Sum and Differences of Odd and Even Powers

Sometimes you need to go back to basics. I found that I was getting confused about the factorisation of general expressions like $x^n+y^n$. 

SUM OF ODD AND EVEN POWERS

It turns out that if $n$ is odd, then factorisation is possible according to the rule:

$$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2- \dots - ab^{2n_1}+b^{2n} \text{ where }n \geq 1$$ As an example, consider $x^7+b^7$ where $${x^7+b^7=\left(x^{6} - x^{5} y + x^{4} y^{2} - x^{3} y^{3} + x^{2} y^{4} - x y^{5} + y^{6}\right)} {\left(x + y\right)}$$ I've seen it stated that $x^n+y^n$ cannot be factored if $n$ is even with $x^2+y^2$ and $x^4+y^4$ cited as examples. See Figure 1.

Figure 1: source

This is certainly true for $n=2$ and $n=4$ and in fact in any power of 2 (8, 16, 32 etc.) but otherwise any composite number will contain an odd factor e.g. 6 = 2 x 3 and thus we can write $x^6+y^6$ as $(x^2)^3+(y^2)^3$ and it is a sum of odd powers! In fact, it is a sum of two cubes:

$$x^6+y^6={\left(x^{4} - x^{2} y^{2} + y^{4}\right)} {\left(x^{2} + y^{2}\right)}$$ Now it might be thought that this is a special case, since any $n$ that is a power of three can be written as a sum of two cubes. So let's consider $x^{10}+y^{10}$. We jump over $x^8+y^8$ because we know that it can't be factored.  $$\ \begin{align} ( x^{10}+y^{10}&=(x^2)^5+(y^2)^5\\ &= {\left(x^{8} - x^{6} y^{2} + x^{4} y^{4} - x^{2} y^{6} + y^{8}\right)} {\left(x^{2} + y^{2}\right)} \end{align}$$ In general, whatever $m$ divides the even index to produce an odd number, then $x^m+y^m$ will appear as a factor (this $m$ must be a multiple of 2). For example, in the case of $x^{48}+y^{48}$, $x^{16}+y^{16}$ must be a factor. $$\begin{align} x^{48}+y^{48} &= (x^{16})^3+(y^{16})^3\\&={\left(x^{32} - x^{16} y^{16} + y^{32}\right)} {\left(x^{16} + y^{16}\right)} \end{align}$$ DIFFERENCE OF ODD AND EVEN POWERS

In the case of $x^n-y^n$, the factorisation is:

$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+ \dots + xy^{n-2} + y^{n-1} \text{ for }n\geq 1$$ If $n$ is odd, then $x-y$ is a factor and if $n$ is even, then $x-y$ and $x+y$ are factors. $$\begin{align} x^7-y^7&={\left(x^{6} + x^{5} y + x^{4} y^{2} + x^{3} y^{3} + x^{2} y^{4} + x y^{5} + y^{6}\right)} {\left(x - y\right)}\\x^8-y^8&={\left(x^{4} + y^{4}\right)} {\left(x^{2} + y^{2}\right)} {\left(x + y\right)} {\left(x - y\right)} \end{align}$$ IN CONCLUSION

• $x^n- y^n$ can always be factored
• $x^n+y^n$ can be factored provided $n$ is not a power of 2.