Combinatorial interpretation of polynomials I'm reading a proof for the following identity:
$$\sum_{k}A_{k}(r,t)A_{n-k}(s,t) = A_{n}(r+s, t), \qquad \text{integer} \ n \geq 0 \tag{1}$$
where $A_{n}(x,t)$ is the $n$th degree polynomial in $x$ that satisfies
$$A_{n}(x,t) = \binom{x-nt}{n}\frac{x}{x-nt}, \qquad \text{for} \ x \neq nt. \tag{2}$$
The author starts off by assuming that $r \neq kt \neq s$ for $0 \leq k \leq n$, since both sides of (1) are polynomials in $r, s, t$.
Question: I understand that we need $r \neq kt$ to prevent division by zero in $A_{k}(r,t) = \binom{r-kt}{k}\frac{r}{r-kt}$, but why are we also allowed to assume $kt \neq s$ for $0 \leq k \leq n$?
Writing $A_{n-k}(s,t)$ in terms of (2), we have
$$A_{n-k}(s,t) = \binom{s-(n-k)t}{n-k}\frac{s}{s-(n-k)t}$$
If $s=kt$, then
$$
A_{n-k}(kt,t) = \binom{kt-(n-k)t}{n-k}\frac{kt}{kt-(n-k)t}
= \binom{2kt-nt}{n-k}\frac{kt}{2kt-nt}
$$
...which doesn't result in division by $0$ if $k=0$ or if $k=n$, so why is $s \neq kt$ needed?
 A: Since $A_k(r, t)$ appears in $(1)$, in order to use the expression $A_k(r, t)=\binom{r-kt}{n}\frac{r}{r-kt}$, you need to assume $r\neq kt$ for all $k\in \{0,1\dots,n\}$ because this is a restriction given in $(2)$.
Since $A_{n-k}(s, t)$ appears in $(1)$, in order to use the expression $A_{n-k}(s, t)=\binom{s-(n-k)t}{n}\frac{s}{s-(n-k)t}$, you need to assume $s\neq (n-k)t$ for all $k\in \{0,1\dots,n\}$, because this is a restriction given in $(2)$.
That is, the author needs to assume

$r\neq kt$ for all $k\in \{0,\dots,n\}$ and $s\neq (n-k)t$ for all $k\in \{0,\dots,n\}$.

The point is, the above is equivalent to assuming

$r\neq kt$ for all $k\in \{0,\dots,n\}$ and $s\neq \color{red}kt$ for all $k\in \{0,\dots,n\}$.

The reason that the second assumptions are equivalent to the first is that, as $k$ runs through all of $\{0,\dots,n\}$, so does $n-k$, so it is OK to replace $(n-k)$ with $k$.
The second set of assumptions can also be written as $r\neq kt\neq s$ for all $k\in \{0,\dots,n\}$.
A: In order to prove
$$ \sum_{k}A_{k}(r,t)A_{n-k}(s,t) = A_{n}(r+s, t), \tag{1} $$
it suffices to define the ordinary generating functions
$$ f(z) := \sum_{n=0}^\infty A_n(x,t)z^n \tag{2}$$ and
$$ g(z) := \sum_{n=0}^\infty (-1)^n\binom{(t+1)(n+1)-1}{n}z^n. \tag{3}$$
Using standard properties of generating functions,
equation $(1)$ is equivalent to
$$ x\int_0^z g(z)\,dz = \log(f(z)). \tag{4} $$
It is elementary to show that $\,A_0(x,t) = 1\,$ and that
$$ A_{n}(x,t) = \binom{x-nt-1}{n-1}\frac{x}{n} \tag{5}$$
for $\,n>0\,$ given the original definition of $\,A_n(x,t)\,$
as an $n$th degree polynomial in $\,x.$ This approach avoids
the need for discussing any possible division by zero.
A: This is a matter of symmetry. Somewhat more detailed the identity is written as
\begin{align*}
\sum_{k=0}^nA_k(r,t)A_{n-k}(s,t)=A_n(r+s,t)\tag{1}
\end{align*}
From the definition
\begin{align*}
A_{n}(x,t) = \binom{x-nt}{n}\frac{x}{x-nt}, \qquad \text{for} \ x \neq nt.\tag{2}
\end{align*}
it follows the term $A_k(r,t)$ has to fulfill the restriction $r\neq kt$ for $0\leq k\leq n$.

On the other hand, changing the order of summation in (1) $k\to n-k$ gives the same identity
\begin{align*}
\sum_{k=0}^nA_{n-k}(r,t)\color{blue}{A_{k}(s,t)}=A_n(r+s,t)\tag{3}
\end{align*}
From the representation (3) we see that also the term $A_{k}(s,t)$ has to fulfill the restriction $s\neq kt$ for $0\leq k\leq n$.

Hint: A thorough introduction in working with sums is given in chapter 2: Sums, especially in section 2.3: Manipulation of Sums in Concrete Mathematics
by R. L. Graham, D. Knuth and O. Patashnik.
