IMO LONGLISTED PROBLEM 1970

[MATHPRITOM]

Prove that,n! can not be the square of any natural number.

[sm.joty]

According to ERDOS theorem, product of $n$ consecutive integers can not be a perfect squire for all $n\geq 2$. So $n!$ can't be squire number.

[sakibtanvir]

Then,Should not you give the prof of erdos theorem??Because it is totally based on it...

[Phlembac Adib Hasan]

Then,Should not you give the prof of erdos theorem??Because it is totally based on it...

Well,I'm giving.(Anyone can easily get it by Google )But, according to seniors, now I also think we should avoid using such advanced theorems.
The product of consecutive integers is never a power

[turash]

we can use ppf!!

[Phlembac Adib Hasan]

নিউটনঃ "আমি জ্ঞানের সাগরে নুরি কুড়াচ্ছি মাত্র।"
মহা বিজ্ঞানী জ্যোতিঃ "আমি কক্সবাজারের উদ্দেশে টিকেট কাটছি মাত্র।"

ইয়ে...মানে...বানানটা ঠিক করলে ভাল হয়।

[nayel]

Hint:

By Bertrand's postulate, there is a prime $p$ between $\frac n2$ and $n$. Now what is the power of $p$ in $n!$?

[nafistiham]

Hint:

By Bertrand's postulate, there is a prime $p$ between $\frac n2$ and $n$. Now what is the power of $p$ in $n!$?

Doesn't it mean there will be a biggest prime $p$ whose power in $n!$ will be $1$ ?

[nayel]

Yes. So how do you conclude from this that $n!$ cannot be a perfect power?

[nafistiham]

So, we can say that, in the perfect power, the power can not be greater than $1$.
because, in a perfect power, the power of any prime factor is a multiple of that power.