Indeed a general form!!

Solving the last problem made you observe that the first 3 perfect numbers are of the \(2^{k-1}(2^{k} - 1)\).

Actually it was proved by Euclid that :

If (\(2^{k}-1)\) is a prime the \(n = 2^{k-1}(2^{k}-1)\) is perfect and every perfect number of this form.

Try to prove it in the comments box :)

You can also see that this implies all even perfect numbers are triangular numbers.

Also no odd perfect number has been found so far.

Note by Eddie The Head
4 years ago

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Let \(2^k-1=M_p\) [short for Mersenne Prime].

The positive divisors of \(n=2^{k-1}(2^k-1)\) are \(1\), \(2\), \(2^2\), \(2^3\) \(\cdots\) \(2^{k-1}\) and \({M_p}, 2M_p, 2^2 M_p, 2^3 M_p, \cdots 2^{k-1} M_p\).

Let's add them all together.

We get \(\sigma(n)=2^k-1 + M_p \times (2^k-1)=(2^k-1)(1+M_p)=(2^k-1)\times 2^k=2\times2^{k-1}(2^k-1)=2n \).

So, \(n\) is a perfect number by definition.

Mursalin Habib - 4 years ago

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When I first found out about perfect numbers (I was 9), I strived to find a pattern between them. After 2 hours only I figured it out but then checked the Internet. Turns out it was discovered by Euclid over 2000 years ago. I still felt proud.

Sharky Kesa - 4 years ago

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It was proven by Euler that every even perfect number must have this form.

Bogdan Simeonov - 4 years ago

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I think you're right ..but the general form was given by Euclid right??

Eddie The Head - 4 years ago

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Yes, Euclid found out that \(\dfrac{p(p+1)}{2}\) was a perfect number for certain prime numbers \(p\). He concluded that those prime numbers had to be of the form \(2^k-1\). These primes are now called Mersenne Primes.

Euler later on proved that all even perfect numbers must have this form.

Mursalin Habib - 4 years ago

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@Mursalin Habib All even prime numbers?

Sharky Kesa - 4 years ago

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@Sharky Kesa Oops! Typo!

Mursalin Habib - 4 years ago

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