Perfect Number Is Sum Of Cube

[tarek like math]

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 $2^{(p-1)}.(2^p-1)$

[tarek like math]

It's an interesting problem hope u guys like it.

[Masum]

Isn't this only a tedious calculation using $P=2^{p-1}(2^p-1)$ and $1^3+2^3+\ldots\ldots+n^3=(\frac{n(n+1)}{2})^2$?

[tarek like math]

Here $n=?$
please post solution if u have.
An additional thing, Does anyone think there r no more perfect number without this form of $2^{p-1}.M_p$

[Masum]

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.

[sm.joty]

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.

Let S = \[1^3+3^3+5^3+\ldots\ldots\ \]
R = \[2^3+4^3+6 ^3+\ldots\ldots\ \]
A = \[1^3+2^3+3^3+4^3+5^3+\ldots\ldots\ \]
Now if $x$ is odd
\[(1^3+2^3+3^3+4^3+5^3+……+x^3)= (1^3+3^3+5^3+……x^3)+(2^3+4^3+……(x-1)^3)-------(1)\]

if $x$ is even then
\[(1^3+2^3+3^3+4^3+5^3+……+x^3)= (1^3+3^3+5^3+……(x-1)^3) +(2^3+4^3+……x^3)-------(2)\]
It is clear that first perfect number 6 is not sum of odd number’s cube. For other perfect numbers (p-1)/2 is always an integer. so n is a even number. So we will use the second equation.
From the second equation we can write, S = A-R
\[=(1^3+2^3+3^3+4^3+5^3+6^3……………+x^3)- (2^3+4^3+6^3………x^3) \]

\[=(1^3+2^3+3^3+4^3+5^3+6^3……………+x^3)- 2^3(1^3+2^3+3^3………(\frac{x}{2})^3)\]
=A - 8a [where \[ a = (1^3+2^3+3^3………x^3) \]
Using this formula : \[1^3+2^3+\ldots\ldots+n^3=(\frac{n(n+1)}{2})^2\]
We can easily show that \[ S= x^2(\frac{x^2-2}{8})\]
Note that here \[ S =(1^3+3^3+5^3+………x^3) \]
Here x is even so we can write x/2=2y
Now we can calculate the sum of y terms of S
So the sum of y terms of S is \[ S = \frac{(2y)^2((2y)^2-2))}{8}
=y^2(2y^2-1) \]
Then if we have n terms
So the final formula is \[1^3+3^3+\ldots\ldots+(2n-1)^3=n^2(2n^2-1)\]
Now we use \[ n=2^{\frac{p-1}{2}}\]
So \[ 1^3+3^3+\ldots\ldots+((2^\frac{p+1}{2})-1)^3 = (2^{\frac{p-1}{2}})^2(2.(2^{\frac{p-1}{2}})^2-1)\]
\[= 2^{(p-1)}.(2^p-1)\]
\[ = \text{Perfect Number}\]



you can also try to prove the formula of The sum of odd numbers by using mathematical induction. i think you can do it ( I couldn't because there I got a big equation which I can't factorize.But it's true that I haven't much concentration also. )

[Masum]

\[=(1^3+2^3+3^3+4^3+5^3+6^3……………+x^3)- 2^3(1^3+2^3+3^3………+\boxed x^3) \]
=A - 8a

Shouldn't it be $\frac x 2$?

[sm.joty]

Oups......
that's a great mistake that I made. So I will edit that.
But, Is my proof correct ?, vaiya

[Shifat]

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????

[Masum]

You can do it without induction. Let, \
Use the fact \[\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}\]
Then, \