Indeed a general form!!

Solving the last problem made you observe that the first 3 perfect numbers are of the 2k1(2k1)2^{k-1}(2^{k} - 1).

Actually it was proved by Euclid that :

If (2k1)2^{k}-1) is a prime the n=2k1(2k1)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
6 years, 12 months ago

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Let 2k1=Mp2^k-1=M_p [short for Mersenne Prime].

The positive divisors of n=2k1(2k1)n=2^{k-1}(2^k-1) are 11, 22, 222^2, 232^3 \cdots 2k12^{k-1} and Mp,2Mp,22Mp,23Mp,2k1Mp{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 σ(n)=2k1+Mp×(2k1)=(2k1)(1+Mp)=(2k1)×2k=2×2k1(2k1)=2n\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, nn is a perfect number by definition.

Mursalin Habib - 6 years, 12 months 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 - 6 years, 12 months ago

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

Bogdan Simeonov - 6 years, 11 months ago

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

Eddie The Head - 6 years, 11 months ago

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Yes, Euclid found out that p(p+1)2\dfrac{p(p+1)}{2} was a perfect number for certain prime numbers pp. He concluded that those prime numbers had to be of the form 2k12^k-1. These primes are now called Mersenne Primes.

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

Mursalin Habib - 6 years, 11 months ago

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@Mursalin Habib All even prime numbers?

Sharky Kesa - 6 years, 11 months ago

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@Sharky Kesa Oops! Typo!

Mursalin Habib - 6 years, 11 months ago

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