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perfect number

number theory
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Or maybe its a conjecture. Taht for n > 3, their are 1 < a < b < c such that 5/n = 1/a + 1/b + 1/c. Do they need to be distinct (ie, can we have 1 <= a <= b <= c)? But most importantely, what is this problem called?

You might be thinking about the Erdős-Straus conjecture, but that's about 4/n, not 5/n. Or you might be thinking about the Sierpiński generalization to all k/n = 1/a + 1/b + 1/c.

For HTML with images mode, this says "Missing cached output, contact an admin," but I can see it just fine in page images mode.

I can see the entry with no problems in both modes.


> I can see the entry with no problems in both modes.


I have noticed that, if the message "missing cached output" comes up for an entry that I own, it always goes away when I rerender.

it takes a long, long time to digest the following argument:

i believe that (O)dd (P)erfect (N)umbers do NOT exist in the same
manner as (E)ven (P)erfect (N)umbers do!

if we allow the generic formula of (k^(m-1))*(k^m - (k-1)) to de-
fine the situation, then we will see that it encompasses both the
proven EPN(n) == (2^(n-1))*(2^n -1) and a closely-related possible
OPN(n) == (3^(n-1))*(3^n -2).

however, we can't be satisfied this approach, since these two in-
stances of a preferred formula don't have comparable 'pole' posi-
tions when 'n' is set equal to zero.

if we let n= 0, then the 'pole' for EPN(0) == (1/2) *(0) remains
unshifted from the origin, and the 'pole' of OPN(0) == (1/3) *(-1)
is shifted by a (-1). comparing the results with respect to per-
fection between these two instances is already suspect from the

after a closer examination, though, the formula (3^(n-1))*(3^(n+1)
-2) turns out to be a more appropriate function but with a little
twist involved; if we let n= 0, the 'pole' of OPN(0) == (1/3) *(1)
is shifted by a (+1). we'll use this fact to later offset our clo-
sest OPN formula by a (-1) to bring the two 'poles' into agreement.

now, a 'proper-incipient' EPN is generated when the exponent 'n' is
less than the value inside the portion of the second parentheses of
its generic formula; n= 1 implies that EPN(1) == (1)*(1), but it's
improper since n= 1 is not less than the second (1); n= 2 implies
that EPN(2) == (2)*(3) and since n= 2< (3), we quickly discover that
6 = 1 *2 *3 = 1 +2 +3 = 6 (is an unshifted 'proper-incipient' char-
acteristic that should be shared by the preferred OPN function).

if n= 1, then the preferred OPN formula gives an OPN(1) == (1)*(7)
and 1< (7). so, the OPN(1) would share the proper 'incipient' qua-
lity of its cousin EPN, but not without an adjustment to its value
with an offset of (-1) to combat the earlier mentioned (+1) shift;
allowing for this little twist on the OPN's existance, the OPN's be-
come not only odd, but a bit quirky!

if we subtract '1' from the summand and allow the use of an 'impro-
per' divisor, then we have OPN(1) == (1)*(7) = (-1) + ((1) +7) = 7
(offsetting it), and the incipient OPN presents itself; but it was
not created using the expected 'EPN' axiom. it's the only OPN attach
ed to k= 3 from the preferred OPN formula, but not the only OPN...

if k= 5 and n= 1, then 21 isn't the next OPN, but...

finally, if all OPN's are == (2^(n-1))*(2^(n+1) -1), or (3^(n-1))*(3
^(n+1) -2), or (5^(n-1)) *(5^(n+1) -4), or (7^(n-1))*(7^(n+1) -6), or
(11^(n-1))*(11^(n+1) -10), or (13^(n-1))* (13^(n+1) -12), etc., then
we can see that the sequence of OPN's intermittently emerge as 3, 7,
(not 21), 43, (not 111), 157, etc.

the OPN's are not governed by the same rules as their cousin EPN's &
are not the expected 'gigantic' numbers that contain a multitude of
factors, but actually a set of individual (must be) prime numbers
that are generated by a continually changing formula and by modify-
ing the calculations of both the product and summations due to a
shift from their respective 'pole' positions.

thus, OPN's could barely exist but not in the usual EPN sense... they
are quirky! at best, the term OPN is a fictional misnomer, because
all possible formulas for an OPN lack the offset that is necessary to
coincide with the zero 'pole' of an already Euler-proven EPN formula
which also share the 'proper-incipient' characteristic.

the definition has to be changed... re-read it!
Bill Bouris

I haven't digested your argument yet, but I'm pretty sure that without a specific example, your arguments lack the force of proof even if they are very convincing.

fan or not,
it's a simple arrangement; without changing the definition for an odd perfect number, the argument for an OPN doesn't present itself.

It's unlikely that a lot of people here will take the time to go through your argument because it is rather difficult to read. It's not clear to me what you are asserting...there is no "definition" of an odd perfect number. A positive integer $n$ is a "perfect number" if $n=\sum_{d\mid n}d-n$. An odd perfect number is then just an odd integer satisfying this condition...there is no "definition" to change. No one is claiming that the odd perfect numbers, if they exist, must adhere to the same kind of rigid formula that the even perfect numbers do.

I started to read it but fell asleep. That probably has more to do with my age than with leavemsg2's ability to write a page-turner.

How many points does he have?

I'm sorry that you don't enjoy my explanation.
my would-be website is...

Yahoo!, Groups, Science, Mathematics, Number Theory,
name... prime_plausibles

you need to re-read the argument; the odd perfect
numbers are off by the -1 shift from the proven Euler-like
summation portion; it's a clear explanation; you just don't
like what it has to offer. enjoy needling someone else.

Clearly the problem is the acceptance of socratic-deductive notation instead of inductive-empirical notation!

I have composed a simple, 1-page algebra proof. It is located at; it took approx. 25 revisions. tell a friend, a math friend, a fiend, a geek, an enemy... it's valid.

I suggest you use latex. It would make your proof legible.

I like the suggestion, but I didn't have to use any complicated symbols to establish this proof. All you have to do is... go to and read about it. thanks, Bill

"let a perfect number N= 1 +f1 +f2 + ... +f(n-1) +fn with the restriction that f1*fn=f2*f(n-1)= fi*fj= ... = N"

28=1+2+4+7+14 is a perfect number and it cannot be expressed as a sum of the above form.

Sorry. Yes it can.

Hi leavemsg2,
I'm afraid you didn't understand ustun's suggestion. It's not a matter of complicated symbols (and LaTeX it is not!) but by the great confusion arising from your proof.
Your starting point is to express a perfect number as
N = 1 + f_1 + f_2 + ... + f_{n-1} + f_n,
with the restriction (???)
f_1.f_n = f_2.f_{n-1} = ... f_i.f_j = ... = N, (1)
1 \leq i < j \leq n, n = 2m even (\leq = less or equal).
So you have
N = 1 + f_1 + f_2 + ... + f_m + f_{m+1} + ... + f_{2m-1} + f_{2m},
with restriction
f_1.f_{2m} = f_2.f_{2m-1} = ... = f_i.f_j = ... = f_m.f_{m+1} = N,
where i + j = 2m + 1, i.e. an odd number. You define,
f_i = q + k_i,
1 \leq i \leq 2m.
You claim q = 1 for all f_i = q + k_i, by induction.
Induction? We don't know, a priori, whether the number of odd perfect numbers are finite or infinite or, simply, they don't exist. Since m is arbitrary, you just only must prove that in
N = 1 + (q + k_1) + (q + k_2) + ... + (q + k_m) + (q + k_{m+1}) + ... + (q + k_{2m-1}) + (q + k_{2m}) = (q + k_m).(q + k_{m+1}),
necessarily q = 1, but it is readily seen that is impossible to prove that one from this equation without any additional conditions. But the most important thing is: what about with those possible odd perfect numbers (if any) not verifying your restriction?
Now, why ustun's suggestion about you using LaTeX? I think it arises because you wrote the following paragraph:
"if fi *fj = fk *fl where a= ki, b= kj, c= kk, and d= kl, then (1 +a)*(1 +b)= (1 +c)*(1 +d)
implies that both a +b +ab = c +d +cd and ab = cd; in the case of even-perfect numbers, let a = 2r +1; so, all a, b, c, and d's are distinct; but in the case of odd-perfect numbers, if a = 2*r, then b(2r +b) = b(c +d) and 2rb = cd which implies that 2rb +bb = bc +bd, and then cd +bb = bc +bd or... b = c; the ki's aren't distinct, and thus, the generic definition of perfect numbers has been violated!"
Here you define (from fi *fj = fk *fl) multiplication of subscripts, i.e. a= ki, b= kj, c= kk, and d= kl, and then you proceed to do certain algebraic manipulations with these product of subscripts, which obviously, requires a lot of explanation.
I don't wish discourage you, but frankly, this is not the way to attack this problem.

I'm sorry that you cannot follow such a straight-forward approach. The matter won't get any simpler than using algebraic equations in a plain text editor. The attack is sufficient and has been reduced to very simple notation... re-read it. Thanks for visiting my website!
bye, I think???,... Bill

Huh! What a good reply! Bye leave!

I corrected the faux pas in notation to avoid further criticism by math persons like yourself. The generic pairings should have been, and were stated in previous revisions as fi *f(n-i+1) = fj *f(n-j+1).
The argument is still valid... pardon me!

Thank you leavemsg2, that was positive. By the way, I'm not a math person, simply an engineer. Here, in PM, we try to find the truth, so simple like that.

Dear Group,

I have finally proven that 'odd-perfect numbers don't exist.'
The proof is only one page and is tied directly into Jacques
Touchard's 1953-paper which states that they must be of the
form 12m +1 or 36q +9; I discovered that they must be of the
form 8*(n^2) +6n +1 from Euler's work. My paper is expressed
in a tongue-in-cheek manner, but it reduces the 355-year-old
problem into only two possible answers-- 45 & 325. The tech-
nique uses algebra alone and can be enjoyed by both math and
non-math persons equally. Please visit the website...www(dot)
oddperfectnumbers(dot)com to see my rigorous solution.


Bill Bouris

sorry, the problem was proposed by Rene Descartes in 1638, so 2011 - 1638 = a 373 y/o problem solved; tell a few peoples... my bad...

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