<s>[INST] Let $n$ be an even perfect number greater than $ 6$. Prove that $n \equiv 4 \pmod 6$. Things I've gotten so far: 
Proven that $2^p-1\equiv \pmod 3$ (for some odd prime)
$n$ is in the form $n =( 2^p-1)(2^{p - 1})$ where $(2^p - 1)$ is prime.$ n > 6$, so $p > 2$.
So I have the congruence $2^p-1 \equiv 1 \pmod 3$, 
and I think I have to prove that $2^p -1 \equiv 4 \pmod 6$ or $-2 \pmod 6$,
but I'm not sure how to accomplish that step. 
 [/INST] Observe that $$x^{ab}-1=(x^a-1)(x^{a(b-1)}+x^{a(b-2)}+\cdots+x^a+1).$$ Hence a number of this form could be prime if and only if $x=2$ and $a=1$ and $b$ is a prime. Here it has shown that an even perfect number is of the form $2^{p-1}(2^p-1),$ where $2^p-1$ is a prime. Hence we only need to consider about prime powers!!
Consider $p=2$ separately (which gives you the prefect number $6$) and note that all the other primes are odd. Let $p=2m+1.$ Binomial theorem gives us, 
$$2^{2m+1}=2(3+1)^m=2(3^m+\cdots+3m+1)=6.3^{m-1}+\cdots+6m+2.$$ Now $n=2^{2m}(2^{2m+1}-1)=2^{2m}(6.3^{m-1}+\cdots+6m+1).$ 
Hence $n \equiv 2^{2m} \pmod 6$ and you can easily prove that $2^{2m} \equiv 4 \pmod 6$ by mathematical induction.
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