Let $f(x,y) := (x+1)\cdots (x+j)-(y+1)\cdots (y+k) \in \mathbb{F}\_{q}$, where $j>k$. Do you know if there is a **quick** way to show that no polynomial of the form $ax+by+c$, with $(a,b)\in \mathbb{F}\_{q} \times \mathbb{F}\_{q}\setminus\{(0,0)\}$, can divide $f(x,y)$?


I originally posted this question here: <https://math.stackexchange.com/questions/1975446/non-divisible-by-a-linear-factor> However, I consider that MO is a more suitable venue for it.


Thanks for your replies, comments, web-links, bibliographical recommendations, etc.



How to prove non-divisibility by a linear factor
Denote $p=ax+by+c$. If $b\ne 0$, we get $y\equiv -(c+ax)/b$ modulo $p$, substituting this into expression for $f(x,y)$ we get some non-zero polynomial in $x$ (it is non-zero as a difference of two polynomials of different degrees), which can not be divisible by $p$. If $b=0,a\ne 0$, substitute $x\equiv -c/a$ to get non-zero polynomial in $y$. 


