Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $m,n$ be positive integers, both odd or both even, with $n\ge m$. I think the following number $$\frac{m+n+2}{2(m+1)}{n\choose m}2^{(n-m)/2}$$ is always an integer, but I have trouble proving it.

share|cite|improve this question
up vote 2 down vote accepted

Split $m+n+2$ as $m+1 + n+1$. You want $$\frac{1}{2} \binom{n}{m} 2^{(n-m)/2}$$ and $$\frac{n+1}{2(m+1)} \binom{n}{m} 2^{(n-m)/2}$$ to be integers. The first one is clear. For the second one, use $\displaystyle \binom{n+1}{m+1} = \frac{n+1}{m+1} \binom{n}{m}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.