I'm hope I'm not mistaken. nAalniaOMliO wrote: Find all positive integers n such that the product $1! \cdot 2! \cdot \cdot \cdot \cdot n!$ is a perfect square Note that $1! \cdot 2! \cdot \cdot \cdot \cdot n!=((n-1)!!)^2 \cdot n!!$, so $n!! $ is perfect square. But it is impossible if $n>4$: if $n$ is odd, then there exists a prime number between $n/2$ and $n$ (by Bertrand postulate) and $v_p(n!!)=1$ for such $p$, so $n!! $ is not a perfect square for $n>4$. And if $n=2k$, we can say that there exists prime $p>2$ between $k/2$ and $k$ and also $v_p((2k)!!)=v_p(k!)=1$. Note that $n=4, 3,2$ are also not works, so the only solution is $n=1$, which is clearly works.

NO_SQUARES wrote: $1! \cdot 2! \cdot \cdot \cdot \cdot n!=((n-1)!!)^2 \cdot n!!$ No, it is false. For instance, take $n=4$. Then $1! \cdot 2! \cdot 3! \cdot 4!=288$, and $(3!!)^2 \cdot 4!!=72$

nAalniaOMliO wrote: NO_SQUARES wrote: $1! \cdot 2! \cdot \cdot \cdot \cdot n!=((n-1)!!)^2 \cdot n!!$ No, it is false. For instance, take $n=4$. Then $1! \cdot 2! \cdot 3! \cdot 4!=288$, and $(3!!)^2 \cdot 4!!=72$ Ahh, I understand there I mistaked, thank you! It all was my trying to say the following statement as briefly as I could: we can divide all numbers on pairs: for $n=4$, say, we can write $(1!)^2 \cdot (3!)^2 \cdot 4!! $ and for $n=5$ – $1 \cdot (2!)^2 \cdot (4!)^2 \cdot 5!! $, so anyway $n!! $ is perfect square.

Hint : Consider the prime such that vp (n!) =2

Hamzaachak wrote: Hint : Consider the prime such that vp (n!) =2 It is even stronger than Bertrand's postulate, how are you going to prove that by elementary means?

pavel kozlov wrote: Hamzaachak wrote: Hint : Consider the prime such that vp (n!) =2 It is even stronger than Bertrand's postulate, how are you going to prove that by elementary means? Yeah Bertrand give you that there is a prime with valuation 2 or 3 and the both of them give the result (between n/2 and n/4) And you can prove n is even by take prime between n and n/2

Hamzaachak wrote: pavel kozlov wrote: Hamzaachak wrote: Hint : Consider the prime such that vp (n!) =2 It is even stronger than Bertrand's postulate, how are you going to prove that by elementary means? Yeah Bertrand give you that there is a prime with valuation 2 or 3 and the both of them give the result (between n/2 and n/4) Valuation 3 doesn't help you in case of odd $n$.

I change my post

Hamzaachak wrote: I change my post Thank you, that makes sense.