MHB Prove that n is not divisible by 105.

  • Thread starter Thread starter lfdahl
  • Start date Start date
AI Thread Summary
The discussion centers on proving that an odd perfect number, denoted as n, cannot be divisible by 105. The number 105 factors into 3, 5, and 7, and the argument suggests examining the properties of divisors and their sums in relation to these prime factors. It highlights that if n were divisible by 105, it would contradict the conditions necessary for n to be an odd perfect number. Additionally, the existence of odd perfect numbers remains an open question in number theory. The conclusion emphasizes the need for further exploration into the properties of odd perfect numbers and their divisibility.
lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Problem: Suppose that a natural number $n$ is an odd perfect number, i.e.:

$n$ is odd and $n$ is equal to the sum of all its positive divisors (including $1$ and excluding $n$).Prove that $n$ is not divisible by $105$.P.S.: To this day, no one knows, whether such a number exists. Here
is a comment on the subject from Wolfram Mathworld.
 
Last edited:
Mathematics news on Phys.org
Suggested solution:
Suppose that $n$ is divisible by $105 = 3 \cdot 5 \cdot 7$. Consider the prime factorization of $n$:

\[n = 3^{\alpha_1}\cdot 5^{\alpha_2 } \cdot 7^{\alpha_3}\cdot p_4^{\alpha_4}\cdot ...\cdot p_k^{\alpha_k},\: \: \: \alpha_1,\alpha_2 , \alpha_3 \geq 1.\]

Let $S(n)$ be the sum of all positive divisors of $n$ (including $1$ and $n$):

\[S(n)= n\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )\left ( 1+\frac{1}{5}+...+\frac{1}{5^{\alpha_2}} \right )\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )\left ( 1+\frac{1}{p_4}+...+\frac{1}{p_4^{\alpha_4}} \right )...\left ( 1+\frac{1}{p_k}+...+\frac{1}{p_k^{\alpha_k}} \right )\]

Since $n$ is an odd perfect number $S(n) = 2n$ and $S(n)$ is not divisible by $4$. Since all primes in the decomposition of $n$ are odd, $\alpha_1$ and $\alpha_3$ are at least $2$, - otherwise

$\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )=\frac{4}{3}$ and $\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )=\frac{8}{7}$ and $S(n)$ is divisible by $4$. Finally:
\[2 = \frac{S(n)}{n}= \left ( 1+\frac{1}{3}+\frac{1}{3^{2}} \right )\left ( 1+\frac{1}{5}\right )\left ( 1+\frac{1}{7}+\frac{1}{7^2} \right )=\frac{13}{9}\cdot \frac{6}{5} \cdot \frac{57}{49}= \frac{4446}{2205}>2.\]
Contradiction.
 
lfdahl said:
Suggested solution:
Suppose that $n$ is divisible by $105 = 3 \cdot 5 \cdot 7$. Consider the prime factorization of $n$:

\[n = 3^{\alpha_1}\cdot 5^{\alpha_2 } \cdot 7^{\alpha_3}\cdot p_4^{\alpha_4}\cdot ...\cdot p_k^{\alpha_k},\: \: \: \alpha_1,\alpha_2 , \alpha_3 \geq 1.\]

Let $S(n)$ be the sum of all positive divisors of $n$ (including $1$ and $n$):

\[S(n)= n\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )\left ( 1+\frac{1}{5}+...+\frac{1}{5^{\alpha_2}} \right )\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )\left ( 1+\frac{1}{p_4}+...+\frac{1}{p_4^{\alpha_4}} \right )...\left ( 1+\frac{1}{p_k}+...+\frac{1}{p_k^{\alpha_k}} \right )\]

Since $n$ is an odd perfect number $S(n) = 2n$ and $S(n)$ is not divisible by $4$. Since all primes in the decomposition of $n$ are odd, $\alpha_1$ and $\alpha_3$ are at least $2$, - otherwise

$\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )=\frac{4}{3}$ and $\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )=\frac{8}{7}$ and $S(n)$ is divisible by $4$. Finally:
\[2 = \frac{S(n)}{n}= \left ( 1+\frac{1}{3}+\frac{1}{3^{2}} \right )\left ( 1+\frac{1}{5}\right )\left ( 1+\frac{1}{7}+\frac{1}{7^2} \right )=\frac{13}{9}\cdot \frac{6}{5} \cdot \frac{57}{49}= \frac{4446}{2205}>2.\]
Contradiction.
why $S(n)=2n\, ?$ ,can you give me an example ?
 
Albert said:
why $S(n)=2n\, ?$ ,can you give me an example ?

Yes: Take $n = 6$, which is an even perfect number, because the sum of its divisors (except for $n$) is:

$1+2+3 = 6$. The proof defines $S(n)$ as the sum of all divisors of $n$ including $n$ itself.

Thus: $S(6) = 1+2+3+6 = 12 = 2 \cdot 6$.

I would like to give you an example with $n$ odd, but this would be a hard task, since math researchers have shown, that the
smallest possible odd perfect number $n > 10^{1500}$.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top