[Number Theory]
Problem: For an integer $n \geq 4$, define $a_{n}$ to be the product of all real numbers that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. Compute

$$
\frac{a_{4}}{a_{5}}+\frac{a_{5}}{a_{6}}+\frac{a_{6}}{a_{7}}+\cdots+\frac{a_{2022}}{a_{2023}} .
$$

Solution: For an integer $n \geq 4$, let $S_{n}$ denote the set of real numbers $x$ that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. (Note that $S_{n}$ is nonempty, as the polynomial $x^{2}+(n-2) x+1$ has a discriminant of $(n-2)^{2}-4$, which is nonnegative for $n \geq 4$.) Then $a_{n}=\prod_{x \in S_{n}} x$.

Suppose that $a, b$, and $c$ are positive integers and $x$ is a real solution to $a x^{2}+b x+c=0$. Then $x$ must be nonzero. (In fact, $x$ must be negative.) Dividing the above equation by $x^{2}$ yields $a+\frac{b}{x}+\frac{c}{x^{2}}=0$, thus $r=\frac{1}{x}$ is a solution to the quadratic equation $c r^{2}+b r+a=0$. This shows that $x \in S_{n}$ if and only if $\frac{1}{x} \in S_{n}$.

One might then think that $a_{n}$ must equal 1, because one can presumably pair up all elements in a given $S_{n}$ into $\left\{x, \frac{1}{x}\right\}$ pairs. But there is a (negative) value of $x$ for which $x=\frac{1}{x}$, namely $x=-1$. Therefore the value of $a_{n}$ depends only on whether $-1 \in S_{n}$. It is readily seen via a parity argument that $-1 \in S_{n}$ if and only if $n$ is even. If $n=2 k$, then the polynomial $x^{2}+k x+(k-1)$ has -1 as a root. (In fact, any quadratic polynomial whose middle coefficient is $k$ and whose coefficients sum to $2 k$ will work.) But if $n=2 k+1$, then $a(-1)^{2}+b(-1)+c=a-b+c=(a+b+c)-2 b=(2 k+1)-2 b$ will be odd, and so $-1 \notin S_{n}$.

Thus $a_{n}=-1$ when $n$ is even, $a_{n}=1$ when $n$ is odd, and finally,

$$
\frac{a_{4}}{a_{5}}+\frac{a_{5}}{a_{6}}+\frac{a_{6}}{a_{7}}+\cdots+\frac{a_{2022}}{a_{2023}}=\underbrace{(-1)+(-1)+(-1)+\cdots+(-1)}_{2019(-1) \mathrm{s}}=-\mathbf{2 0 1 9} .
$$

Final answer: -2019