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.

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 $${M_p}, 2M_p, 2^2 M_p, 2^3 M_p, \cdots 2^{k-1} M_p$$.

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.

- 4 years ago

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.

- 4 years ago

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

- 4 years ago

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

- 4 years ago

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.

- 4 years ago

All even prime numbers?

- 4 years ago

Oops! Typo!

- 4 years ago