IMO LONGLISTED PROBLEM 1970

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IMO LONGLISTED PROBLEM 1970
Prove that,n! can not be the square of any natural number.
Re: IMO LONGLISTED PROBLEM 1970
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.
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Re: IMO LONGLISTED PROBLEM 1970
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
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.sakibtanvir wrote:Then,Should not you give the prof of erdos theorem??Because it is totally based on it...
The product of consecutive integers is never a power
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Re: IMO LONGLISTED PROBLEM 1970
we can use ppf!!
 Phlembac Adib Hasan
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Re: IMO LONGLISTED PROBLEM 1970
ইয়ে...মানে...বানানটা ঠিক করলে ভাল হয়।sm.joty vaia wrote:নিউটনঃ "আমি জ্ঞানের সাগরে নুরি কুড়াচ্ছি মাত্র।"
মহা বিজ্ঞানী জ্যোতিঃ "আমি কক্সবাজারের উদ্দেশে টিকেট কাটছি মাত্র।"
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Re: IMO LONGLISTED PROBLEM 1970
Hint:
"Everything should be made as simple as possible, but not simpler."  Albert Einstein
 nafistiham
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Re: IMO LONGLISTED PROBLEM 1970
nayel wrote:Hint:
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: IMO LONGLISTED PROBLEM 1970
Yes. So how do you conclude from this that $n!$ cannot be a perfect power?
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 nafistiham
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Re: IMO LONGLISTED PROBLEM 1970
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.
because, in a perfect power, the power of any prime factor is a multiple of that power.
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
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