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Old 2013-08-10, 14:40   #1
jcrombie
 
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"Jonathan"
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Default Bi-curvious 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
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Old 2013-08-10, 16:13   #2
BudgieJane
 
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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.
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Old 2013-08-10, 17:52   #3
jcrombie
 
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Quote:
Originally Posted by BudgieJane View Post
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.
Yes, it is quite easy to verify.

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
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Old 2013-08-10, 20:23   #4
R.D. Silverman
 
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Quote:
Originally Posted by jcrombie View Post
Yes, it is quite easy to verify.

From my perspective,
You don't know enough math. You are not entitled to even have a
perspective, as evidenced by the following piece of nonsense:

Quote:
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.
This prior bit is total nonsense.

As for the "has anyone else noticed..." this is like asking a baker
if he ever put his hands in a bag of flour.
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Old 2013-08-10, 21:52   #5
jcrombie
 
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Quote:
Originally Posted by R.D. Silverman View Post
You don't know enough math.
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?

Quote:
Originally Posted by R.D. Silverman View Post
This prior bit is total nonsense.
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.

Quote:
Originally Posted by R.D. Silverman View Post
As for the "has anyone else noticed..." this is like asking a baker
if he ever put his hands in a bag of flour.
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.
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Old 2013-08-10, 22:06   #6
R.D. Silverman
 
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Quote:
Originally Posted by jcrombie View Post
Sorry, but I'm of the opinion, that number factorization should be accessible to all,
Knowledge of math requires study. And your "opinion" comes from
almost total ignorance.

Quote:
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.
This is nonsense. 11^3-1 DOES NOT HAVE an Aurifeullian factorization.
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.
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Old 2013-08-10, 22:07   #7
chalsall
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Quote:
Originally Posted by R.D. Silverman View Post
You are not entitled to even have a perspective...
Mr. Silverman... With all due respect, you need to learn how to manage humans.
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Old 2013-08-10, 22:38   #8
jcrombie
 
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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 2013-08-10 at 22:39 Reason: typo
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Old 2013-08-10, 23:35   #9
Batalov
 
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You are only digging yourself deeper, man.
The simple identity that you started with, in post #1, is true -- but trivial.

The follow-up in post #3 has no generality.
Take a pencil and paper and repeat your exercise for the factorization of 129 +1 :
(12^9+1)/(12^3+1) = 1657 · 1801 = "12,9L" · "12,9M"
What can you factor using this? Hint: 1727^3-1, not 11^9-1...
But if you were given 1727^3-1 to factor, you could do the same factorization by observing what 1727 was, i.e. 12^3-1.

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.
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Old 2013-08-10, 23:52   #10
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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.
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Old 2013-08-11, 00:01   #11
jcrombie
 
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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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