How do I factorize $a^n-b^n$?

Question

$a^n - b^n$

If $n$ is a negative natural number?

And is the first proof on this website correct?

http://www.matheprisma.uni-wuppertal.de/Module/Ableitung/Beweise1.htm

My math teacher says it only works if $n$ is a positive, natural number.

Thanks in advance!

Answer 1

All we need is that $n$ is an integer, being positive is not necessary.

The easiest way to see that this works for all integers equally well, is to observe that if you have negative exponents $a^{-n}-b^{-n}$, you can simply multiply this term by $a^nb^n$ to instead compute

$$a^nb^n(a^{-n}-b^{-n})=a^nb^na^{-n}-a^nb^nb^{-n}=b^n-a^n.$$

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You can also check this identity directly by going the other way around.

\begin{align} (a-b)&(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}) \\ &=a(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})-b(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})\\ &=(a^n+\color{blue}{a^{n-1}b}+\cdots+a^2b^{n-2}+\color{red}{ab^{n-1}}) -(\color{blue}{a^{n-1}b}+a^{n-2}b^2+\cdots+\color{red}{ab^{n-1}}+b^{n}) \end{align}

If you investigate both terms (the ones in the parenthesis in the last line) carefully, you can see that each term in the left parenthesis appears again in the right one $-$ with two exceptions: the terms $a^n$ and $b^n$ are unique. I colored two of these terms for you. The minus between the terms then makes most of this vanish and it only remains $a^n-b^n$. Note that we have not used any property of $n$ except that it is a whole number.

If you then divide both sides by $a-b$ you obtain

$$\frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}.$$

Answer 2

Consider the Lucas sequence $\, U_n(a,b) = (a^n - b^n) / (a - b) \,$ where one of its many properties is that $\, U_n(a,b) = - (ab)^n U_{-n}(a,b) = -(ab)^{-1} U_{-n}(1/a,1/b). \,$ If $\, n>0 \,$ then $\, U_n \,$ is a polynomial in $\, a,b \,$ and if $\, n<0 \,$ then $\, U_n \,$ is a polynomial in $\, 1/a,1/b. \,$ In both cases, $\, a^n-b^n \,$ factors.