Why can't we factor $a^n + b^n$ for $n$ even in same way as $a^n + b^n$ for $n$ odd?
For $n$ odd we have $a^n + b^n = (a + b)(a^{n - 1} - a^{n-2}b + \ldots - ab^{n - 2} + b^{n - 1})$. Not sure why we can't do the same for $a^n + b^n$ when $n$ is even.
If $n$ is even, the second factor no longer has a $b^{n-1}$ term; it’s
$$a^{n-1}-a^{n-2}b+\ldots\color{red}+ab^{n-2}\color{red}-b^{n-1}\,,$$
and when you multiply that by $a+b$ you get $a^n-b^n$, not $a^n+b^n$.
You can often factor $a^n+b^n$ even when $n$ is even if $n$ has a odd factor.But we can't do it the same way for $n$ odd case.
It is because $a+b$ doesn't divide $a^n+b^n$ in general for even $n$.(use $a\equiv -b \pmod{a+b}$ )
You could do it by using complex numbers. For instance $$ a^2 + b^2 = a^2 - (ib)^2 = (a +ib)(a-ib) $$ and more generally, $$ a^n + b^n = (a-z_1b)(a-z_2b) \dotsm (a-z_nb) $$ where $z_1, \ldots, z_n$ are the complex roots of the equation $z^n = -1$.
