Perfect Number Is Sum Of Cube
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if $(2^p-1)$ is mersenne prime then $2^{(p-1)}.(2^p-1)$ is perfect number. prove they are sum of odd number's cube. if it sum of 1~nth odd number's cube then $n=2^{\frac{p-1}{2}}$
Hint: calculate sum of n odd numbers where $n=2^{\frac{p-1}{2}}$ then show result as $2^{(p-1)}.(2^p-1)$
Hint: calculate sum of n odd numbers where $n=2^{\frac{p-1}{2}}$ then show result as $2^{(p-1)}.(2^p-1)$
Last edited by Masum on Sat May 07, 2011 1:40 pm, edited 1 time in total.
Reason: Use proper Latex
Reason: Use proper Latex
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: Perfect Number Is Sum Of Cube
It's an interesting problem hope u guys like it.
Re: Perfect Number Is Sum Of Cube
Isn't this only a tedious calculation using $P=2^{p-1}(2^p-1)$ and $1^3+2^3+\ldots\ldots+n^3=(\frac{n(n+1)}{2})^2$?
One one thing is neutral in the universe, that is $0$.
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- Posts:56
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Re: Perfect Number Is Sum Of Cube
Here $n=?$
please post solution if u have.
An additional thing, Does anyone think there r no more perfect number without this form of $2^{p-1}.M_p$
please post solution if u have.
An additional thing, Does anyone think there r no more perfect number without this form of $2^{p-1}.M_p$
Last edited by Masum on Fri May 13, 2011 10:46 am, edited 1 time in total.
Reason: why don't you use LaTeX properly? Use curly braces to enclose whole power
Reason: why don't you use LaTeX properly? Use curly braces to enclose whole power
Re: Perfect Number Is Sum Of Cube
I don't know this (though unfortunately no one knows so far). But all even perfect numbers are of the form $2^{p-1}(2^p-1)$ where $M_p$ is prime.
One one thing is neutral in the universe, that is $0$.
Re: Perfect Number Is Sum Of Cube
hi vaiya . I think I have a solution of my own. See it and cheek it . If you can see any error as soon as possible inform me. I'm not well in English so try to understand it.
Last edited by sm.joty on Sun Aug 21, 2011 1:32 am, edited 1 time in total.
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
Re: Perfect Number Is Sum Of Cube
Shouldn't it be $\frac x 2$?sm.joty wrote: \[=(1^3+2^3+3^3+4^3+5^3+6^3……………+x^3)- 2^3(1^3+2^3+3^3………+\boxed x^3) \]
=A - 8a
One one thing is neutral in the universe, that is $0$.
Re: Perfect Number Is Sum Of Cube
Oups......
that's a great mistake that I made. So I will edit that.
But, Is my proof correct ?, vaiya
that's a great mistake that I made. So I will edit that.
But, Is my proof correct ?, vaiya
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
Re: Perfect Number Is Sum Of Cube
we know \[1^3+2^3+3^3+............(2n-1)^3= n^2(2n^2-1)\]
eta asholei induction diye prove kora jai, Inter er math boi er dharay eta aseo.....akhon n er value boshalei ans chole ashe.......... BTW amar ques holo nth odd number cube jodi perfect hoy tahole $n=2^{\frac{p-1}2}$.. eta ki kore prove kora jay????
eta asholei induction diye prove kora jai, Inter er math boi er dharay eta aseo.....akhon n er value boshalei ans chole ashe.......... BTW amar ques holo nth odd number cube jodi perfect hoy tahole $n=2^{\frac{p-1}2}$.. eta ki kore prove kora jay????
Last edited by Masum on Mon Aug 22, 2011 10:01 pm, edited 1 time in total.
Reason: LaTeXified and don't use bengali in english
Reason: LaTeXified and don't use bengali in english
Re: Perfect Number Is Sum Of Cube
You can do it without induction. Let, \
Use the fact \[\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}\]
Then, \
Use the fact \[\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}\]
Then, \
One one thing is neutral in the universe, that is $0$.