20130810, 14:40  #1 
"Jonathan"
Jul 2010
In a tangled web...
2·107 Posts 
Bicurvious Identity
Does anyone have a reference to the following identity?
(a^3  b^3)/(a  b) = ((a + b)^3 + b^3)/(a + 2b) So far, the few mathematicians I know are not familiar with it. Jonathan 
20130810, 16:13  #2 
"Jane Sullivan"
Jan 2011
Beckenham, UK
2^{4}×3×5 Posts 
What do you mean "have a reference to"?
Simple long division gives (a^3  b^3)/(a  b) = a^2 + ab + b^2. Multiplying this by a + 2b gives a^3 + 3(a^2)b + 3ab^2 + 2b^3 = (a+b)^3 + b^3. So your identity can be verified by direct calculation. 
20130810, 17:52  #3  
"Jonathan"
Jul 2010
In a tangled web...
2·107 Posts 
Quote:
From my perspective, where one is always on the lookout for factorization shortcuts, this would allow (to take a really simple example) to apply the Aurifeuillian L,M factors of 12^3 + 1 (7L,19M) to the primitive portion of 11^3  1. As far I know, no one has mentioned this previously. Bases which are perfect powers can also be used in this fashion. Cheers, Jonathan 

20130810, 20:23  #4  
Nov 2003
2^{2}·5·373 Posts 
You don't know enough math. You are not entitled to even have a
perspective, as evidenced by the following piece of nonsense: Quote:
As for the "has anyone else noticed..." this is like asking a baker if he ever put his hands in a bag of flour. 

20130810, 21:52  #5 
"Jonathan"
Jul 2010
In a tangled web...
2×107 Posts 
Sorry, but I'm of the opinion, that number factorization should be accessible to all, regardless of educational background or acquired knowledge. This is I believe the philosophy behind GIMPS where anyone can install the client and derive satisfaction from any results or from just knowing they are contributing to some greater project. If my question reveals that I'm a member of the 99.99% of the world population that didn't know the answer, then maybe you could just say the answer, nicely?
The great thing I appreciate (from my perspective) about mathematics is you're absolutely right or absolutely wrong. I still see an Aurifeuillian factor freebie to 11^3  1. So, that's an emphatic yes. And therefore we really can get free factors which would seem to contradict your statement that I was talking nonsense. 
20130810, 22:06  #6  
Nov 2003
2^{2}×5×373 Posts 
Quote:
almost total ignorance. Quote:
Every time you open your mouth you look more foolish. Especially foolish is using terminology and concepts that YOU DO NOT UNDERSTAND. You clearly fail to understand Aurifeullian identities. You are a clear example of Dunning & Kruger in action. 

20130810, 22:07  #7 
If I May
"Chris Halsall"
Sep 2002
Barbados
2×11×431 Posts 

20130810, 22:38  #8 
"Jonathan"
Jul 2010
In a tangled web...
2·107 Posts 
OK, I'm starting to think there is a disconnect going on here.
Let us pick a = 11 and b = 1 LHS = (a^3  b^3)/(a  b) = (11^3  1^3)/(11  1) = (1331  1)/10 = 133 RHS = ((11 + 1)^3 + 1^3)/(11 + 2) = (12^3 + 1)/(12+1) = 133 Now, 12,3L = 7 12,3M = 19 I predict we can transfer for free the Aurifeuillian factors 7 & 19 to the remaining unfactored composite of 11^3  1 which is 133. Last fiddled with by jcrombie on 20130810 at 22:39 Reason: typo 
20130810, 23:35  #9 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}·17·61 Posts 
You are only digging yourself deeper, man.
The simple identity that you started with, in post #1, is true  but trivial. The followup in post #3 has no generality. Take a pencil and paper and repeat your exercise for the factorization of 12^{9} +1 : (12^9+1)/(12^3+1) = 1657 · 1801 = "12,9L" · "12,9M" What can you factor using this? Hint: 1727^31, not 11^91... But if you were given 1727^31 to factor, you could do the same factorization by observing what 1727 was, i.e. 12^31. Now, what is the use for a known, say, 12,201L · 12,201M or for 12,669L · 12,669M factorization? The only form you arrive to every time is a^2+ab+b^2, so you learn nothing new. 
20130810, 23:52  #10 
May 2013
East. Always East.
6BF_{16} Posts 
R. D. Silverman: If you're ever on the lookout for a soulmate, I can refer you to Davieddy. You'll be able to yell at kids on your lawn together.

20130811, 00:01  #11 
"Jonathan"
Jul 2010
In a tangled web...
2·107 Posts 
This where my poor communications skills are making themselves manifest.
@Batalov  this definitely does not generalize to higher powers and is not about Aurifeuillian factors per se. Let's look at the well known identities for cubic binomials: a^3 + b^3 = (a + b)(a^2  ab + b^2) a^3  b^3 = ( a  b)(a^2 + ab + b^2) Let's call the "a^2  ab + b^2" and "a^2 + ab + b^2" parts the primitive parts. Denote that with prim(). The identity shows that prim( a^3  b^3 ) = prim( (a+b)^3 + b^3 ) 
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