# Find all values of for which the ratio is an integer

Find all values of $n$ for which, $$\dfrac{(\dfrac{n+3}{2}) \cdots n}{(\dfrac{n-1}{2})!}$$

is an integer.

I have tried the problem for some primes. Each time it seemed true. But I still haven't made any progress even towards the proof (or disproof) of the case when $n$ is a prime, except from some trivial observations which gave me some weaker congruence. A complete proof for in general $n$ is required.

• I'm not quite sure what the dots are on top. Are we increasing of $n+3, n+4, \ldots n+n$ as the numerators? Further, how are we to interpret the bottom if $n$ is not odd? (Or must $n$ always be odd?) If this is the correct interpretation (which feels most natural to me), then half the numerator terms will look like $n + 2k$, which is then odd, and thus not divisible by $2$. – davidlowryduda Jun 27 '14 at 17:43
• Do you intend for $n$ to be odd and for the numerator to be $\frac{n+3}{2}\cdot\frac{n+5}{2}\cdot\frac{n+7}{2}\cdots n$? – rogerl Jun 27 '14 at 17:43

I assume $n = 2k + 1$ is an odd integer, please advise in case. Then $$\dfrac{(\dfrac{n+3}{2}) \cdots n}{(\dfrac{n-1}{2})!} = \frac{(k+2) (k+3) \cdots (2k+1)}{k!} = \frac{(2k+1)!}{k! (k+1)!} = \binom{2k+1}{k}.$$

The combinatorial approach is best. However, one can prove by induction the following result:

Lemma: Let $n$ be a positive integer. The product of any $n$ consecutive positive integers is divisible by $n!$.

The argument is mildly tricky, since it involves a double induction.

Assume by way of induction hypothesis that $n!$ divides the product of any $n$ consecutive positive integers. We show that $(n+1)!$ divides the product of any $n+1$ consecutive positive integers.

So we want to show that for any $m\ge 1$, $(n+1)!$ divides $m(m+1)\cdots(m+n)$. We prove this by induction on $m$. The result is evidently true at $m=1$. Suppose the result is true at $m$. We show it is true at $m+1$.

So we want to show that $(n+1)!$ divides $(m+1)(m+2)\cdots (m+1+n)$. Note that $$(m+1)(m+2)\cdots (m+1+n)-m(m+1)\cdots(m+n)=(n+1)\left[(m+1)(m+2)\cdots(m+n)\right].$$

By the (outer) induction hypothesis, we have that $n!$ divides $(m+1)(m+2)\cdots(m+n)$, so $(n+1)!$ divides $(n+1)[(m+1)(m+2)\cdots(m+n)]$. Since $m(m+1)\cdots(m+n)$ is divisible by $(n+1)!$, it follows that so is $(m+1)(m+2)\cdots (m+1+n)$.