Let $[n]\_q=1+q+\dots +q^{n-1}$, $ {[n]\_q}! =[1]\_q [2]\_q \dots [n]\_q$ and $\binom{n}{j}\_q = \frac{[n]\_q!}{[j]\_q![n-j]\_q!}$ be the usual $q$-notation.


Consider the polynomials $p\_n(q,r,x)= \sum\_{j=0}^n q^{r\binom{j}{2}}\binom{n}{j}\_qx^j$ and let $d\_{n,r}(q)=\Delta\_{x} p\_n(q,r,x)$ be their [discriminants](https://en.wikipedia.org/wiki/Discriminant).


The well-known identity $p\_n(q,1,x) =(1+x)(1+qx)\dots(1+q^{n-1}x)$ implies that
$$d\_{n,1}(q)=\Delta\_{x} p\_n(q,1,x)=(1-q)^{2\binom{n}{2}}q^{2\binom{n}{3}}\Big(\prod\_{j=0}^{n-1}[j]\_q!\Big)^2.$$
This gives $ d\_{n,1}(q)=q^{n(n-1)^2} d\_{n,1}(q^{-1}).$
Computations suggest that more generally the following holds for integers $r$:
$$ d\_{n,1+r}(q)=q^{n(n-1)^2} d\_{n,1-r}(q^{-1}).$$



> 
> **QUESTION.** Is this a known result? If yes, where can I find it, if no, any idea how to prove it?
> 
> 
> 



Discriminants of some $q$-analogs of $(1+x)^n$
This is true.


We have
\begin{align\*} 
p\_n (q^{-1}, 1-r, x) &= \sum\_{j=0}^n q^{ (r-1) \binom{j}{2}} \binom{n}{j}\_{q^{-1}} x^j \\
&= \sum\_{j=0}^n q^{ (r-1) \binom{j}{2}} q^{-j (n-j)} \binom{n}{j}\_{q} x^j \\ 
&=\sum\_{j=0}^n q^{ (r+1) \binom{j}{2}} q^{-j (n-j) - j (j-1)} \binom{n}{j}\_{q} x^j \\
&= \sum\_{j=0}^n q^{ (r+1) \binom{j}{2}} \binom{n}{j}\_{q} (x q^{-(n-1)} )^j \\
&= p\_n(q, 1+r, x q^{-(n-1)}).
\end{align\*}


and changing variables $x \to \lambda x$ in a polynomial multiplies the discriminant by $\lambda^{ n (n-1)}$.


