# How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?

By Cauchy identity, $${}_1\phi_0(a;—;q,z)=\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_{\infty}}{(z;q)_\infty},\quad|z|<1,|q|<1,$$ we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=(1-z)^{-a-b},$ $${}_1\phi_0(a;—;q,z){}_1\phi_0(b;—;q,az)={}_1\phi_0(ab;—;q,z),$$ which is $$\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n\sum_{m\geq0}\frac{(b;q)_m}{(q;q)_m}(az)^m=\sum_{n\geq0}\frac{(ab;q)_n}{(q;q)_n}z^n.$$

Comparing the coefficients of $z$ in the both side of equation above, the identity in the title can be established as $$\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}.$$

However, what I really concern is that there is any possible for us to prove this identity directly without utilizing Cauchy identity. Maybe we can use mathematical induction on $n$ or some other combinatorial or algebraic proofs?

## 2 Answers

Here's a proof that doesn't involve Cauchy's identity. I will begin by stating two elementary lemmas.

Lemma 1: Let $f \in \mathbb{C}[x,y,z]$ be a polynomial. If $f$ vanishes on the set $$\{ (a,a^i,a^j) : a \in \mathbb{C} , i,j \in \mathbb{N}_{>0} \},$$ then $f$ is identically 0.

Lemma 2: Let $\binom{n}{k}_q :=\frac{(q;q)_n}{(q;q)_k (q;q)_{n-k}}$ denote the $q$-binomial coefficients. Then the following Vandermonde-like identity holds: $$\sum_{i} \binom{i+A-1}{i}_q \binom{n-i+B-1}{n-i}_q q^{A(n-i)} = \binom{n+A+B-1}{n}_q.$$

Given these two lemmas, we proceed as follows: Let $$S(q,a,b) = (q;q)_n \sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i} - (ab;q)_n.$$ Your identity is equivalent to $S$ vanishing identically. Since $S$ is polynomial in its variables, by Lemma 1 it suffices to show that $$(*) S(q,q^A, q^B) = 0$$ for all $q,A,B$. Since $(q^A;q)_i =\frac{(q;q)_{i+A-1}}{(q;q)_{A-1}}$, $(*)$ is equivalent to $$\sum_{i=0}^{n} \frac{(q;q)_{i+A-1}}{(q;q)_{A-1} (q;q)_i} \frac{(q;q)_{n-i+B-1}}{(q;q)_{B-1} (q;q)_{n-i}} q^{A(n-i)} = \frac{(q;q)_{n+A+B-1}}{(q;q)_{A+B-1} (q;q)_{n}},$$ or, in $q$-nomial notation, $$\sum_{i} \binom{i+A-1}{i}_q \binom{n-i+B-1}{n-i}_q q^{A(n-i)} = \binom{n+A+B-1}{n}_q,$$ which is true by Lemma 2.

Lemma 1 may be proved, for instance, using the Combinatorial Nullstellensatz. Lemma 2 is proved just like the q-Vandermonde identity, only that the proof uses the identity $\prod_{k=0}^{n-1} (1-q^k t)^{-1} = \sum_{k} \binom{n+k-1}{k}_q t^k$ instead of $\prod_{k=0}^{n-1} (1+q^k t) = \sum_{k} \binom{n}{k}_q q^{\binom{k}{2}} t^k$.

Below is a simplification of the argument by Ofir Gorodetsky.

At first, we denote $a=1/c$ and multiply both sides by $c^n (q;q)_n$, we get an equivalent identity $$\sum_{i=0}^n \binom{n}i_q(c-1)(c-q)\dots (c-q^{i-1})(1-b)(1-bq)\dots (1-bq^{n-i-1})=\\ =(c-b)(c-qb)\dots (c-q^{n-1}b).$$ Consider both parts as polynomials in $b,c$ of degree at most $n$. It suffices to check that they take the same values on a "triangle" $\{c=q^k,b=q^{-m}:k,m\in \{0,1,2\dots,n\}, k+m\leqslant n\}$ (this is a useful lemma in Combinatorial Nullstellensatz spirit, it may be deduced from CN itself). But this is almost obvious: if $k+m<n$, both parts take zero values (any summand in LHS takes zero value), and if $k+m=n$, there is unique non-zero summand in LHS corresponding to $i=k$, and it is straightforward that it equals RHS.

Note: taking homogeneous (in $b,c$) parts of this identity we get $q$-binomial theorem. Specific summands in LHS are not homogeneous, but the total sum somehow is.

• I see. By plugging negative powers of $q$, instead of positive ones as I did, one gets a much simpler proof. Very nice! – Ofir Gorodetsky Oct 26 '16 at 5:34