Question

Trying to work out a proof that there are no odd perfect numbers that are of the form 25*n where gcd(25,n)=1.

Answer 1

I had been reading about this thing called the Goormaghtigh conjecture and found this to be true:

\[1+5+25 = 1+2+4+8+16\]
in other words,
\[\sigma(25)=\sigma(16)\]
and realized what if we multiply by some number that is relatively prime to both?
\[\sigma(25n)=\sigma(16n)\]

My original false proof was that since:

\[16*(2^4-1)\]
is not a perfect number (which is the only possible candidate)
then \[25*(2^4-1)\]
is also not a perfect number and neither is any choice of n for 25*n.

But this is clearly wrong!

Answer 2

link to the tiny thing I was reading: https://en.wikipedia.org/wiki/Goormaghtigh_conjecture

Eh, I guess it's probably impossible to prove, oh well.

Answer 3

http://unsolvedproblems.org/index_files/OddPerfect.htm ;-;

Answer 4

I'm not trying to prove the general case, just a special case of odd numbers that can't be perfect numbers! For example, no prime number is a perfect number is very easy to show!

\[\sigma(p) = 1+p \ne 2p\]

Answer 5

yes

Answer 6

n can't be prime and it also can't end with 5 or 0

Answer 7

\(\Large \sigma(5^{ 2n })=\dfrac{5^{2n+1}-1}{5-1}\)

we just need to show this is odd (wew!)

Answer 8

Ohhh actually that's brilliant

Answer 9

I think that also shows \(p^2\) is never a perfect number

Answer 10

We could even say in general \(\sigma((2n+1)^{2k})\) is never a perfect number

Answer 11

I think...?

Answer 12

**
I think that also shows \(p^n\) is never a perfect number

Answer 13

Hmmm how so? I guess I should explain my reasoning for my claim and see if I run into trouble in my proof, one sec.

Answer 14

25*n

2 of the factors will be 5 and 1
5+1=6

but to be perfect the sum of all factors shuld be 25+n
25+n=6 +(sum of factors of n except 1)
18+n=sum of factors of n

does there exist any such n whos sum of factors is more than its magnitude?

Answer 15

\((2n+1)^{2k}\) is an odd square raised to any power, so all of its prime factors must be odd primes raised to even powers as well, so we can write it as:

\[\sigma((2n+1)^{2k})=\sigma(p_1^{2a})\sigma(p_2^{2b})\cdots\]
Then we see that:

\[\sigma(p^2)=1+p+p^2\] which is odd+odd+odd = odd and every higher power will always add an even to this and even +odd = odd so all of this stuff will be odd... which basically proves it.

Answer 16

this is kinda simple :3 hi5 @Kainui

Answer 17

Now we just have to prove there are no perfect numbers of the form \((2n+1)^{2k+1}\) and we proved there are no odd perfect numbers... LOL

Answer 18

wait no :(

Answer 19

hahahaha
all odds means 2n+1

Answer 20

Ok so that would be when k=0 then! hahah xD

Also @imqwerty I'm not sure I follow your reasoning could you explain that more what you're doing?

Answer 21

Specifically here:
"but to be perfect the sum of all factors shuld be 25+n"

I am thinking the sum of all factors should be (1+5+25)*(factors of n) = 2*25*n

Answer 22

sry i made a mistake there

Answer 23

see this- http://prntscr.com/943w0z

Answer 24

Hmmm are these conjectures? I am pretty sure these are not all true, but I'm not sure.

Answer 25

http://www-history.mcs.st-and.ac.uk/HistTopics/Perfect_numbers.html

Answer 26

"We will see how these assertions have stood the test of time as we carry on with our discussions, but let us say at this point that assertions (1) and (3) are false while, as stated, (2), (4) and (5) are still open questions."

Answer 27

If there were infinitely many perfect numbers and it was also proven there were only even perfect numbers, then that would prove there are infinitely many Mersenne primes which is cool. :P

Answer 28

can there be any number whos sum of factors except 1 and itself is more than it?

Answer 29

my method is based on this :)
if we can prove this its done

Answer 30

In terms of the divisor function we can write that condition as do any n satisfy this:

\[n > \sigma(n)-n-1\]
in other words,
\[2n+1 > \sigma(n)\]

it looks like this means every abundant number satisfies this condition. Quick check using 12 the first abundant number:

\[2+3+4+6 > 12\]
Yep, so this is a counter example I guess

Answer 31

also now I realize that ganeshie's idea is really more general than what I said I don't know what Iw as thinking lol

Answer 32

I just realized something fun:
\[d |n\]
This just means d divides n, then if this is true then:

\[\sigma(p^{d-1}) | \sigma(p^{n-1})\]
which might be useful to play around with