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 $1 \leq k \leq N-1$.

Proposed by: Romania
$\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ 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 a proper sequence $a_{0}, a_{1}, \ldots, a_{N}$, because from 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 $N=n$.