mersenneforum.org Solutions to a^2-ab+b^2 = 3^n
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2017-02-09, 04:08 #1 carpetpool     "Sam" Nov 2016 311 Posts 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
2017-02-09, 06:12   #2
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

22×2,281 Posts

Quote:
 Originally Posted by carpetpool ... except (1, 1) and (-2, 1)
How 'bout (2, -1)?

 2017-02-09, 06:41 #3 LaurV Romulan Interpreter     Jun 2011 Thailand 210368 Posts 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

 Thread Tools

 Similar Threads Thread Thread Starter Forum Replies Last Post EdH Linux 16 2016-03-18 17:20 flouran Math 20 2009-09-08 05:48 Vijay Math 6 2005-04-14 05:19 jtavares Factoring 3 2005-04-11 19:25 Orgasmic Troll Puzzles 12 2003-07-16 09:36

All times are UTC. The time now is 09:01.

Sun Sep 20 09:01:05 UTC 2020 up 10 days, 6:12, 0 users, load averages: 0.84, 1.24, 1.38

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.