Since RHS is divisible by $4$ whereas LHS is divisible by $2$ only we see no such pair exists. Nice and short problem, I must say. Cheers, grammarphobic.

Unexpectedly, the problem was easy. But I guess it can be transformed into a nice one. For example, we can try $2^2m?=n(n+1)(n+2)(n+3)$. Because, the case $2m?$ is trivial for a similar reason: right side would be divisible by $8$. Since there are two consecutive even integers.

joybangla wrote: For $m\in \mathbb{N}$ define $m?$ be the product of first $m$ primes. Determine if there exists positive integers $m,n$ with the following property : \[ m?=n(n+1)(n+2)(n+3) \] Proposed by Matko Ljulj By contradiction, assume that exists an $m$ and an $n$ such that works: Since they are both positive integers $n(n+1)(n+2)(n+3) \equiv 0 \pmod 4$. Note that the only pair number and prime is $2$ and then $m?=2 \cdot 3 \cdots p$ where $p$ is the $m$ order prime and this means $m? \equiv 2 \pmod 4$. Hence $2 \equiv 0 \pmod 4 \; \text{contradiction!!}$. That means no such integers, then we are done