Find all natural solutions for $\binom mn=1984$

Question

Find all positive integers $m$ and $n$ such that $${m \choose n}= 1984$$

My approach:

It is easy to define $m=1984$ and $n=1$ or $1983$. But how to show that there are no other solutions or, if any, then how to find them?

Answer 1

Note that $31\mid 1984$, hence we need $m\ge 31$. Then if $3\le n\le m-3$, we have ${m\choose n}\ge {m\choose 3}=\frac{m(m-1)(m-2)}{6}\ge {31\choose 3}=4495$. Remains the case $n=2$ (or $n=m-2)$ and $1984=\frac{m(m-1)}{2}$. Since this expression in $m$ is increasing and ${63\choose 2}=1953<1984<2016={64\choose 2}$, no solution with $n=2$ (or $n=m-2$) exists. We conclude that the two solutions you found are all there is.