
Problem 1. Suppose $r \geq 2$ is an integer, and let $m_{1}, n_{1}, m_{2}, n_{2}, \cdots, m_{r}, n_{r}$ be $2 r$ integers such that

$$
\left|m_{i} n_{j}-m_{j} n_{i}\right|=1
$$

for any two integers $i$ and $j$ satisfying $1 \leq i<j \leq r$. Determine the maximum possible value of $r$.


Solution. Let $m_{1}, n_{1}, m_{2}, n_{2}$ be integers satisfying $m_{1} n_{2}-m_{2} n_{1}= \pm 1$. By changing the signs of $m_{2}, n_{2}$ if need be, we may assume that

$$
m_{1} n_{2}-m_{2} n_{1}=1
$$

If $m_{3}, n_{3}$ are integers satisfying $m_{1} n_{3}-m_{3} n_{1}= \pm 1$, again we may assume (by changing their signs if necessary) that

$$
m_{1} n_{3}-m_{3} n_{1}=1
$$

So $m_{1}\left(n_{2}-n_{3}\right)=n_{1}\left(m_{2}-m_{3}\right)$.

As $m_{1}, n_{1}$ are relatively prime, $m_{1}$ divides $m_{2}-m_{3}$; say, $m_{2}-m_{3}=m_{1} a$ for some integer $a$. Thus, we get $n_{2}-n_{3}=n_{1} a$. In other words,

$$
m_{3}=m_{2}-m_{1} a, n_{3}=n_{2}-n_{1} a
$$

Now, if $m_{2} n_{3}-n_{2} m_{3}= \pm 1$, we get

$$
\pm 1=m_{2}\left(n_{2}-n_{1} a\right)-n_{2}\left(m_{2}-m_{1} a\right)=\left(m_{1} n_{2}-m_{2} n_{1}\right) a=a
$$

Thus, $m_{3}=m_{2}-m_{1} a=m_{2} \pm m_{1}, n_{3}=n_{2}-n_{1} a=n_{2} \pm n_{1}$.

Now if we were to have another pair of integers $m_{4}, n_{4}$ such that

$$
m_{1} n_{4}-n_{1} m_{4}= \pm 1
$$

we may assume that $m_{1} n_{4}-n_{1} m_{4}=1$. As seen above, $m_{4}=m_{2} \mp m_{1}, n_{4}=n_{2} \mp n_{1}$. But then

$$
m_{3} n_{4}-n_{3} m_{4}=\left(m_{2} \pm m_{1}\right)\left(n_{2} \mp n_{1}\right)-\left(n_{2} \pm n_{1}\right)\left(m_{2} \mp m_{1}\right)= \pm 2
$$

Therefore, there can be only 3 pairs of such integers.

That there do exist many sets of 3 pairs is easy to see; for instance, $(1,0),(1,1),(0,1)$ is such a triple.

Alternate Solution. It is clear that $r$ can be 3 due to the valid solution $m_{1}=1, n_{1}=$ $1, m_{2}=1, n_{2}=2, m_{3}=2, n_{3}=3$.

If possible, let $r>3$. We observe that:

$$
\begin{aligned}
& m_{1} n_{2} n_{3}-m_{2} n_{1} n_{3}= \pm n_{3} \\
& m_{2} n_{3} n_{1}-m_{3} n_{2} n_{1}= \pm n_{1} \\
& m_{3} n_{1} n_{2}-m_{1} n_{3} n_{2}= \pm n_{2}
\end{aligned}
$$

Adding, we get $\pm n_{1} \pm n_{2} \pm n_{3}=0$; which forces at least one of $n_{1}, n_{2}, n_{3}$ to be even; WLOG let $n_{1}$ be even.

Repeating the argument for indices $2,3,4$, we deduce that at least one of $n_{2}, n_{3}, n_{4}$ is even; WLOG let $n_{2}$ be even. This leads to a contradiction, since $\left|m_{1} n_{2}-m_{2} n_{1}\right|=1$ cannot be even. Hence $r>3$ is not possible, as claimed.
