TITLE: Discussion on even and odd perfect numbers.
QUESTION [8 upvotes]: First of all thank you so much for answering my previous post. These are few interesting problems drawn from Prof. Gandhi lecture notes. kindly discuss:
1) If $n$ is even perfect number then $(8n +1)$ is always a perfect square.
2) Every odd perfect number has at least three different prime factors. This is by observation we can understand that. But, how to prove mathematically?
3) Every even perfect number (other than 6) can be expressible as sum of consecutive odd cubes.
Thanks in advance.

REPLY [6 votes]: As Andre mentioned, the even perfect numbers have the form $2^{p-1} (2^{p} -1 )$.
Claim: The sum of the first $2^{(p-1)/2}$ odd cubes is equal to this perfect number.
Proof: We know that the sum of the first $n$ odd cubes is:
$\sum_{i=1}^{n} (2i-1)^3 = \sum_{i=1}^{2n} i^3 - \sum_{i=1}^{n} (2i)^3 \\= \left[ \frac {(2n)(2n+1)}{2} \right] ^2 - 8\left[ \frac {n(n+1)}{2} \right]^2 = \frac {n^2 [16n^2 + 16n + 4 - 8(n^2 + 2n+1)]}{4}\\ = n^2(2n^2 - 1) $
Now substitute $n = 2^{(p-1)/2}$.