# 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. 6 years, 5 months 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.

- 6 years, 5 months 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.

- 6 years, 5 months ago

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

- 6 years, 5 months ago

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

- 6 years, 5 months 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.

- 6 years, 5 months ago

All even prime numbers?

- 6 years, 5 months ago

Oops! Typo!

- 6 years, 5 months ago