What is the Expansion of X^n + Y^n When n is Even?

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SUMMARY

The expansion of \(x^n + y^n\) when \(n\) is even cannot be factored over the reals, except in the trivial case where \(x\) and \(y\) are both zero. Unlike the case for odd \(n\), where the expression can be factored as \((x+y)(x^{n-1}-x^{n-2}y+\ldots+y^{n-1})\), the even case requires the use of complex numbers for factorization. Specifically, \(x^n + y^n\) can be expressed using complex roots, leading to the formulation \(x^n + y^n = \prod_k (x - \exp(i\pi k/n)y)\).

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What is the expansion of xn +yn , when is even ??/
 
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I don't see anything that can be expanded.
 
elibj123 said:
I don't see anything that can be expanded.

i mean that xn - yn can be written as (x-y)(xn-1 +xn-2y ...+yn-1 )
similarly what can xn +yn be written as ?
 
Try alternating signs, and it becomes straightforward.
 
I think you need to think about zeros
x^n+y^n=0
x^n=-y^n
x=y\cdot\exp(i\pi k/n)
\therefore x^n+y^n=\prod_k (x-\exp(i\pi k/n)y)

Occationally combining a subset of these factors together will give you a real solution.

Now you need to think when... :)
 
Last edited:
phymatter said:
What is the expansion of xn +yn , when is even ??/

phymatter said:
i mean that xn - yn can be written as (x-y)(xn-1 +xn-2y ...+yn-1 )
similarly what can xn +yn be written as ?

Then you mean what are the factors :smile:

If n is odd, you can factor it as so:

x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-x^{n-4}y^3+...-xy^{n-2}+y^{n-1})

However, if n is even, then x^n+y^n\neq 0 except for in the trivial case of x,y=0. This means you can't factor it over the reals. You'll need to use complex numbers. You could convert it into a few different ways, such as x^n-i^2y^n and take difference of two squares, or, if you want to follow the same factorizing process as above, take x^n+(iy)^n and take two cases, when [itexi^n[/itex] is equal to 1, and when equal to -1.
 

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