Mathematical Reflections 2013, Issue 2 - Problem J262

Problem:
Find all positive integers $m, n$ such that $${m+1 \choose n}={n \choose m+1}.$$

Proposed by Roberto Bosch Cabrera.

Solution:
If $m+1 \geq n$, we have ${m+1 \choose n}>0$. If $n<m+1$, then ${n \choose m+1}=0$, so $n \geq m+1$, i.e. $n=m+1$.
If $m+1 < n$, then ${m+1 \choose n}=0$, so it must be ${n \choose m+1}=0$, which gives $n < m+1$, a contradiction. Hence all positive integers which satisfies the given equation are the consecutive positive integers $m,m+1$.