Show $n\equiv 1 \pmod 9$ for Even Perfect Numbers $n>6$

[alexmahone]

Show that $n\equiv 1 \pmod 9$ for all even perfect numbers $n>6$.

 

[Euge]

Hi Alexmahone,

Please check back to make sure your statement is the intended one, because as it stands it is false. For instance, $4^2 = 16$ is not congruent to $1\pmod{9}$ and $12^2 = 144 = 16\cdot 9$ is not congruent to $1\pmod{9}$.

 

[alexmahone]

Hi Alexmahone,

Please check back to make sure your statement is the intended one, because as it stands it is false. For instance, $4^2 = 16$ is not congruent to $1\pmod{9}$ and $12^2 = 144 = 16\cdot 9$ is not congruent to $1\pmod{9}$.

I said "perfect numbers", not "perfect squares".

A perfect number is a number that is equal to the sum of its proper divisors. Perfect number - Wikipedia, the free encyclopedia
Eg, $6=1+2+3$
$28=1+2+4+7+14$

 

[Euge]

Sorry about that, I misread it as perfect squares. Let's try again. Every even perfect number greater than $6$ can be written in the form

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

where $p$ is an odd prime. Show that $2^p - 2$ and $2^{p-1}-1$ are divisible by $3$, using the fact that $p$ is odd. Then deduce that the above form is congruent to $1\pmod{9}$. More explicitly, you can express the above form as $1 + 9t_{(2^p-2)/3}$ where $ t_k $ represents the $k$-th triangular number.

In the future, please do not simply state the question --- show what work you've done or where you're stuck so that we may help you more effectively. :)