Let $p(x,y)$, $c(x,y)$ and $d(x,y)$ be two variable polynomials with integer coefficients which satisfy $p(x,y)=c(x,y)\cdot d(x,y)$.
Given $m, n$ positive prime numbers and given $e(x,y)$ another polynomial which grade is lower than $p(x,y)$'s grade. If $e(x,y)\not=c(x,y)$ and $e(x,y)\not=d(x,y)$.
I know it's clear that $\frac{p(m,n)}{c(m,n)}=d(m,n)$ and $\frac{p(m,n)}{d(m,n)}=c(m,n)$, both, represent a integer number for any $m$ and $n$ primes.
But, is possible to have $\frac{p(m,n)}{e(m,n)}$ as an integer number?
The question is focused on a evaluative method applied to polynomial with given possible factors, to determine which one is the correct factor. For example:
One factor of $p(x,y)$ corresponds to:
- $c(x,y)$
- $e(x,y)$
- $c(x,y)+e(x,y)$
- $d(x,y)+e(x,y)$
Using the primes $m$ and $n$, we get that the only answer which results in an integer is 1. Because, 2., 3. and 4. results on a non integer number. It is possible to get 2. as another integer, but changing the prime numbers for a new evaluation, let say $m'$ and $n'$, and they gives the result as before, alternatives 1. and 2. with integer results. Can this always happen for many pairs of primes even if $e(x,y)\not\mid p(x,y)$?