S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Thu, 23 Oct 2014 07:38:37 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 8 posts ] 
Author Message
 Post subject: perfect cube
PostPosted: Sun, 18 Sep 2011 20:54:59 UTC 
Offline
Member

Joined: Sun, 23 Mar 2008 10:55:47 UTC
Posts: 29
Find all x for which x!+3x-9 is perfect cube of an integer.

edited


Last edited by mathemagics on Sun, 18 Sep 2011 21:03:45 UTC, edited 1 time in total.

Top
 Profile  
 
 Post subject: Re: perfect cube
PostPosted: Sun, 18 Sep 2011 20:59:42 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14147
Location: Austin, TX
mathemagics wrote:
Find x for which x!+3x-9 is perfect cube of an integer.


This is easy just by inspection that x=4 works. Topic moved.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: perfect cube
PostPosted: Sun, 18 Sep 2011 21:03:11 UTC 
Offline
Member

Joined: Sun, 23 Mar 2008 10:55:47 UTC
Posts: 29
Sorry! I mean find every x for which x!+3x-9 is perfect cube


Top
 Profile  
 
 Post subject: Re: perfect cube
PostPosted: Sun, 18 Sep 2011 21:08:20 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14147
Location: Austin, TX
mathemagics wrote:
Sorry! I mean find every x for which x!+3x-9 is perfect cube


OK, topic moved back.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: perfect cube
PostPosted: Sat, 29 Oct 2011 13:55:29 UTC 
Offline
Member

Joined: Wed, 4 Jun 2008 21:01:47 UTC
Posts: 12
x=2; x=4 ; x=6 are the only solutions

For x>6 x! + 3x - 9 == 5mod7

All cubes are 0mod7, 1mod7 or 6mod7

QED

_________________
"The Amusing Math Genius" is an interesting title. Why?


Top
 Profile  
 
 Post subject: Re: perfect cube
PostPosted: Sat, 29 Oct 2011 14:46:20 UTC 
Online
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6935
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
NickMcG wrote:
x=2; x=4 ; x=6 are the only solutions

For x>6 x! + 3x - 9 == 5mod7

All cubes are 0mod7, 1mod7 or 6mod7

QED


Why does x!+3x-9\equiv 5\pmod{7}? For example, 10!+3(10)-9\equiv 0\pmod{7}.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
 Post subject: Re: perfect cube
PostPosted: Sat, 29 Oct 2011 20:35:43 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14147
Location: Austin, TX
Indeed, as outermeasure has pointed out, for any x>7, we are left with a linear congruence modulo 7, so we can make it be whatever we feel like by picking x properly.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: perfect cube
PostPosted: Sun, 30 Oct 2011 08:18:34 UTC 
Online
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6935
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Shadow wrote:
Indeed, as outermeasure has pointed out, for any x>7, we are left with a linear congruence modulo 7, so we can make it be whatever we feel like by picking x properly.


Indeed, if you want to use mod, you probably need some heavy analytic number theory machinery to help you pick a prime p_x<x which is 1 (mod 3), for every sufficiently large x, for which 3x-9 is not a cubic residue mod p_x. Heuristically such prime p_x should exist for all sufficiently large x
Spoiler:
If the cubic residues are uniformly distributed and independent, the probability of fixed x which doesn't have such prime p_x would be about (\frac{1}{3})^{x/(2\log x)}, and so the expected number of such x is finite.

but you get into something very nasty very early on and AFAIK we don't know the behaviour of the cubic Gauss sum to say cubic residues are close enough to randomly distributed for the heuristic argument to work.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 8 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA