TITLE: Unexpected Proofs Using Generating Functions
QUESTION [38 upvotes]: I recently came across this beautiful proof by Erdős that uses generating functions in a unique way:

Let $S = \{a_1, \cdots, a_n \}$ be a finite set of positive integers such that no two subsets of $S$ have the same sum. Prove that $$\sum_{i=1}^n \frac{1}{a_i} < 2.$$

Question: Are there any more examples of surprising or unexpected proofs using generating functions that this community is aware of? 
(Please refrain from posting answers that are widely known such as change making, closed form for Fibonacci, etc.) 
The proof of the above theorem:
Proof:
Suppose $0< x < 1$. We have 
$$\prod_{i=1}^n (1 + x^{a_i}) < \sum_{i = 0}^{\infty} x^i = \frac{1}{1-x}.$$ Thus, 
$$\begin{align*}
\sum_{i=1}^n \log(1+x^{a_i}) &< - \log(1-x) \\
\sum_{i=1}^n \int_0^1 \frac{\log(1+x^{a_i})}{x} \ dx &< - \int_0^1 \frac{\log(1-x)}x \ dx .
\end{align*}$$
Putting $x^{a_i} = y$, we obtain 
$$\begin{align*}
\sum_{i=1}^{n} \frac{1}{a_i} \int_0^1 \frac{\log(1+y)}{y} \ dy < - \int_0^1 \frac{\log(1-x)}{x} \ dx 
\end{align*}$$ i.e.,
$$\sum_{i=1}^n \frac{1}{a_i} \left( \frac{\pi^2}{12} \right) < \frac{\pi^2}6.$$
Thus, $\sum_{i=1}^n \frac{1}{a_i} < 2$ and the theorem is proved.

REPLY [5 votes]: In the same vein as the example of @JimmyK4542, although different. 
Is it possible that the sum of the outcomes of two "flawed" dices with usual numbers $\{1,2, \cdots 6\}$ on their sides has a uniform distribution on $\{2, \cdots 12\}$ ? The answer is negative.
Proof: We should have, with evident notations
$$\tag{1}(p_1x+p_2x^2+\cdots p_6x^6)(p'_1x+p'_2x^2+\cdots p'_6x^6)=\frac{1}{11}(x^2+x^3+\cdots x^{12})$$
Factoring $x^2$, $(1)$ is equivalent to:
$$\tag{2}(p_1+p_2x+\cdots p_6x^5)(p'_1+p'_2x+\cdots p'_6x^5)=\frac{1}{11}(1+x+\cdots x^{10})$$
But this is impossible because 

on the LHS of $(2)$, we have the product of two odd degree polynomials, each one having at least one real root, whereas
the roots of the polynomial in the RHS, which can be written under the form $\frac{1-x^{11}}{1-x}$, are the $e^{2i\pi k/11}$, k=1..10$, with no real root among them. Contradiction.

Remark: insted of "flawed dices", it is probably better to give a context of two "roulettes" with unequal sectors as represented on the graphics below: