Exact factorization of $a^m+b^m$ and $a^m-b^m$

Question

Let $a,b$ be two indeterminates. I read here that $a^m + b^m$ is divisible by $(a+b)$ if $m$ is odd and $a^m - b^m$ is divisible by $(a+b)$ if $m$ is even. I am interested in an exact expression $f(a,b)$ such that $a^m + b^m=(a+b)f(a,b)$ if $m$ is odd and the same for the case of even exponents. Can someone provide such a $f(a,b)$ for both cases? I would be very thankful.

Answer 1

If $m$ is odd, one has $a^m+b^m = (a+b)(a^{m-1}-a^{m-2}b+\dots-ab^{m-2}+b^{m-1})$, where in the last factor, the powers of $a$ and $b$ add to $m-1$ in every term.

If $m$ is even, $m=2l$ for an integer $l$. We get $a^m-b^m=a^{2l}-b^{2l}=(a^l)^2-(b^l)^2=(a^l+b^l)(a^l-b^l)$. If $l$ is even, one can keep repeating this trick on the second factor until one gets the factor $(a^k+b^k)(a^k-b^k)$ where $k$ is odd (since one always gets an odd number after dividing by 2 for a finite amount of times). Then one can use the first argument.

Edit: It is hard to explicitely write down the second factorization for arbitrary even $m$, since one does not know how many times one can divide $m$ by 2 before getting an odd number, but given an explicit $m$, it should be clear how you should find the factorization.

Answer 2

You can perform long division (eg divide $x^m + 1$ for $m$ odd and $x^m - 1$ for $m$ even by $x + 1$, where $x = a/b$) to get $a^m + b^m = (a + b)(a^{m - 1} - a^{m-2} b + \dots + b^{m-1})$ when $m$ is odd and $a^m - b^m = (a + b)(a^{m - 1} - a^{m-2} b + \dots - b^{m - 1})$ when $m$ is even. Once you have the formulas it's even easier to check by multiplying out.