Determine all $\displaystyle {m, n}\in\mathbb{N}$ such that $\displaystyle \displaystyle \binom{n+1}{m+1} = \binom{n+1}{m} = \frac{5}{3} \binom{n+1}{m-1}$.
We can first get rid of the trivial cases where all are 0; ie, m-1 > n+1. Then I believe the first equation implies that n is even and m=n/2. Then since the sequence of C(n+1,m)/C(n+1,m-1) for n = {2,4,6,...} and m=n/2 is strictly decreasing, the unique other solution is given by (n,m) = (6,3).