# Thread: Positive integers in a binomial relation

1. ## Positive integers in a binomial relation

Determine all $\displaystyle {m, n}\in\mathbb{N}$ such that $\displaystyle \displaystyle \binom{n+1}{m+1} = \binom{n+1}{m} = \frac{5}{3} \binom{n+1}{m-1}$.

2. Originally Posted by Demandeur
Determine all $\displaystyle {m, n}\in\mathbb{N}$ such that $\displaystyle \displaystyle \binom{n+1}{m+1} = \binom{m+1}{m} = \frac{5}{3} \binom{n+1}{m-1}$.
Should there be an n in your second binomial coefficient?

3. Originally Posted by chiph588@
Should there be an n in your second binomial coefficient?
In fact, Cliph, yes. Thank you. I've edited now. I'm sorry for any trouble which that might have caused. Strange enough, I remember double-checking.

4. Originally Posted by Demandeur
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).