The product of composite numbers in the range $[m+2, 2m+2]$ is not less than $m!$.

Question

If $m \in \mathbb{N}$ is a natural number, prove that

$$ m! \leq \prod\limits_{\begin{align}k = &m+2 \\ k \text{ not }& \text{prime.} \end{align}}^{2m+1}k.$$

Answer 1

Note that the binomial coefficient $\binom{2m+1}m$ is an integer, and $$\binom{2m+1}m=\frac{(2m+1)!}{m!(m+1)!}=\frac 1{m!}\prod_{k=m+2}^{2m+1}k$$ Here comes the number-theory part: we know $m!(m+1)!$ divides $(2m+1)!$, and since no primes $p>m+1$ is divisible by the donominator $m!(m+1)!$, we see that $m!(m+1)!$ divides $(2m+1)!/(\text{products of all primes between m+2 and 2m+1})$. The result follows.

Answer 2

There has been an edit to the question that makes this answer hard to understand. The original version of the question asked: $$ \frac{\displaystyle\prod_{k=m+2\\k\text{ not prime}}^{2m+1}k}{m!}\geq 1. $$

Sketch:

• Assume that $p$ is a prime such that $p\leq m$.

• The number of times that $p$ divides $m!$ is computed by: $$ \sum_{i=1}^\infty\left\lfloor\frac{m}{p^i}\right\rfloor $$

• The number of times that $p$ divides the numerator is $$ \sum_{i=1}^\infty\left\lfloor\frac{2m+1}{p^i}\right\rfloor-\left\lfloor\frac{m+1}{p^i}\right\rfloor $$ It is the same as the number of times that $p$ divides $\frac{(2m+1)!}{(m+1)!}$ since this quotient, when compared to the given quotient, only includes extra primes and $p$ can't divide any of those.

• Observe that the number of powers of $p$ in the quotient is $$ \sum_{i=1}^\infty\left\lfloor\frac{2m+1}{p^i}\right\rfloor-\left\lfloor\frac{m+1}{p^i}\right\rfloor-\left\lfloor\frac{m}{p^i}\right\rfloor $$

• Since each term in the sum is nonnegative, some nonnegative number of $p$'s divide the quotient. Therefore, the given quotient is an integer (which is greater than or equal to $1$).