mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Blogorrhea > carpetpool

Reply
 
Thread Tools
Old 2017-02-09, 04:08   #1
carpetpool
 
carpetpool's Avatar
 
"Sam"
Nov 2016

2·3·53 Posts
Post Solutions to a^2-ab+b^2 = 3^n

The basic algebra rules for factoring sums and differences of powers are here. For a^n+-b^n when n is prime, (a^n+-b^n)/(a+-b) = 0 or 1 (mod n). If (a^n+-b^n)/(a+-b) = 0 (mod n), let n^k be the highest power of n dividing (a^n+-b^n)/(a+-b). Then (a^n+-b^n)/((a+-b)*n^k) = 1 (mod n). Would it ever be the case that (a, b) > 1, k > 1, (a^n+-b^n)/((a+-b)*n^k) = 1?

For n = 3, we are looking to find

Solutions to a^2+ab+b^2 = 3^n (I messed up in the question title btw, this is what I meant to ask.)

The only known solutions with integers are

1^2+1*1+1^2 = 3^1
2^2-2*1+1^2 = 3^1

It is probably the case that no more solutions with (a, b) > 0 exists other than these two. If you know of one, or have a proof that no solutions to a^2+ab+b^2 = 3^n exist except (1, 1) and (-2, 1), please post arguments here.

Last fiddled with by carpetpool on 2017-02-09 at 04:47
carpetpool is offline   Reply With Quote
Old 2017-02-09, 06:12   #2
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

23C616 Posts
Default

Quote:
Originally Posted by carpetpool View Post
... except (1, 1) and (-2, 1)
How 'bout (2, -1)?
Batalov is offline   Reply With Quote
Old 2017-02-09, 06:41   #3
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
Jun 2011
Thailand

22·7·11·29 Posts
Default

Tip: you will first note that the expressions are symmetric in a and b. Then, the two are equivalent, as \(a^2+ab+b^2\) is \((a+b)^2-ab\) therefore if you substitute \(a=c+b\) you will have \((c+b)^2-(c+b)b+b^2=c^2+2cb+b^2-cb-b^2+b^2=c^2+cb+b^2\). So, if there is a solution for plus side, then there is a solution for the minus side if you increase \(a\) with \(b\).

Given that, think about the modularity of a and b to 3. The plus expression can only be a multiple of 3 if a and b are both either 0, 1, or 2, (mod 3). For (0, 1), (0, 2), or (1, 2) you always get \((a+b)^2-ab=1\) (mod 3). (others are symmetrical). For the minus expression, the only valid groups are (0, 0) and (1, 2) (with symmetry) otherwise again, all the other combinations result in 1 (mod 3).

Now, for both cases, if (a, b)=(0, 0) (mod 3), then you can divide by 3 on both sides of your equation, and you get a smaller solution (see infinite descent method).

So, you only have to study the plus cases when (a, b)=(1, 1), or (2, 2), and the minus case when (a, b)=(1, 2). Due to plus/minus equivalence, you have only to study a single case. Now, tell us the solutions... [Hint 2: the left side is always 3 (mod 9) therefore no solution with n>=2 can exist ]

Last fiddled with by LaurV on 2017-02-09 at 07:07
LaurV is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Fedora 22 (or, Gnome 3) Annoyances - are there solutions? EdH Linux 16 2016-03-18 17:20
Number of Solutions to d(p) flouran Math 20 2009-09-08 05:48
Possible solutions to an equation: Vijay Math 6 2005-04-14 05:19
Looking for solutions to w^2-n*x^2=z^2 jtavares Factoring 3 2005-04-11 19:25
Puzzles without solutions Orgasmic Troll Puzzles 12 2003-07-16 09:36

All times are UTC. The time now is 14:54.

Thu Nov 26 14:54:36 UTC 2020 up 77 days, 12:05, 5 users, load averages: 1.22, 1.17, 1.26

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.