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.