Reply v Quote
Frenzied Member

(x - y)*(xn-1 + xn-2*y + . . .
You are watching: X^n-y^n
+ yn-1) = xn - ynFor example, the following.(x - y)*(x + y) = x2 - y2(x - y)*(x2 + x*y + y2) = x3 - y3(x - y)*(x3 + x2*y + x*y2 + y3) = x4 - y4Sorry if there are any kind of typo"s in the above.
Live lengthy & prosper.The Dinosaur from primitive era prior to computers. Eschew obfuscation!If a billion people believe a foolish idea, the is still a foolish idea!VB.net 2010 Express64Bit & 32Bit home windows 7 & home windows XP. I operation 4 operating systems on a solitary PC.
Reply through Quote
what Guv claimed was true, but it does not prove that x-y is a aspect of x^n-y^nyou could use the aspect theorem: if x-a is a factor of polynomial f(x), then f(a)=0f(x):=x^n-y^nand we view that f(y)=0 in other words x-y is a aspect of f(x)=x^n-y^n
Massey RuleZ! ^-^__Cheers!__^-^ Massey RuleZ!Did you understand that...The probability the a arbitrarily rational number has actually an also denominator is 1/3 (Salamin and Gosper 1972)? This an outcome is separately verified by me (2002)!
BugzPodder: really cute.I was considering posting multiplied out versions the the following.x*(xn-1 + xn-2*y + . . . + yn-1)-y*(xn-1 + xn-2*y + . . . + yn-1)It is straightforward to display that every thing cancels out but the following.
See more: Sweeping Second Hand Watch With Sweep Second Hand Watch? You Asked Us: Which Watches Sweep
mods work nicely too:x == y (mod
Originally post by sql_lall mods work-related nicely too:x == y (mod
but the course...now that i think the it, since we are talking about factors, ns guess that suggests integers...oh, and possibly you could use induction:let p(n) = xn - yn(x-y) is a aspect of p(1)
Quick NavigationMaths ForumTopSite AreasForumsVisual BasicVisual simple .NETVisual basic 6 and also EarlierGames and Graphics Programming.NET and MoreGeneraltastecraftedmcd.com Coding ContestsOther Languagestastecraftedmcd.com CodeBanktastecraftedmcd.com UtilityBankProjectsJobsCommunityGeneral conversation / Chit Chat