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Old 2009-11-10, 17:25   #1
grandpascorpion
 
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Default Useful Identity?

I was fiddling around with WolframAlpha the other day and came across this identity:

(x^(n+3)+x^n + x + 1) = (x+1)*(x^(n+2)-x^(n+1)+x^n+1)

I hadn't seen it before and I was wondering if it's already known/in use for factoring larger numbers.

This can be rearranged as (x^3+1)x^n+x+1

So, a more general identity could be derived from

(x^(2m+1) + 1)x^n+ax+a
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Old 2009-11-10, 19:44   #2
R.D. Silverman
 
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Quote:
Originally Posted by grandpascorpion View Post
I was fiddling around with WolframAlpha the other day and came across this identity:

(x^(n+3)+x^n + x + 1) = (x+1)*(x^(n+2)-x^(n+1)+x^n+1)

I hadn't seen it before and I was wondering if it's already known/in use for factoring larger numbers.

This can be rearranged as (x^3+1)x^n+x+1

So, a more general identity could be derived from

(x^(2m+1) + 1)x^n+ax+a

It is trivial stuff from 1st/2nd year algebra. x^(n+3) + x^n =
x^n(x^3+1). x^3+1 is trivially divisible by x+1.

It is already known in the sense that it is so trivial that no mathematician
would ever write it down or try to claim it as a "discovery". It is like
the fact that 1+1=2.
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Old 2009-11-10, 20:44   #3
grandpascorpion
 
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Fair enough, it's trivial, too trivial to label it an identity even. I was just wondering if it could be useful like say the Aurifeuillian factorizations are.
Specifically, if it was useful in factoring a large number relatively near a perfect power.

And, obviously, x^3 + 1 is a trivial bit of algebra. That's why I pointed out a general version of this IN THE NEXT LINE.
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