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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.

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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 $y$.

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  • $\begingroup$ In the last line, it would have to be x congruent to (-c-by)/a, right... $\endgroup$
    – Jamai-Con
    Commented Oct 19, 2016 at 16:41
  • $\begingroup$ In the last line, it would have to be x congruent to (-c-by)/a, right? $\endgroup$
    – Jamai-Con
    Commented Oct 19, 2016 at 16:43
  • $\begingroup$ But $b=0$ in the case under consideration. $\endgroup$
    – Fedor Petrov
    Commented Oct 19, 2016 at 17:15
  • $\begingroup$ Duh! You are absolutely right. $\endgroup$
    – Jamai-Con
    Commented Oct 19, 2016 at 17:17

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