Let $R(a,b) = c \times a^q + d\times b^p$ be a two variables polynomial in $\mathbb{Z}[a,b]$, that is we have $c,d \in \mathbb{Z}^2$ and $q,p \in \mathbb{N}^2$.
For which $c,d,p$ and $q$ can we tell if there exists two polynomials $Q$ and $P$ in $\mathbb{Z}[a,b]$ with each having a degree being greater or equal to $1$ such that $R(a,b) = Q(a,b)P(a,b)$?
This question is motivated by the fact that we have such formulas for:
$a^{n+1} - b^{n+1} = (a-b)\sum_{k=0}^{n}{a^kb^{n-k}} \\ a^{n+1} + b^{n+1} = (a+b)\sum_{k=0}^n{(-1)^k a^k b^{n-k}} \text{ with $n+1$ odd} \\ a^4 + 4b^4 = (a^2 - 2ab + 2b^2)(a^2 + 2ab + 2b^2)$
And moreover combinations of these ones. It's clear as an example that if a coefficient $c$ is it self a $n$-th power of an integer, we have $a^n - cb^n$ that is then factorizable. It's also clear that if $q$ and $p$ have an odd common divisor $k$, i.e $q = kq'$ and $p = kp'$ then we can also factorize this using formula for $a^k + b^k$. We even have $a^{n+1} + b^{n+1} = a^{n+1} - (-b)^{n+1}$ for $n$ odd, the factorization can be derived from the formula of $a^n - b^n$.
Questions directly linked:
- Is there any other factorization possible for $R$ with some setting of its parameters, which does not correspond to the above ones and theirs combinations?
- If not, can we prove it? Or giving proofs for specific polynomials? (proving such a one is not factorizable)
- If $R$ is in $\mathbb{Z}[a,b]$ but we allow $P$ and $Q$ to be on other rings, for example allowing radicals or algebraic coefficients, can we get new factorizations?
- Same question for when $R$ is on $\mathbb{R}[a,b]$, i.e $c$ and $d$ can be real.
This seems like a question that a lot could've asked, furthermore being something known. I'm then sorry if this has already been answered, I didn't manage to find it here.