Before I gain flamed, correct this is a homework question, although its not being significant as I"m learning this food on the side, and will take it it in september.The inquiries is:Show the (x - y) is a factor of xn - yn. Ns don"t have actually a proviso how...Thanks

Frenzied Member Join date Jul 1999Location Huntingdon Valley, PA 19006Posts 1,151
(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.
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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
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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.

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mods work nicely too:x == y (mod ) ->easy to see, simply take y from every side=> xn - yn (mod )== yn - yn (mod )== 0 (mod )=> is a aspect of xn - yn P.S, i know that mods are only supposed to be for integers, but every one of the logic used here works for reals too.
Originally post by sql_lall mods work-related nicely too:x == y (mod ) ->easy to see, just take y from each side=> xn - yn (mod )== yn - yn (mod )== 0 (mod )=> is a aspect of xn - yn P.S, i understand that mods room only an alleged to be because that integers, but all of the logic used below works because that reals too.

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) p(k)-p(k-1)= xn - yn - xn-1 + yn-1 = (x-y) (something else)now, every u gotta perform is find the "something else"
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