Expansion of $x^n-y^n$

Question

Studying polynomials I couldn't find a way to expand $x^n-y^n$ as a product of other polynomials. Now of course we know that $$x^4-y^4=(x^2+y^2)(x^2-y^2)=(x^2+y^2)(x+y)(x-y)$$ and I came up with this:

Given $x^n-y^n$, if $n=2^k$ ($n$ and $k$ are natural numbers) then $$x^n-y^n=(x-y) \prod_{j=0}^{k-1}(x^{2^j}+y^{2^j})$$ However I couldn't came up with the expansion when $n$ is odd.

Is the formula correct? Is there a formula for $n$ odd?

Answer 1

Notice that you can write your polynomial as $x^n((y/x)^n-1)$, so we can look for factorizations of $t^n-1$ and then substitute.

Over the integers or the rational numbers, we have a complete answer. They are certain polynomials $\Phi_k$, called cyclotomic polynomials, one for eachtive integer $k$. There are irreducible and have the property that

$$t^n-1=\prod_{\text{$d$ a divisor of $n$}}\Phi_d(t).$$

It follows from this that

$$y^n-x^n=x^n\prod_{\text{$d$ a divisor of $n$}}\Phi_d(y/x).$$

The degree of $\Phi_d$ is $\phi(d)$, the number of positive integers less than $d$ and coprime with $d$. Since

$$\sum_{\text{$d$ a divisor of $n$}}\phi(d)=n,$$

we can rewrite the above product as

$$y^n-x^n=\prod_{\text{$d$ a divisor of $n$}}x^{\phi(d)}\Phi_d(y/x)$$

and now each factor is a polynomial with integer coefficients. One can check that these factors are irreducible in $\mathbb Z[x,y]$. Sadly, there is no simple description of these cyclotomic polynomials: they are what they are.

If instead you are interested infactorizations over the complex numbers, we can proceed as follows: we know that

$$t^n-1=\prod_{k=0}^{n-1}(t-e^{2\pi i k/n})$$

so that

$$y^n-x^n=x^n\prod_{k=0}^{n-1}(y/x)-e^{2\pi i k/n}=\prod_{k=0}^{n-1}(y-e^{2\pi i k/n}x)$$

and each factor in this last product is irreducible in $\mathbb C[x,y]$.

Over the real numbers there is a similar factorization, with roughly half the number of terms, which someone motivated enough can write down easily.

Answer 2

This may be of some help:

If $n$ is odd, $$(x^n\pm a^n)\div(x\pm a)=\sum_{k=1}^n(\mp a)^{k-1}x^{n-k}.$$ If $n$ is even, $$(x^n-a^n)\div(x\pm a)=\sum_{k=1}^n (\mp a)^{k-1}x^{n-k}.$$ If $n$ is even, $$f(x)=x^n+a^n\text{ has no real roots.}$$

Answer 3

Note:

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

But we can actually go deeper than that.

Suppose we know a factor n is t. Then it follows that:

$$x^n-y^n = (x^t-y^t)(y^{n-t} + x^ty^{n-2t} +... x^{n-2t}y^t +x^{n-t}$$

So for a given n, we find the largest non trivial factor of it. And then decompose, and recursively split up every difference we encounter along the way.

But... If you wanted even higher degrees of factorization. I claim that over the complex numbers every such polynomial can be factored into a sequence of monomials. Namely (x-uy). Where the u are each of the possible nth roots of unity.