Curious polynomial factorization identity

Question

Let $j \in {\mathbb N}$ and $q \in {\mathbb N}$ be positive integers and let $n$ be a parameter. Define: \begin{equation} Q_j:=3 \left\lfloor \frac{j-2}{2}\right\rfloor +2 \left\lfloor \frac{j-1}{2}\right\rfloor +4 \end{equation}

Then the following identity holds: \begin{eqnarray} &&\frac{1}{\left(-j+n+Q_j\right)^{1\cdot 1_{j \bmod 2=0} + 2 \cdot 1_{j \bmod 2=1}}}\\ &&\left(\right.\\ &&\left. \frac{(5 j+2 q-3)^{(3)}}{Q_j+q} \right.\\ &&\left. \frac{6 (5 j+2 q-1)^2 }{Q_j+q+1}\cdot (-j+n-q)^{(1)}+\right.\\ &&\left.12 \frac{(5 j+2 q) }{Q_j+q+2} \cdot (-j+n-q-1)^{(2)} +\right.\\ &&\left. \frac{8 }{Q_j+q+3} \cdot (-j+n-q-2)^{(3)}\right.\\ &&\left. \right)=\\ && \begin{array}{r} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!-4 (15 j+4 q-4)+8 Q_j+\frac{6 \left(25 j^2+20 j q-10 j+4 q^2-4 q+1\right)}{Q_j+q+1}+ \left(\frac{60 j+24 q}{Q_j+q+2}-8\right) \cdot (-j+n-q)+ \frac{8 }{Q_j+q+3} \cdot (-j+n-q-1)^{(2)} & \mbox{if $j$ is even} \\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\frac{60 j+24 q}{Q_j+q+2}-16+\frac{8 }{Q_j+q+3}\cdot (-j+n-q) & \mbox{if $j$ is odd} \end{array} \end{eqnarray} Note that the left hand side of the identity above is a rational function in $n$ and the right hand side is a polynomial in $n$. As such the identity states that the numerator of the left hand side is always divisible by the denominator whatever values for $j$ and $q$ we choose. We discovered this identity by inserting random values for $j$ and $q$ into the lhs and then noticing that the result is always a polynomial (meaning no fraction) in $n$.

In[360]:= Clear[Q]; 
Q[j_] := 4 + 3 Floor[1/2 (-2 + j)] + 2 Floor[1/2 (-1 + j)];
j = RandomInteger[{0, 50}];
q = RandomInteger[{0, 50}]; n =.;
eX = 1/Pochhammer[n - j + Q[j] + 1, 
     If[Mod[j, 2] == 0, 1, 
      2]] (Pochhammer[2 q + 5 j - 3, 3] 1/(q + Q[j]) (1) +
      6 (-1 + 5 j + 2 q)^2 1/(1 + q + Q[j]) (n + -j - q - 0) +
      (60 j + 24 q) 1/(2 + q + Q[j]) (n + -j - q - 1) (n - j - q) +
      8 1/(
       3 + q + Q[j]) (n + -j - q - 2) (n - j - q - 1) (n - j - 
         q)) - (If[
     Mod[j, 2] == 
      0, (-4 (-4 + 15 j + 4 q) + 8 Q[j] + (
        6 (1 - 10 j + 25 j^2 - 4 q + 20 j q + 4 q^2))/(
        1 + q + Q[j])) + (-8 + (60 j + 24 q)/(2 + q + Q[j])) (n - j - 
         q) +  8/(3 + q + Q[j]) (n - j - q) (n - j - q - 1), -16 + (
      60 j + 24 q)/(2 + q + Q[j]) + 8/(3 + q + Q[j]) (n - j - q)]);
Simplify[eX]

Out[364]= 0

How would you go about proving this identity? Is there any combinatorial interpretation of this result ?

Answer 1

Using Mathematica for odd and even $j$ separately it is possible to reduce the expression to the following more readable form: $$ \begin{cases} 4\left(\frac{(3j+2n+2)(3 j+2 n+4)}{5 j+2 q+4}-\frac{6(3j+2n+2)}{5 j+2 q+2}+\frac{3}{5 j+2 q}\right),& j\bmod2=0;\\ 8\left(\frac{3j+2n+3}{5 j+2 q+3}-\frac{3}{5 j+2 q+1}\right),& j\bmod2=1. \end{cases} $$

Answer 2

Rewrite $$Q_j = 3 \left\lfloor \frac{j-2}{2}\right\rfloor +2 \left\lfloor \frac{j-1}{2}\right\rfloor +4 = %3 floor(j/2 - 1) +2 floor(j/2 - 1/2) +4 = \begin{cases} \frac52j - 1 & \textrm{ if $j$ is even} \\ \frac52j - \frac32 & \textrm{ if $j$ is odd} \end{cases}$$

Then separate into two identities, one for even $j$ and one for odd $j$. The algebra is ugly but algorithmic, so you can ask Mathematica to simplify it.