Factoring:Show that (x-y) is a factor of x^n-y^n

[nishantp]

Before I get flamed, yes this is a homework question, although its not being marked as I'm learning this course on the side, and will take it in september.

The questions is:

Show that (x - y) is a factor of xⁿ - yⁿ. I don't have a clue how...

Thanks

[Guv]

(x - y)*(x^n-1 + x^n-2*y + . . . + y^n-1) = xⁿ - yⁿ

For 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 - y⁴

Sorry if there are any typo's in the above.

[nishantp]

Thanks

[bugzpodder]

what Guv said was true, but it does not prove that x-y is a factor of x^n-y^n

you could use the factor theorem: if x-a is a factor of polynomial f(x), then f(a)=0

f(x):=x^n-y^n

and we see that f(y)=0 ie x-y is a factor of f(x)=x^n-y^n

[Guv]

BugzPodder: Very cute.

I was considering posting multiplied out versions of the following.

• x*(x^n-1 + x^n-2*y + . . . + y^n-1)
  
  -y*(x^n-1 + x^n-2*y + . . . + y^n-1)

It is easy to show that every thing cancels out but the following.

• xⁿ - yⁿ

Isn't that a lot more work and not so elegant as your post?

[sql_lall]

mods work nicely too:

x == y (mod [x-y]) ->easy to see, just take y from each side

=> xⁿ - yⁿ (mod [x-y])
== yⁿ - yⁿ (mod [x-y])
== 0 (mod [x-y])

=> [x-y] is a factor of xⁿ - yⁿ

P.S, i know that mods are only supposed to be for integers, but all of the logic used here works for reals too.

[DiGiTaIErRoR]

Originally posted by sql_lall
mods work nicely too:

x == y (mod [x-y]) ->easy to see, just take y from each side

=> xⁿ - yⁿ (mod [x-y])
== yⁿ - yⁿ (mod [x-y])
== 0 (mod [x-y])

=> [x-y] is a factor of xⁿ - yⁿ

P.S, i know that mods are only supposed to be for integers, but all of the logic used here works for reals too.

You and mods.

[sql_lall]

but of course...

now that i think of it, because we are talking about factors, i guess that implies integers...

oh, and possibly you could use induction:
let p(n) = xⁿ - yⁿ

(x-y) is a factor of p(1) [as p(1) IS (x-y)]

p(k)-p(k-1)
= xⁿ - yⁿ - x^n-1 + y^n-1
= (x-y) (something else)
now, all u gotta do is find the "something else"