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:$$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}{b}^2-\cdots -a{b}^{2n_1}+b^{2n}\text{ where }n\ge 1$$As an example, consider $x^7+b^7$ where$$x^7+b^7=(x^6-x^{5}y+x^4{y}^2-x^3{y}^3+x^2{y}^4-x{y}^5+y^6)(x+y)$$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:$$x^6+y^6=(x^4-x^2{y}^2+y^4)(x^2+y^2)$$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. $$\begin{array}{rl}(x^{10}+y^{10}& =(x^2{)}^5+(y^2{)}^5\\ & =(x^8-x^6{y}^2+x^4{y}^4-x^2{y}^6+y^8)(x^2+y^2)\end{array}$$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.$$\begin{array}{rl}{x}^{48}+y^{48}& =(x^{16}{)}^3+(y^{16}{)}^3\\ & =(x^{32}-x^{16}{y}^{16}+y^{32})(x^{16}+y^{16})\end{array}$$DIFFERENCE OF ODD AND EVEN POWERS

In the case of $x^n-y^n$, the factorisation is:$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots +x{y}^{n-2}+y^{n-1}\text{ for }n\ge 1$$If $n$ is odd, then $x-y$ is a factor and if $n$ is even, then $x-y$ and $x+y$ are factors.$$\begin{array}{rl}{x}^7-y^7& =(x^6+x^{5}y+x^4{y}^2+x^3{y}^3+x^2{y}^4+x{y}^5+y^6)(x-y)\\ x^8-y^8& =(x^4+y^4)(x^2+y^2)(x+y)(x-y)\end{array}$$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.