Divisibility of perfect numbers

Question

I am new at the Number theory, I have a question that;

n is an even perfect number without 28, for all of the other even perfect numbers, prove that n = 1 or -1 (mod7). Actually, I don't know where to start, Is there anyone to help me? Thanks in advance.

Answer 1

It is well known that every even perfect number is of the form $$n=2^{p-1}(2^p-1),$$ for $2^p-1$ a Mersenne prime. Except for the case $p=3$ ($n=28$), $2^p-1\neq7$, so that $2^p\not\equiv1\pmod7$, and $p\not\equiv0\pmod3$. We have two other cases:

• If $p\equiv1\pmod3$, $$n\equiv 1\cdot1\equiv1\pmod7.$$
• If $p\equiv2\pmod3$, $$n\equiv2\cdot3\equiv-1\pmod7.$$

This establishes what we wished to prove. $\blacksquare$

Answer 2

An even perfect $n\ne 28$ is of the form $2^{p-1}(2^p-1)$ with $p,\,2^p-1$ both prime and $p\ne3$, so $3\nmid p$ and modulo $7$ we have$$2^{p-1}\in\{1,\,2\}\implies2^{p-1}(2^p-1)\in\{1(1\times2-1),\,2(2\times2-1)\}=\{1,\,6\}.$$