sorry for posting here, but it's just too good to pass up!

even, quasi-, & odd perfect numbers all have a great deal in common!
the convention sigma(n) = n for the even perfect numbers must stand
as I'm only concerned with the 'proper' divisors of 'n'.

they all have to pass this test, or their cover is blown!:

IS sigma(x) > (y) -2 ???, followed by some adjustment until both the

positive and negative counterparts are on opposing sides, and sigma

can be taken AGAIN on both sides; 'x' and 'y' can be different; it

doesn't matter.

three definitions...

if sigma(n) = n, then 'n' is considered to be an even perfect number
some 'evens' slip through the test as you already know...

if sigma(n^2) = n^2 +1, then 'n^2' is to be a quasi-perfect number;
n^2 has to be an odd perfect square, and 'n'>3 to see the result.

finally, if sigma(2n+1) = 2n+1, then '2n+1' is an odd perfect number
I'll show you that odd perfect numbers CAN'T exist!

watch... how the test applies to each case to achieve some results:


for evens...
if sigma(n)= n, then 'n' is considered to be an even perfect number.
some of them squeak through!

if sigma(n)= n, then sigma(n) > (n) -2 ???; adding 1 TBS & sigma AGAIN

leads to... sigma(sigma(n)+1) > sigma(n -1); which reduces to...

sigma(n +1) > sigma(n -1) by substitution of definition again.


without any hesitation, n = 5, 27, etc. pass the test, and Euler saw

their true formula; these examples can be easily exposed & verified.


for quasi's...
if sigma(n^2)= n^2 +1 for an odd 'n'>3, then 'n^2' is considered to
be a quasi-perfect number! (please excuse the early substitution...
it reads better, later).

if sigma(n^2)= n^2 +1, then sigma(n^2) > (n^2 +1) -2 ???; along with

sigma AGAIN leads to... sigma(sigma(n^2)) > sigma(n^2 -1); which re-

duces to sigma(n^2 +1) > sigma(n^2 -1) by subst. of definition again.


with some hesitation, (n^2 -1) has the form '8m' and (n^2 +1) has

the form '8m +2' for the same 'm'; clearly, sigma(n^2 -1) is more a-

bundant than sigma(n^2 +1), --- but only when an odd 'n'> 3 ---

hence, by contradiction, a quasi-perfect number cannot exist.


for odds...
if sigma(2n+1)= 2n +1 for some 'n', then '2n+1' is considered to be
an odd perfect number! I'll show you why there aren't any...

if sigma(2n +1)= 2n +1, then sigma(2n+1) > (2n+1) -2 ???; and sigma

AGAIN leads to... sigma(sigma(2n+1)) > sigma(2n-1); which reduces to

sigma(2n +1) > sigma(2n -1) by substitution of definition again.


clearly, sigma(2n+1) is ALWAYS more abundant than sigma(2n -1) which

means that for every 'n', sigma(2n+1) = 2n+1; if you evaluate it, you

quickly realize that the original definition cannot possibly be right

for every 'n' and we have a false-positive answer! (tricky?)


hence, by contradiction, an odd perfect number CAN'T exist.


enjoy!