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 Register FAQ Search Today's Posts Mark Forums Read 2009-11-10, 17:25 #1 grandpascorpion   Jan 2005 Transdniestr 503 Posts 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   2009-11-10, 19:44   #2
R.D. Silverman

Nov 2003

22×5×373 Posts Quote:
 Originally Posted by grandpascorpion 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.   2009-11-10, 20:44 #3 grandpascorpion   Jan 2005 Transdniestr 50310 Posts 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.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post jcrombie Miscellaneous Math 51 2013-09-09 18:51 only_human Software 2 2012-05-11 13:43 ewmayer Lounge 12 2010-02-04 21:26 SPWorley Math 6 2009-08-28 18:02 fivemack Factoring 4 2008-03-04 05:04

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