## IMO LONGLISTED PROBLEM 1970

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MATHPRITOM
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IMO LONGLISTED PROBLEM 1970
Prove that,n! can not be the square of any natural number.

sm.joty
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### 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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তবুও এগিয়ে যেতে হবে.........
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sakibtanvir
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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...
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

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### Re: IMO LONGLISTED PROBLEM 1970

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

we can use ppf!!

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### Re: IMO LONGLISTED PROBLEM 1970

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

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

Yes. So how do you conclude from this that $n!$ cannot be a perfect power?
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

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