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

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The expansion of x^n + y^n when n is even cannot be factored over the reals, as it equals zero only in the trivial case of x, y = 0. Unlike the odd case, where it can be expressed as a product of factors, the even case requires complex numbers for a proper expansion. By considering x^n + (iy)^n, one can explore different factorization methods based on the values of i^n. The discussion emphasizes the need to utilize complex solutions to understand the behavior of the expression. Ultimately, the even case presents unique challenges that differ significantly from the odd case.
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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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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