MHB Prove that n is not divisible by 105.

[lfdahl]

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.

 

[lfdahl]

Suggested solution:

Spoiler

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.

 

[Albert1]

Suggested solution:

Spoiler

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.

Spoiler

why $S(n)=2n\, ?$ ,can you give me an example ?

 

[lfdahl]

Spoiler

why $S(n)=2n\, ?$ ,can you give me an example ?

Spoiler

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 $n > 10^{1500}$.