and a positive integer $k$ such that $sum_j=0^k p^j = 55$ (to yield a factor with denominator $55$). No such pair exists, so the search strategies I've come up with thus far are exhausted.
Hopefully, one of the math sages here can take this further from here.
answered Jul 27 at 5:00
Nominal Animal
5,6952414
1
This is the way to the jungle. For example, the equality $2^11-1=23cdot89$ requires consideration of the variant $k_0=3$ etc.
– Yuri Negometyanov
Jul 29 at 3:52
@YuriNegometyanov: At first glance, using $2^10$ does indeed look promising. Unfortunately, however, $frac89^3sigma(89^3)cdotfrac2^10sigma(2^10)$ is already below the target ratio of $frac792115840$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$.
– jpvee
Jul 30 at 8:36
I do believe my efforts stalled, because rewriting $eqrefNA2$ as $eqrefNA3$ places an additional constraint on the coefficients. I really should rewrite $eqrefNA2$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $eqrefNA3$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)?
– Nominal Animal
Jul 30 at 9:45
@NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist.
– Yuri Negometyanov
Jul 30 at 10:55
1
@YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3).
– jpvee
Jul 30 at 11:36
 |Â
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This is not an answer either: I have written a program doing a non-exhaustive search, and so far, no solutions have come up for the Mathieu group $M_11$; however, the program did find a solution e.g. for the larger Mathieu group $M_22$ of order $443520=2^7cdot3^2cdot5cdot7cdot11$:
Let $n=55009909630=2cdot5cdot13cdot79cdot109cdot157cdot313$, then since $$
beginalign
(|M_22|+1)cdotsigma(n)&=443521cdotsigma(2cdot5cdot13cdot79cdot109cdot157cdot313) \
&=13cdot109cdot313cdot3cdot6cdot14cdot80cdot110cdot158cdot314 \
&=2^9cdot3^2cdot5^2cdot7cdot11cdot13cdot79cdot109cdot157cdot313 \
&=2cdot|M_22|cdot ntext,endalign$$
$M_22timesmathbbZ_55009909630$ must be immaculate.
answered Jul 27 at 12:16
jpvee
2,48421326
add a comment |Â
3 Answers
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3 Answers
3
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active
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active
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up vote
4
down vote
accepted
$$mathbfcolorgreenFixed version$$
$$mathbfcolorbrownConstraints of the task$$
The issue equation is
$$89^2sigma(n)=2^53^25^111^1n.tag1$$
Let WLOG
$$n=2^A3^B5^C11^D89^U+2m,quad (A,B,C,D,U)inmathbb Ncup 0 ,quad gcd(m, 2cdot3cdot5cdot11cdot89)=1,tag2$$
then $(1)$ takes the form of
$$89^2cdotdfrac2^A+1-11 cdotdfrac3^B+1-12 cdotdfrac5^C+1-14cdotdfrac11^D+1-110cdotdfrac89^U+3-188cdotsigma(m) =$$
$$2^A+53^B+25^C+111^D+189^U+2m,tag3$$
or
$$dfrac2^A+1-12^A+1cdotdfrac3^B+1-13^B+1cdotdfrac5^C+1-15^C+1cdotdfrac11^D+1-111^D+1cdotdfrac89^U+3-188cdot89^Ucdotsigma(m) = 2^8cdot3cdot5cdotm,tag4$$
$$left(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac89^3-89^-U88cdotsigma(m) = 3840m.tag5$$
Easy to see that
$$dfrac89^3-89^-U88 ge dfrac89^3-188 = 8011.$$
At the same time, $sigma(m)$ is the sum of divisors, so
$$sigma(m)ge 1+m.$$
Thus,
$$8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 3840.tag6$$
Similarly,
begincases
8011cdotleft(1-dfrac2^-A2right)cdotdfrac23cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right)cdotdfrac23cdotdfrac1011 le3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)cdotdfrac23cdotdfrac45 le 3840\
8011cdotleft(1-dfrac2^-12right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac23 le 3840,tag7\
endcases
or
begincases
8011cdotleft(1-dfrac2^-A2right) le 7920\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right) le 5280\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right) le 6336\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)le 7200\
8011cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 7680,quad A=1, B=0.tag8\
endcases
The system $(8)$ has the solution
$$left[
beginaligned
&A=0\
&A=1, B=0, C=D=1\
&A=1, B=0, C>2\
&A=2, B=C=0\
&5ge Age3, B=C=D=0,
endalignedright.tag9\$$
or
$$left[
beginalign
&2 not| nhspace40pt\
&((n=2m)vee(n=8m)vee(n=16m)vee(n=32m)vee(n=110m))\ &quadwedge(gcd(2cdot3cdot5cdot11,m)=1)\
&(n=50m)wedge(gcd(2cdot3, m)=1)\
&(n=4m)wedge(gcd(2cdot3cdot5,m)=1)\
endalignright..tag10
$$
$$mathbfcolorbrownApplying of the constraints$$
Obtained constraints $(10)$ allow to filter the possible solutions.
At first, becames impossible the case
$$n=2^1089^3k.$$
Secondly, if
$$n=89^2cdot8011cdot2003cdot167k,quad gcd(89cdot167cdot2003cdot8011,k) = 1,$$
with the equation
$$28sigma(k)=55k,$$
then
begincases
49sigma(m)=55m,text if k=4m, gcd(2cdot3cdot5,m)=1 \
21sigma(m)=22m,text if k=8m, gcd(2cdot3cdot5cdot11,m)=1 \
31cdot7sigma(m)=4cdot55mtext if k=16m, gcd(2cdot3cdot5cdot11,m)=1 \
21^2sigma(m)=8cdot55mtext if k=32m, gcd(2cdot3cdot5cdot11,m)=1 .tag11
endcases
Taking in account that
$$sigma(7)=2^3,quadsigma(49)= 2^2cdot5^2,$$
the system $(11)$ really can not be satisfied.
So I think that the issue equation $mathbfcolorbrownhas not solutions in the positive integer numbers.$
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
1
$m$ not divisible by 2, 3, 5, 11, and 89?
– qwr
Jul 29 at 4:50
1
How did you get from $89^2cdotldotscdotfrac89^U+1-188$ to $frac89^U+3-188$?
– jpvee
Jul 29 at 6:50
1
@YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^11cdot3cdot5cdot11cdot m$.
– jpvee
Jul 30 at 11:35
1
@YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :)
– jpvee
Jul 30 at 12:19
2
@YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^11$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3).
– jpvee
Jul 30 at 13:14
 |Â
show 14 more comments
up vote
4
down vote
accepted
$$mathbfcolorgreenFixed version$$
$$mathbfcolorbrownConstraints of the task$$
The issue equation is
$$89^2sigma(n)=2^53^25^111^1n.tag1$$
Let WLOG
$$n=2^A3^B5^C11^D89^U+2m,quad (A,B,C,D,U)inmathbb Ncup 0 ,quad gcd(m, 2cdot3cdot5cdot11cdot89)=1,tag2$$
then $(1)$ takes the form of
$$89^2cdotdfrac2^A+1-11 cdotdfrac3^B+1-12 cdotdfrac5^C+1-14cdotdfrac11^D+1-110cdotdfrac89^U+3-188cdotsigma(m) =$$
$$2^A+53^B+25^C+111^D+189^U+2m,tag3$$
or
$$dfrac2^A+1-12^A+1cdotdfrac3^B+1-13^B+1cdotdfrac5^C+1-15^C+1cdotdfrac11^D+1-111^D+1cdotdfrac89^U+3-188cdot89^Ucdotsigma(m) = 2^8cdot3cdot5cdotm,tag4$$
$$left(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac89^3-89^-U88cdotsigma(m) = 3840m.tag5$$
Easy to see that
$$dfrac89^3-89^-U88 ge dfrac89^3-188 = 8011.$$
At the same time, $sigma(m)$ is the sum of divisors, so
$$sigma(m)ge 1+m.$$
Thus,
$$8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 3840.tag6$$
Similarly,
begincases
8011cdotleft(1-dfrac2^-A2right)cdotdfrac23cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right)cdotdfrac23cdotdfrac1011 le3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)cdotdfrac23cdotdfrac45 le 3840\
8011cdotleft(1-dfrac2^-12right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac23 le 3840,tag7\
endcases
or
begincases
8011cdotleft(1-dfrac2^-A2right) le 7920\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right) le 5280\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right) le 6336\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)le 7200\
8011cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 7680,quad A=1, B=0.tag8\
endcases
The system $(8)$ has the solution
$$left[
beginaligned
&A=0\
&A=1, B=0, C=D=1\
&A=1, B=0, C>2\
&A=2, B=C=0\
&5ge Age3, B=C=D=0,
endalignedright.tag9\$$
or
$$left[
beginalign
&2 not| nhspace40pt\
&((n=2m)vee(n=8m)vee(n=16m)vee(n=32m)vee(n=110m))\ &quadwedge(gcd(2cdot3cdot5cdot11,m)=1)\
&(n=50m)wedge(gcd(2cdot3, m)=1)\
&(n=4m)wedge(gcd(2cdot3cdot5,m)=1)\
endalignright..tag10
$$
$$mathbfcolorbrownApplying of the constraints$$
Obtained constraints $(10)$ allow to filter the possible solutions.
At first, becames impossible the case
$$n=2^1089^3k.$$
Secondly, if
$$n=89^2cdot8011cdot2003cdot167k,quad gcd(89cdot167cdot2003cdot8011,k) = 1,$$
with the equation
$$28sigma(k)=55k,$$
then
begincases
49sigma(m)=55m,text if k=4m, gcd(2cdot3cdot5,m)=1 \
21sigma(m)=22m,text if k=8m, gcd(2cdot3cdot5cdot11,m)=1 \
31cdot7sigma(m)=4cdot55mtext if k=16m, gcd(2cdot3cdot5cdot11,m)=1 \
21^2sigma(m)=8cdot55mtext if k=32m, gcd(2cdot3cdot5cdot11,m)=1 .tag11
endcases
Taking in account that
$$sigma(7)=2^3,quadsigma(49)= 2^2cdot5^2,$$
the system $(11)$ really can not be satisfied.
So I think that the issue equation $mathbfcolorbrownhas not solutions in the positive integer numbers.$
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
1
$m$ not divisible by 2, 3, 5, 11, and 89?
– qwr
Jul 29 at 4:50
1
How did you get from $89^2cdotldotscdotfrac89^U+1-188$ to $frac89^U+3-188$?
– jpvee
Jul 29 at 6:50
1
@YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^11cdot3cdot5cdot11cdot m$.
– jpvee
Jul 30 at 11:35
1
@YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :)
– jpvee
Jul 30 at 12:19
2
@YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^11$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3).
– jpvee
Jul 30 at 13:14
 |Â
show 14 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$$mathbfcolorgreenFixed version$$
$$mathbfcolorbrownConstraints of the task$$
The issue equation is
$$89^2sigma(n)=2^53^25^111^1n.tag1$$
Let WLOG
$$n=2^A3^B5^C11^D89^U+2m,quad (A,B,C,D,U)inmathbb Ncup 0 ,quad gcd(m, 2cdot3cdot5cdot11cdot89)=1,tag2$$
then $(1)$ takes the form of
$$89^2cdotdfrac2^A+1-11 cdotdfrac3^B+1-12 cdotdfrac5^C+1-14cdotdfrac11^D+1-110cdotdfrac89^U+3-188cdotsigma(m) =$$
$$2^A+53^B+25^C+111^D+189^U+2m,tag3$$
or
$$dfrac2^A+1-12^A+1cdotdfrac3^B+1-13^B+1cdotdfrac5^C+1-15^C+1cdotdfrac11^D+1-111^D+1cdotdfrac89^U+3-188cdot89^Ucdotsigma(m) = 2^8cdot3cdot5cdotm,tag4$$
$$left(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac89^3-89^-U88cdotsigma(m) = 3840m.tag5$$
Easy to see that
$$dfrac89^3-89^-U88 ge dfrac89^3-188 = 8011.$$
At the same time, $sigma(m)$ is the sum of divisors, so
$$sigma(m)ge 1+m.$$
Thus,
$$8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 3840.tag6$$
Similarly,
begincases
8011cdotleft(1-dfrac2^-A2right)cdotdfrac23cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right)cdotdfrac23cdotdfrac1011 le3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)cdotdfrac23cdotdfrac45 le 3840\
8011cdotleft(1-dfrac2^-12right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac23 le 3840,tag7\
endcases
or
begincases
8011cdotleft(1-dfrac2^-A2right) le 7920\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right) le 5280\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right) le 6336\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)le 7200\
8011cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 7680,quad A=1, B=0.tag8\
endcases
The system $(8)$ has the solution
$$left[
beginaligned
&A=0\
&A=1, B=0, C=D=1\
&A=1, B=0, C>2\
&A=2, B=C=0\
&5ge Age3, B=C=D=0,
endalignedright.tag9\$$
or
$$left[
beginalign
&2 not| nhspace40pt\
&((n=2m)vee(n=8m)vee(n=16m)vee(n=32m)vee(n=110m))\ &quadwedge(gcd(2cdot3cdot5cdot11,m)=1)\
&(n=50m)wedge(gcd(2cdot3, m)=1)\
&(n=4m)wedge(gcd(2cdot3cdot5,m)=1)\
endalignright..tag10
$$
$$mathbfcolorbrownApplying of the constraints$$
Obtained constraints $(10)$ allow to filter the possible solutions.
At first, becames impossible the case
$$n=2^1089^3k.$$
Secondly, if
$$n=89^2cdot8011cdot2003cdot167k,quad gcd(89cdot167cdot2003cdot8011,k) = 1,$$
with the equation
$$28sigma(k)=55k,$$
then
begincases
49sigma(m)=55m,text if k=4m, gcd(2cdot3cdot5,m)=1 \
21sigma(m)=22m,text if k=8m, gcd(2cdot3cdot5cdot11,m)=1 \
31cdot7sigma(m)=4cdot55mtext if k=16m, gcd(2cdot3cdot5cdot11,m)=1 \
21^2sigma(m)=8cdot55mtext if k=32m, gcd(2cdot3cdot5cdot11,m)=1 .tag11
endcases
Taking in account that
$$sigma(7)=2^3,quadsigma(49)= 2^2cdot5^2,$$
the system $(11)$ really can not be satisfied.
So I think that the issue equation $mathbfcolorbrownhas not solutions in the positive integer numbers.$
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
$$mathbfcolorgreenFixed version$$
$$mathbfcolorbrownConstraints of the task$$
The issue equation is
$$89^2sigma(n)=2^53^25^111^1n.tag1$$
Let WLOG
$$n=2^A3^B5^C11^D89^U+2m,quad (A,B,C,D,U)inmathbb Ncup 0 ,quad gcd(m, 2cdot3cdot5cdot11cdot89)=1,tag2$$
then $(1)$ takes the form of
$$89^2cdotdfrac2^A+1-11 cdotdfrac3^B+1-12 cdotdfrac5^C+1-14cdotdfrac11^D+1-110cdotdfrac89^U+3-188cdotsigma(m) =$$
$$2^A+53^B+25^C+111^D+189^U+2m,tag3$$
or
$$dfrac2^A+1-12^A+1cdotdfrac3^B+1-13^B+1cdotdfrac5^C+1-15^C+1cdotdfrac11^D+1-111^D+1cdotdfrac89^U+3-188cdot89^Ucdotsigma(m) = 2^8cdot3cdot5cdotm,tag4$$
$$left(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac89^3-89^-U88cdotsigma(m) = 3840m.tag5$$
Easy to see that
$$dfrac89^3-89^-U88 ge dfrac89^3-188 = 8011.$$
At the same time, $sigma(m)$ is the sum of divisors, so
$$sigma(m)ge 1+m.$$
Thus,
$$8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 3840.tag6$$
Similarly,
begincases
8011cdotleft(1-dfrac2^-A2right)cdotdfrac23cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right)cdotdfrac45cdotdfrac1011 le 3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right)cdotdfrac23cdotdfrac1011 le3840\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)cdotdfrac23cdotdfrac45 le 3840\
8011cdotleft(1-dfrac2^-12right)cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right)cdotdfrac23 le 3840,tag7\
endcases
or
begincases
8011cdotleft(1-dfrac2^-A2right) le 7920\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac3^-B3right) le 5280\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac5^-C5right) le 6336\
8011cdotleft(1-dfrac2^-A2right)cdotleft(1-dfrac11^-D11right)le 7200\
8011cdotleft(1-dfrac5^-C5right)cdotleft(1-dfrac11^-D11right) le 7680,quad A=1, B=0.tag8\
endcases
The system $(8)$ has the solution
$$left[
beginaligned
&A=0\
&A=1, B=0, C=D=1\
&A=1, B=0, C>2\
&A=2, B=C=0\
&5ge Age3, B=C=D=0,
endalignedright.tag9\$$
or
$$left[
beginalign
&2 not| nhspace40pt\
&((n=2m)vee(n=8m)vee(n=16m)vee(n=32m)vee(n=110m))\ &quadwedge(gcd(2cdot3cdot5cdot11,m)=1)\
&(n=50m)wedge(gcd(2cdot3, m)=1)\
&(n=4m)wedge(gcd(2cdot3cdot5,m)=1)\
endalignright..tag10
$$
$$mathbfcolorbrownApplying of the constraints$$
Obtained constraints $(10)$ allow to filter the possible solutions.
At first, becames impossible the case
$$n=2^1089^3k.$$
Secondly, if
$$n=89^2cdot8011cdot2003cdot167k,quad gcd(89cdot167cdot2003cdot8011,k) = 1,$$
with the equation
$$28sigma(k)=55k,$$
then
begincases
49sigma(m)=55m,text if k=4m, gcd(2cdot3cdot5,m)=1 \
21sigma(m)=22m,text if k=8m, gcd(2cdot3cdot5cdot11,m)=1 \
31cdot7sigma(m)=4cdot55mtext if k=16m, gcd(2cdot3cdot5cdot11,m)=1 \
21^2sigma(m)=8cdot55mtext if k=32m, gcd(2cdot3cdot5cdot11,m)=1 .tag11
endcases
Taking in account that
$$sigma(7)=2^3,quadsigma(49)= 2^2cdot5^2,$$
the system $(11)$ really can not be satisfied.
So I think that the issue equation $mathbfcolorbrownhas not solutions in the positive integer numbers.$
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
edited Aug 1 at 9:55
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
answered Jul 28 at 13:12
Yuri Negometyanov
9,3031523
9,3031523
1
$m$ not divisible by 2, 3, 5, 11, and 89?
– qwr
Jul 29 at 4:50
1
How did you get from $89^2cdotldotscdotfrac89^U+1-188$ to $frac89^U+3-188$?
– jpvee
Jul 29 at 6:50
1
@YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^11cdot3cdot5cdot11cdot m$.
– jpvee
Jul 30 at 11:35
1
@YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :)
– jpvee
Jul 30 at 12:19
2
@YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^11$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3).
– jpvee
Jul 30 at 13:14
 |Â
show 14 more comments
1
$m$ not divisible by 2, 3, 5, 11, and 89?
– qwr
Jul 29 at 4:50
1
How did you get from $89^2cdotldotscdotfrac89^U+1-188$ to $frac89^U+3-188$?
– jpvee
Jul 29 at 6:50
1
@YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^11cdot3cdot5cdot11cdot m$.
– jpvee
Jul 30 at 11:35
1
@YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :)
– jpvee
Jul 30 at 12:19
2
@YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^11$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3).
– jpvee
Jul 30 at 13:14
1
1
$m$ not divisible by 2, 3, 5, 11, and 89?
– qwr
Jul 29 at 4:50
$m$ not divisible by 2, 3, 5, 11, and 89?
– qwr
Jul 29 at 4:50
1
1
How did you get from $89^2cdotldotscdotfrac89^U+1-188$ to $frac89^U+3-188$?
– jpvee
Jul 29 at 6:50
How did you get from $89^2cdotldotscdotfrac89^U+1-188$ to $frac89^U+3-188$?
– jpvee
Jul 29 at 6:50
1
1
@YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^11cdot3cdot5cdot11cdot m$.
– jpvee
Jul 30 at 11:35
@YuriNegometyanov: I believe that equation (3) is still wrong. I think the right-hand side should read $2^11cdot3cdot5cdot11cdot m$.
– jpvee
Jul 30 at 11:35
1
1
@YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :)
– jpvee
Jul 30 at 12:19
@YuriNegometyanov: I like the style of your comments - very poetic! But I don't understand what you are trying to say :)
– jpvee
Jul 30 at 12:19
2
2
@YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^11$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3).
– jpvee
Jul 30 at 13:14
@YuriNegometyanov: Thanks for the clarification. Then please tell me what you did to the $88$ in the denominator in the equation before (3) - I see that the $8$ has become part of the $2^11$ on the right-hand side of (3), but I do not see what became of the remaining $11$ of the $88=8cdot 11$. Also the same question for the $10$ in the denominator. And finally, the $m$ vanished from the upper equation completely when going to (3).
– jpvee
Jul 30 at 13:14
 |Â
show 14 more comments
up vote
7
down vote
This is not an answer, but an extended comment, that should help anyone interested in using brute-force numerical search to find the solution.
The problem is to find $n in mathbbN$ for which
$$7921 sigma(n) = 15840 n tag1labelNA1$$
where $sigma(n)$ is the sum of all divisors of $n$, as defined in the Wikipedia divisor function article, and as a sequence in OEIS A000203.
Because $sigma(n) = n + 1$ if $n$ is a prime, and
$$7921 (n + 1) ne 15840 n, quad n in mathbbN$$
we already know there is no prime solution to $eqrefNA1$.
Consider the prime factorization of $n$. Let $p_i in mathbbN$ be nonrepeating primes ($p_i = p_j$ if and only if $i = j$), and $1 le k_i in mathbbN$. Then,
$$n = prod_i=0^N-1 p_i^k_i$$
and
$$sigma(n) = prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1$$
because $sigma(p^k) = sum_j=0^k p^j = (p^k+1-1)/(p-1)$ when $p$ is a prime.
We can now rewrite the problem $eqrefNA1$ as
$$7921 prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1 = 15840 prod_i=0^N-1 p_i^k_i tag2labelNA2$$
Rearranging the terms yields
$$prod_i=0^N-1 frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = frac792115840 = fracnsigma(n) = frac89^22^5 cdot 3^2 cdot 5 cdot 11
tag3labelNA3$$
Note the term
$$f_i = frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = fracp_i^k_isum_j=0^k_i p^j, quad frac12 lt f_i lt 1
tag4labelNA4$$
i.e.,
$$beginarrayll
f_i = fracp_ip_i + 1, & k_i = 1 \
f_i = fracp_i^2p_i^2 + p_i + 1, & k_i = 2 \
f_i = fracp_i^3p_i^3 + p_i^2 + p_i + 1 , & k_i = 3 \
f_i = fracp_i^k_ip_i^k_i + p_i^k_i-1 + dots + p_i + 1 & \
endarray$$
Thus, the numerical search problem is now reduced to find the set of terms $f_i$ based on primes $p_i$ and their positive powers $k_i$, so that the product $$prod_i=0^N-1 f_i = frac792115840$$
In particular, because $f_i lt 1$, a particular set can be rejected immediately if the product falls below the target ratio.
For example, if $p_0 = 89$, $k_0 = 2$, to eliminate the prime factor in the numerator. Repeating, that leads to $p_1 = 8011$, $k_1 = 1$; $p_2 = 2003$, $k_2 = 1$; and $p_3 = 167$, $k_3 = 1$, to get us to a result with a composite numerator and a denominator:
$$beginarrayl
n & fracnsigma(n) & frac15840 n7921 sigma(n) \
hline
89^2 cdot 8011 cdot 2003 cdot 167
& frac79218064 = frac79212^7 cdot 3^2 cdot 7
& frac5528 = frac5 cdot 112^2 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 7
& frac79219216 = frac79212^10 cdot 3^2
& frac5532 = frac5 cdot 112^5
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2
& frac792112096 = frac79212^6 cdot 3^3 cdot 7
& frac5542 = frac5 cdot 112 cdot 3 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2 cdot 7
& frac792113824 = frac79212^9 cdot 3^3
& frac5548 = frac5 cdot 112^4 cdot 3
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2^2
& frac792114112 = frac79212^5 cdot 3^2 cdot 7^2
& frac5549 = frac5 cdot 117^2
\ endarray$$
If you append $p_5 = 7$, $k_5 = 1$ or $k_5 = 2$ to $n$ in the final row above, the rightmost field drops below 1 (to $55/56$ for $k_5 = 1$, and to $55/57$ for $k_5 = 2$), leading nowhere. Similarly, appending $p_6 = 3$, $k_6 = 1$ or $k_6 = 2$ to $n$ in the second-to-last row (to $55/64$ for $k_6 = 1$, and to $165/208$ for $k_6 = 2$) leads nowhere.
It looks to (very non-mathematician) me that an exhaustive search over primes $p$ is possible, due to terms $f_i$ having a power of a prime in the numerator, as specified in $eqrefNA4$. Whether an exhaustive search is possible or not is an open question (and is important for those looking for proof), but the efficient numerical brute force search strategies are straightforward; especially if one is looking for some other ratios than $fracnsigma(n) = frac792115840$.
To continue the search above, I'd need a prime $p$ and a positive integer $k$ such that $sum_j=0^k p^j = 55$ (to yield a factor with denominator $55$). No such pair exists, so the search strategies I've come up with thus far are exhausted.
Hopefully, one of the math sages here can take this further from here.
answered Jul 27 at 5:00
Nominal Animal
5,6952414
1
This is the way to the jungle. For example, the equality $2^11-1=23cdot89$ requires consideration of the variant $k_0=3$ etc.
– Yuri Negometyanov
Jul 29 at 3:52
@YuriNegometyanov: At first glance, using $2^10$ does indeed look promising. Unfortunately, however, $frac89^3sigma(89^3)cdotfrac2^10sigma(2^10)$ is already below the target ratio of $frac792115840$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$.
– jpvee
Jul 30 at 8:36
I do believe my efforts stalled, because rewriting $eqrefNA2$ as $eqrefNA3$ places an additional constraint on the coefficients. I really should rewrite $eqrefNA2$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $eqrefNA3$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)?
– Nominal Animal
Jul 30 at 9:45
@NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist.
– Yuri Negometyanov
Jul 30 at 10:55
1
@YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3).
– jpvee
Jul 30 at 11:36
 |Â
show 2 more comments
up vote
7
down vote
This is not an answer, but an extended comment, that should help anyone interested in using brute-force numerical search to find the solution.
The problem is to find $n in mathbbN$ for which
$$7921 sigma(n) = 15840 n tag1labelNA1$$
where $sigma(n)$ is the sum of all divisors of $n$, as defined in the Wikipedia divisor function article, and as a sequence in OEIS A000203.
Because $sigma(n) = n + 1$ if $n$ is a prime, and
$$7921 (n + 1) ne 15840 n, quad n in mathbbN$$
we already know there is no prime solution to $eqrefNA1$.
Consider the prime factorization of $n$. Let $p_i in mathbbN$ be nonrepeating primes ($p_i = p_j$ if and only if $i = j$), and $1 le k_i in mathbbN$. Then,
$$n = prod_i=0^N-1 p_i^k_i$$
and
$$sigma(n) = prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1$$
because $sigma(p^k) = sum_j=0^k p^j = (p^k+1-1)/(p-1)$ when $p$ is a prime.
We can now rewrite the problem $eqrefNA1$ as
$$7921 prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1 = 15840 prod_i=0^N-1 p_i^k_i tag2labelNA2$$
Rearranging the terms yields
$$prod_i=0^N-1 frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = frac792115840 = fracnsigma(n) = frac89^22^5 cdot 3^2 cdot 5 cdot 11
tag3labelNA3$$
Note the term
$$f_i = frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = fracp_i^k_isum_j=0^k_i p^j, quad frac12 lt f_i lt 1
tag4labelNA4$$
i.e.,
$$beginarrayll
f_i = fracp_ip_i + 1, & k_i = 1 \
f_i = fracp_i^2p_i^2 + p_i + 1, & k_i = 2 \
f_i = fracp_i^3p_i^3 + p_i^2 + p_i + 1 , & k_i = 3 \
f_i = fracp_i^k_ip_i^k_i + p_i^k_i-1 + dots + p_i + 1 & \
endarray$$
Thus, the numerical search problem is now reduced to find the set of terms $f_i$ based on primes $p_i$ and their positive powers $k_i$, so that the product $$prod_i=0^N-1 f_i = frac792115840$$
In particular, because $f_i lt 1$, a particular set can be rejected immediately if the product falls below the target ratio.
For example, if $p_0 = 89$, $k_0 = 2$, to eliminate the prime factor in the numerator. Repeating, that leads to $p_1 = 8011$, $k_1 = 1$; $p_2 = 2003$, $k_2 = 1$; and $p_3 = 167$, $k_3 = 1$, to get us to a result with a composite numerator and a denominator:
$$beginarrayl
n & fracnsigma(n) & frac15840 n7921 sigma(n) \
hline
89^2 cdot 8011 cdot 2003 cdot 167
& frac79218064 = frac79212^7 cdot 3^2 cdot 7
& frac5528 = frac5 cdot 112^2 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 7
& frac79219216 = frac79212^10 cdot 3^2
& frac5532 = frac5 cdot 112^5
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2
& frac792112096 = frac79212^6 cdot 3^3 cdot 7
& frac5542 = frac5 cdot 112 cdot 3 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2 cdot 7
& frac792113824 = frac79212^9 cdot 3^3
& frac5548 = frac5 cdot 112^4 cdot 3
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2^2
& frac792114112 = frac79212^5 cdot 3^2 cdot 7^2
& frac5549 = frac5 cdot 117^2
\ endarray$$
If you append $p_5 = 7$, $k_5 = 1$ or $k_5 = 2$ to $n$ in the final row above, the rightmost field drops below 1 (to $55/56$ for $k_5 = 1$, and to $55/57$ for $k_5 = 2$), leading nowhere. Similarly, appending $p_6 = 3$, $k_6 = 1$ or $k_6 = 2$ to $n$ in the second-to-last row (to $55/64$ for $k_6 = 1$, and to $165/208$ for $k_6 = 2$) leads nowhere.
It looks to (very non-mathematician) me that an exhaustive search over primes $p$ is possible, due to terms $f_i$ having a power of a prime in the numerator, as specified in $eqrefNA4$. Whether an exhaustive search is possible or not is an open question (and is important for those looking for proof), but the efficient numerical brute force search strategies are straightforward; especially if one is looking for some other ratios than $fracnsigma(n) = frac792115840$.
To continue the search above, I'd need a prime $p$ and a positive integer $k$ such that $sum_j=0^k p^j = 55$ (to yield a factor with denominator $55$). No such pair exists, so the search strategies I've come up with thus far are exhausted.
Hopefully, one of the math sages here can take this further from here.
answered Jul 27 at 5:00
Nominal Animal
5,6952414
1
This is the way to the jungle. For example, the equality $2^11-1=23cdot89$ requires consideration of the variant $k_0=3$ etc.
– Yuri Negometyanov
Jul 29 at 3:52
@YuriNegometyanov: At first glance, using $2^10$ does indeed look promising. Unfortunately, however, $frac89^3sigma(89^3)cdotfrac2^10sigma(2^10)$ is already below the target ratio of $frac792115840$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$.
– jpvee
Jul 30 at 8:36
I do believe my efforts stalled, because rewriting $eqrefNA2$ as $eqrefNA3$ places an additional constraint on the coefficients. I really should rewrite $eqrefNA2$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $eqrefNA3$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)?
– Nominal Animal
Jul 30 at 9:45
@NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist.
– Yuri Negometyanov
Jul 30 at 10:55
1
@YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3).
– jpvee
Jul 30 at 11:36
 |Â
show 2 more comments
up vote
7
down vote
up vote
7
down vote
This is not an answer, but an extended comment, that should help anyone interested in using brute-force numerical search to find the solution.
The problem is to find $n in mathbbN$ for which
$$7921 sigma(n) = 15840 n tag1labelNA1$$
where $sigma(n)$ is the sum of all divisors of $n$, as defined in the Wikipedia divisor function article, and as a sequence in OEIS A000203.
Because $sigma(n) = n + 1$ if $n$ is a prime, and
$$7921 (n + 1) ne 15840 n, quad n in mathbbN$$
we already know there is no prime solution to $eqrefNA1$.
Consider the prime factorization of $n$. Let $p_i in mathbbN$ be nonrepeating primes ($p_i = p_j$ if and only if $i = j$), and $1 le k_i in mathbbN$. Then,
$$n = prod_i=0^N-1 p_i^k_i$$
and
$$sigma(n) = prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1$$
because $sigma(p^k) = sum_j=0^k p^j = (p^k+1-1)/(p-1)$ when $p$ is a prime.
We can now rewrite the problem $eqrefNA1$ as
$$7921 prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1 = 15840 prod_i=0^N-1 p_i^k_i tag2labelNA2$$
Rearranging the terms yields
$$prod_i=0^N-1 frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = frac792115840 = fracnsigma(n) = frac89^22^5 cdot 3^2 cdot 5 cdot 11
tag3labelNA3$$
Note the term
$$f_i = frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = fracp_i^k_isum_j=0^k_i p^j, quad frac12 lt f_i lt 1
tag4labelNA4$$
i.e.,
$$beginarrayll
f_i = fracp_ip_i + 1, & k_i = 1 \
f_i = fracp_i^2p_i^2 + p_i + 1, & k_i = 2 \
f_i = fracp_i^3p_i^3 + p_i^2 + p_i + 1 , & k_i = 3 \
f_i = fracp_i^k_ip_i^k_i + p_i^k_i-1 + dots + p_i + 1 & \
endarray$$
Thus, the numerical search problem is now reduced to find the set of terms $f_i$ based on primes $p_i$ and their positive powers $k_i$, so that the product $$prod_i=0^N-1 f_i = frac792115840$$
In particular, because $f_i lt 1$, a particular set can be rejected immediately if the product falls below the target ratio.
For example, if $p_0 = 89$, $k_0 = 2$, to eliminate the prime factor in the numerator. Repeating, that leads to $p_1 = 8011$, $k_1 = 1$; $p_2 = 2003$, $k_2 = 1$; and $p_3 = 167$, $k_3 = 1$, to get us to a result with a composite numerator and a denominator:
$$beginarrayl
n & fracnsigma(n) & frac15840 n7921 sigma(n) \
hline
89^2 cdot 8011 cdot 2003 cdot 167
& frac79218064 = frac79212^7 cdot 3^2 cdot 7
& frac5528 = frac5 cdot 112^2 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 7
& frac79219216 = frac79212^10 cdot 3^2
& frac5532 = frac5 cdot 112^5
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2
& frac792112096 = frac79212^6 cdot 3^3 cdot 7
& frac5542 = frac5 cdot 112 cdot 3 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2 cdot 7
& frac792113824 = frac79212^9 cdot 3^3
& frac5548 = frac5 cdot 112^4 cdot 3
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2^2
& frac792114112 = frac79212^5 cdot 3^2 cdot 7^2
& frac5549 = frac5 cdot 117^2
\ endarray$$
If you append $p_5 = 7$, $k_5 = 1$ or $k_5 = 2$ to $n$ in the final row above, the rightmost field drops below 1 (to $55/56$ for $k_5 = 1$, and to $55/57$ for $k_5 = 2$), leading nowhere. Similarly, appending $p_6 = 3$, $k_6 = 1$ or $k_6 = 2$ to $n$ in the second-to-last row (to $55/64$ for $k_6 = 1$, and to $165/208$ for $k_6 = 2$) leads nowhere.
It looks to (very non-mathematician) me that an exhaustive search over primes $p$ is possible, due to terms $f_i$ having a power of a prime in the numerator, as specified in $eqrefNA4$. Whether an exhaustive search is possible or not is an open question (and is important for those looking for proof), but the efficient numerical brute force search strategies are straightforward; especially if one is looking for some other ratios than $fracnsigma(n) = frac792115840$.
To continue the search above, I'd need a prime $p$ and a positive integer $k$ such that $sum_j=0^k p^j = 55$ (to yield a factor with denominator $55$). No such pair exists, so the search strategies I've come up with thus far are exhausted.
Hopefully, one of the math sages here can take this further from here.
answered Jul 27 at 5:00
Nominal Animal
5,6952414
This is not an answer, but an extended comment, that should help anyone interested in using brute-force numerical search to find the solution.
The problem is to find $n in mathbbN$ for which
$$7921 sigma(n) = 15840 n tag1labelNA1$$
where $sigma(n)$ is the sum of all divisors of $n$, as defined in the Wikipedia divisor function article, and as a sequence in OEIS A000203.
Because $sigma(n) = n + 1$ if $n$ is a prime, and
$$7921 (n + 1) ne 15840 n, quad n in mathbbN$$
we already know there is no prime solution to $eqrefNA1$.
Consider the prime factorization of $n$. Let $p_i in mathbbN$ be nonrepeating primes ($p_i = p_j$ if and only if $i = j$), and $1 le k_i in mathbbN$. Then,
$$n = prod_i=0^N-1 p_i^k_i$$
and
$$sigma(n) = prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1$$
because $sigma(p^k) = sum_j=0^k p^j = (p^k+1-1)/(p-1)$ when $p$ is a prime.
We can now rewrite the problem $eqrefNA1$ as
$$7921 prod_i=0^N-1 fracp_i^k_i+1 - 1p_i - 1 = 15840 prod_i=0^N-1 p_i^k_i tag2labelNA2$$
Rearranging the terms yields
$$prod_i=0^N-1 frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = frac792115840 = fracnsigma(n) = frac89^22^5 cdot 3^2 cdot 5 cdot 11
tag3labelNA3$$
Note the term
$$f_i = frac p_i^k_i + 1 - p_i^k_i p_i^k_i + 1 - 1 = fracp_i^k_isum_j=0^k_i p^j, quad frac12 lt f_i lt 1
tag4labelNA4$$
i.e.,
$$beginarrayll
f_i = fracp_ip_i + 1, & k_i = 1 \
f_i = fracp_i^2p_i^2 + p_i + 1, & k_i = 2 \
f_i = fracp_i^3p_i^3 + p_i^2 + p_i + 1 , & k_i = 3 \
f_i = fracp_i^k_ip_i^k_i + p_i^k_i-1 + dots + p_i + 1 & \
endarray$$
Thus, the numerical search problem is now reduced to find the set of terms $f_i$ based on primes $p_i$ and their positive powers $k_i$, so that the product $$prod_i=0^N-1 f_i = frac792115840$$
In particular, because $f_i lt 1$, a particular set can be rejected immediately if the product falls below the target ratio.
For example, if $p_0 = 89$, $k_0 = 2$, to eliminate the prime factor in the numerator. Repeating, that leads to $p_1 = 8011$, $k_1 = 1$; $p_2 = 2003$, $k_2 = 1$; and $p_3 = 167$, $k_3 = 1$, to get us to a result with a composite numerator and a denominator:
$$beginarrayl
n & fracnsigma(n) & frac15840 n7921 sigma(n) \
hline
89^2 cdot 8011 cdot 2003 cdot 167
& frac79218064 = frac79212^7 cdot 3^2 cdot 7
& frac5528 = frac5 cdot 112^2 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 7
& frac79219216 = frac79212^10 cdot 3^2
& frac5532 = frac5 cdot 112^5
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2
& frac792112096 = frac79212^6 cdot 3^3 cdot 7
& frac5542 = frac5 cdot 112 cdot 3 cdot 7
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2 cdot 7
& frac792113824 = frac79212^9 cdot 3^3
& frac5548 = frac5 cdot 112^4 cdot 3
\ 89^2 cdot 8011 cdot 2003 cdot 167 cdot 2^2
& frac792114112 = frac79212^5 cdot 3^2 cdot 7^2
& frac5549 = frac5 cdot 117^2
\ endarray$$
If you append $p_5 = 7$, $k_5 = 1$ or $k_5 = 2$ to $n$ in the final row above, the rightmost field drops below 1 (to $55/56$ for $k_5 = 1$, and to $55/57$ for $k_5 = 2$), leading nowhere. Similarly, appending $p_6 = 3$, $k_6 = 1$ or $k_6 = 2$ to $n$ in the second-to-last row (to $55/64$ for $k_6 = 1$, and to $165/208$ for $k_6 = 2$) leads nowhere.
It looks to (very non-mathematician) me that an exhaustive search over primes $p$ is possible, due to terms $f_i$ having a power of a prime in the numerator, as specified in $eqrefNA4$. Whether an exhaustive search is possible or not is an open question (and is important for those looking for proof), but the efficient numerical brute force search strategies are straightforward; especially if one is looking for some other ratios than $fracnsigma(n) = frac792115840$.
To continue the search above, I'd need a prime $p$ and a positive integer $k$ such that $sum_j=0^k p^j = 55$ (to yield a factor with denominator $55$). No such pair exists, so the search strategies I've come up with thus far are exhausted.
Hopefully, one of the math sages here can take this further from here.
answered Jul 27 at 5:00
Nominal Animal
5,6952414
edited Jul 27 at 5:09
answered Jul 27 at 5:00
Nominal Animal
5,6952414
answered Jul 27 at 5:00
Nominal Animal
5,6952414
answered Jul 27 at 5:00
Nominal Animal
5,6952414
5,6952414
1
This is the way to the jungle. For example, the equality $2^11-1=23cdot89$ requires consideration of the variant $k_0=3$ etc.
– Yuri Negometyanov
Jul 29 at 3:52
@YuriNegometyanov: At first glance, using $2^10$ does indeed look promising. Unfortunately, however, $frac89^3sigma(89^3)cdotfrac2^10sigma(2^10)$ is already below the target ratio of $frac792115840$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$.
– jpvee
Jul 30 at 8:36
I do believe my efforts stalled, because rewriting $eqrefNA2$ as $eqrefNA3$ places an additional constraint on the coefficients. I really should rewrite $eqrefNA2$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $eqrefNA3$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)?
– Nominal Animal
Jul 30 at 9:45
@NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist.
– Yuri Negometyanov
Jul 30 at 10:55
1
@YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3).
– jpvee
Jul 30 at 11:36
 |Â
show 2 more comments
1
This is the way to the jungle. For example, the equality $2^11-1=23cdot89$ requires consideration of the variant $k_0=3$ etc.
– Yuri Negometyanov
Jul 29 at 3:52
@YuriNegometyanov: At first glance, using $2^10$ does indeed look promising. Unfortunately, however, $frac89^3sigma(89^3)cdotfrac2^10sigma(2^10)$ is already below the target ratio of $frac792115840$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$.
– jpvee
Jul 30 at 8:36
I do believe my efforts stalled, because rewriting $eqrefNA2$ as $eqrefNA3$ places an additional constraint on the coefficients. I really should rewrite $eqrefNA2$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $eqrefNA3$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)?
– Nominal Animal
Jul 30 at 9:45
@NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist.
– Yuri Negometyanov
Jul 30 at 10:55
1
@YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3).
– jpvee
Jul 30 at 11:36
1
1
This is the way to the jungle. For example, the equality $2^11-1=23cdot89$ requires consideration of the variant $k_0=3$ etc.
– Yuri Negometyanov
Jul 29 at 3:52
This is the way to the jungle. For example, the equality $2^11-1=23cdot89$ requires consideration of the variant $k_0=3$ etc.
– Yuri Negometyanov
Jul 29 at 3:52
@YuriNegometyanov: At first glance, using $2^10$ does indeed look promising. Unfortunately, however, $frac89^3sigma(89^3)cdotfrac2^10sigma(2^10)$ is already below the target ratio of $frac792115840$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$.
– jpvee
Jul 30 at 8:36
@YuriNegometyanov: At first glance, using $2^10$ does indeed look promising. Unfortunately, however, $frac89^3sigma(89^3)cdotfrac2^10sigma(2^10)$ is already below the target ratio of $frac792115840$. In fact, if you use $k_0=3$, then the largest power of $2$ that may appear in $n$ is $2^5$.
– jpvee
Jul 30 at 8:36
I do believe my efforts stalled, because rewriting $eqrefNA2$ as $eqrefNA3$ places an additional constraint on the coefficients. I really should rewrite $eqrefNA2$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $eqrefNA3$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)?
– Nominal Animal
Jul 30 at 9:45
I do believe my efforts stalled, because rewriting $eqrefNA2$ as $eqrefNA3$ places an additional constraint on the coefficients. I really should rewrite $eqrefNA2$ to use another index variable, say $j$, on the right side. Similarly, the indexing in the numerator and denominator in $eqrefNA3$ do not need to be the same. @YuriNegometyanov: Is this what you meant by the way to the jungle (off the track)?
– Nominal Animal
Jul 30 at 9:45
@NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist.
– Yuri Negometyanov
Jul 30 at 10:55
@NominalAnimal I mean the jungles of variants in the situation where solution doesn't exist.
– Yuri Negometyanov
Jul 30 at 10:55
1
1
@YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3).
– jpvee
Jul 30 at 11:36
@YuriNegometyanov: I don't think your proof works - see my comment regarding the right-hand side in equation (3).
– jpvee
Jul 30 at 11:36
 |Â
show 2 more comments
up vote
2
down vote
This is not an answer either: I have written a program doing a non-exhaustive search, and so far, no solutions have come up for the Mathieu group $M_11$; however, the program did find a solution e.g. for the larger Mathieu group $M_22$ of order $443520=2^7cdot3^2cdot5cdot7cdot11$:
Let $n=55009909630=2cdot5cdot13cdot79cdot109cdot157cdot313$, then since $$
beginalign
(|M_22|+1)cdotsigma(n)&=443521cdotsigma(2cdot5cdot13cdot79cdot109cdot157cdot313) \
&=13cdot109cdot313cdot3cdot6cdot14cdot80cdot110cdot158cdot314 \
&=2^9cdot3^2cdot5^2cdot7cdot11cdot13cdot79cdot109cdot157cdot313 \
&=2cdot|M_22|cdot ntext,endalign$$
$M_22timesmathbbZ_55009909630$ must be immaculate.
answered Jul 27 at 12:16
jpvee
2,48421326
add a comment |Â
up vote
2
down vote
This is not an answer either: I have written a program doing a non-exhaustive search, and so far, no solutions have come up for the Mathieu group $M_11$; however, the program did find a solution e.g. for the larger Mathieu group $M_22$ of order $443520=2^7cdot3^2cdot5cdot7cdot11$:
Let $n=55009909630=2cdot5cdot13cdot79cdot109cdot157cdot313$, then since $$
beginalign
(|M_22|+1)cdotsigma(n)&=443521cdotsigma(2cdot5cdot13cdot79cdot109cdot157cdot313) \
&=13cdot109cdot313cdot3cdot6cdot14cdot80cdot110cdot158cdot314 \
&=2^9cdot3^2cdot5^2cdot7cdot11cdot13cdot79cdot109cdot157cdot313 \
&=2cdot|M_22|cdot ntext,endalign$$
$M_22timesmathbbZ_55009909630$ must be immaculate.
answered Jul 27 at 12:16
jpvee
2,48421326
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is not an answer either: I have written a program doing a non-exhaustive search, and so far, no solutions have come up for the Mathieu group $M_11$; however, the program did find a solution e.g. for the larger Mathieu group $M_22$ of order $443520=2^7cdot3^2cdot5cdot7cdot11$:
Let $n=55009909630=2cdot5cdot13cdot79cdot109cdot157cdot313$, then since $$
beginalign
(|M_22|+1)cdotsigma(n)&=443521cdotsigma(2cdot5cdot13cdot79cdot109cdot157cdot313) \
&=13cdot109cdot313cdot3cdot6cdot14cdot80cdot110cdot158cdot314 \
&=2^9cdot3^2cdot5^2cdot7cdot11cdot13cdot79cdot109cdot157cdot313 \
&=2cdot|M_22|cdot ntext,endalign$$
$M_22timesmathbbZ_55009909630$ must be immaculate.
answered Jul 27 at 12:16
jpvee
2,48421326
This is not an answer either: I have written a program doing a non-exhaustive search, and so far, no solutions have come up for the Mathieu group $M_11$; however, the program did find a solution e.g. for the larger Mathieu group $M_22$ of order $443520=2^7cdot3^2cdot5cdot7cdot11$:
Let $n=55009909630=2cdot5cdot13cdot79cdot109cdot157cdot313$, then since $$
beginalign
(|M_22|+1)cdotsigma(n)&=443521cdotsigma(2cdot5cdot13cdot79cdot109cdot157cdot313) \
&=13cdot109cdot313cdot3cdot6cdot14cdot80cdot110cdot158cdot314 \
&=2^9cdot3^2cdot5^2cdot7cdot11cdot13cdot79cdot109cdot157cdot313 \
&=2cdot|M_22|cdot ntext,endalign$$
$M_22timesmathbbZ_55009909630$ must be immaculate.
answered Jul 27 at 12:16
jpvee
2,48421326
answered Jul 27 at 12:16
jpvee
2,48421326
answered Jul 27 at 12:16
jpvee
2,48421326
answered Jul 27 at 12:16
jpvee
2,48421326
2,48421326
add a comment |Â
add a comment |Â
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10
This seems like such a random question to ask until you see the entirely reasonable motivation.
– Arthur
Jul 20 at 9:13
@Arthur Tbh I saw a motivation at the start, this being essentially a question of how close $sigma(n)$ can be to $2n$.
– fretty
Jul 20 at 9:54
2
An explantion why it is hopeless to search such a number with brute force, would be nice.
– Peter
Jul 20 at 11:25
6
I figured out a positive integer $n$ with $$7921sigma(n)=2cdot 15840cdot n$$ $$n=563432025678613816032$$ Not sure whether this helps
– Peter
Jul 20 at 12:17
1
I have tried hard to find a solution, but I failed. My conjecture is that there is no solution, but a proof will be very difficult considering that it is , for example , an open question whether there is an odd perfect number.
– Peter
Jul 21 at 16:46