Problem (Algebra):
Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that

(1) $a_{0}+a_{1}=-\frac{1}{n}$, and

(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.

Solution:
$\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ is equivalent to $\left(a_{k}+a_{k-1}+1\right)\left(a_{k}+a_{k+1}-1\right)=-1$. Let $b_{k}=a_{k}+a_{k+1}$. Thus we need $b_{0}, b_{1}, \ldots$ the following way: $b_{0}=-\frac{1}{n}$ and $\left(b_{k-1}+1\right)\left(b_{k}-1\right)=-1$. There is a proper sequence $b_{0}, b_{1}, \ldots, b_{N-1}$ if and only if there is proper sequence $a_{0}, a_{1}, \ldots, a_{N}$, because from a a proper $\left(a_{k}\right)$ sequence we can get a proper $\left(b_{k}\right)$ sequence with $b_{k}=a_{k}+a_{k+1}$ for $k=0,1, \ldots, N-1$ and from a proper $\left(b_{k}\right)$ sequence we can get a proper $\left(a_{k}\right)$ sequence by arbitrarily setting $a_{0}$ and then inductively defining $a_{k}=b_{k-1}-a_{k-1}$ for $k=1,2, \ldots, N$.

We prove by induction that $b_{k}=-\frac{1}{n-k}$ for $k<n$. This is true for $k=0$, as $b_{0}=-\frac{1}{n}$ and

$$
b_{k}=1-\frac{1}{b_{k-1}+1}=1-\frac{1}{1-\frac{1}{n-k+1}}=-\frac{1}{n-k}
$$

for $k<n$. Thus there is a proper sequence $b_{0}, b_{1}, \ldots, b_{n-1}$, but it can't be continued, because $b_{n-1}+1=$ 0 so there is no $b_{n}$ for which $\left(b_{n-1}+1\right)\left(b_{n}-1\right)=-1$.

Therefore the longest proper sequence $\left(b_{k}\right)$ is $n$-long, so the longest proper sequence $\left(a_{k}\right)$ is $n+1$ long, so $N=n$.

The required maximum is $N=n$.

To rule out the case $N \geq n+1$, it is clearly sufficient to rule out the case $N=n+1$.

Assume for contradiction that $a_{0}, a_{1}, \ldots, a_{n+1}$ are real numbers satisfying both conditions in the statement. It is sufficient to show that $a_{k}+a_{k+1}=0$ for some $k \leq n$, because then $a_{k-1}-a_{k+1}=0$ so $a_{k+1}=a_{k-1}$, therefore $a_{k-1}+a_{k}=0$ and so on, by backwards recursion we get that $a_{j}+a_{j+1}=0$ for all $0 \leq j \leq k$, but this is a contradiction with $a_{0}+a_{1}=-\frac{1}{n}$.

To prove that $a_{k}+a_{k+1}=0$ for some $k \leq n$, assume that $a_{k}+a_{k+1} \neq 0$ for all $k \leq n$, to rewrite the second condition in the statement in the form

$$
\frac{1}{a_{k}+a_{k+1}}-\frac{1}{a_{k-1}+a_{k}}=1, \quad k=1, \ldots, n
$$

and sum both sides over the full range from $k=1$ to $n$. This gives

$$
\frac{1}{a_{n}+a_{n+1}}-\frac{1}{a_{0}+a_{1}}=n
$$

As $a_{0}+a_{1}=-\frac{1}{n}$, this means that $\frac{1}{a_{n}+a_{n+1}}=0$, which is a contradiction. Consequently, $N \leq n$.

To provide $n+1$ real numbers satisfying both conditions in the statement, fix $a_{0}$ and go through the telescoping procedure above to obtain

$$
a_{k}=(-1)^{k} a_{0}+\sum_{j=1}^{k} \frac{(-1)^{k-j+1}}{n-j+1}, \quad k=1, \ldots, n
$$

This concludes the proof.

Final answer: $N=n$