IMO LONGLISTED PROBLEM 1970

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MATHPRITOM
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IMO LONGLISTED PROBLEM 1970

Unread post by MATHPRITOM » Sun May 01, 2011 12:54 pm

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

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sm.joty
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by sm.joty » Tue Mar 13, 2012 12:13 am

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. :D
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by sakibtanvir » Tue Mar 13, 2012 7:58 pm

Then,Should not you give the prof of erdos theorem??Because it is totally based on it...
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Phlembac Adib Hasan
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by Phlembac Adib Hasan » Wed Mar 21, 2012 9:52 pm

sakibtanvir wrote: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
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turash
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by turash » Fri Mar 23, 2012 12:30 am

we can use ppf!!

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Re: IMO LONGLISTED PROBLEM 1970

Unread post by Phlembac Adib Hasan » Fri Mar 23, 2012 12:03 pm

sm.joty vaia wrote:নিউটনঃ "আমি জ্ঞানের সাগরে নুরি কুড়াচ্ছি মাত্র।"
মহা বিজ্ঞানী জ্যোতিঃ "আমি কক্সবাজারের উদ্দেশে টিকেট কাটছি মাত্র।"
ইয়ে...মানে...বানানটা ঠিক করলে ভাল হয়। :|
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nayel
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by nayel » Sat Mar 24, 2012 10:19 am

Hint:
By Bertrand's postulate, there is a prime $p$ between $\frac n2$ and $n$. Now what is the power of $p$ in $n!$?
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by nafistiham » Sat Mar 24, 2012 11:24 pm

nayel wrote: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 $1$ ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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nayel
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by nayel » Sun Mar 25, 2012 1:19 am

Yes. So how do you conclude from this that $n!$ cannot be a perfect power?
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Re: IMO LONGLISTED PROBLEM 1970

Unread post by nafistiham » Mon Mar 26, 2012 12:19 am

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.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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