# Thread: The function o(n)

1. ## The function o(n)

Hello all, I have questions about the function o(n) as well. o(n) is the sum of the divisors of n including n itself.

1a. First I need the find the n for which o(n) = 15, and then prove that n is unique. How do you accomplish this ?

1b. Similarly, how can I prove that no solutions exist to the equation o(n) = 17?

2. I don't have time to solve this symbolically but I will tell you this:

$\sigma(n)>n$ so we only have finitely many numbers to check.

Scanning through we get $\sigma(8)=15$ and no numbers satisfy $\sigma(n)=17$.

3. I'm sorry, but if someone can provide a little clearer and more thorough explanation for both of my questions, I'd really appreciate it!

4. Originally Posted by 1337h4x
I'm sorry, but if someone can provide a little clearer and more thorough explanation for both of my questions, I'd really appreciate it!
What's unclear to you?

5. Originally Posted by chiph588@
I don't have time to solve this symbolically but I will tell you this:

$\sigma(n)>n$ so we only have finitely many numbers to check.

Scanning through we get $\sigma(8)=15$ and no numbers satisfy $\sigma(n)=17$.
What about this? Let $n\in\mathbb{N}$ be arbitrary. Then, $n=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$ and so $\sigma(n)=\prod_{j=1}^{m}\frac{p_m^{\alpha_m+1}-1}{p_m-1}$. So, if $\sigma(n)=17$ it follows that $17=\frac{p_m^{\alpha_m+1}-1}{p-1}$ from it's factorization properties. But, this means that $17=1+\cdots+p_m^{\alpha_m}$ and thus $16=p_m^1+\cdots+p_m^{\alpha_m}$ but this means that $16=p_m(1+\cdots+p_m^{\alpha_m})$ and thus $p_m=2\implies 8=1+\cdots+p_m^{\alpha_m-1}$ which is ridicuoulous why?

6. Originally Posted by Drexel28
What about this? Let $n\in\mathbb{N}$ be arbitrary. Then, $n=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$ and so $\sigma(n)=\prod_{j=1}^{m}\frac{p_m^{\alpha_m+1}-1}{p_m-1}$. So, if $\sigma(n)=17$ it follows that $17=\frac{p_m^{\alpha_m+1}-1}{p-1}$ from it's factorization properties. But, this means that $17=1+\cdots+p_m^{\alpha_m}$ and thus $16=p_m^1+\cdots+p_m^{\alpha_m}$ but this means that $16=p_m(1+\cdots+p_m^{\alpha_m})$ and thus $p_m=2\implies 8=1+\cdots+p_m^{\alpha_m-1}$ which is ridicuoulous why?
I believe this argument shows there is no $n$ such that $\sigma(n)=2^m+1$ where $m>1$ and $2^m+1$ is prime.

7. Originally Posted by chiph588@
I believe this argument shows there is no $n$ such that $\sigma(n)=2^m+1$ where $m>1$ and $2^m+1$ is prime.
Agreed!

8. For the other problem try this:

$15=3\cdot5$ and the other solution shows there is no $n$ such that $\sigma(n)=5$.

This means if $\sigma(n)=15$ then $n$ must be a prime power, otherwise if $n=ab$ where $(a,b)=1$ then $\sigma(n)=\sigma(a)\sigma(b)=3\cdot5$ which means $\sigma(b)=5$, which is impossible.

So $n=p^m$ a prime power means $15=p^m+\cdots+1$. Thus $14=p^m+\cdots+p=p\cdot(p^{m-1}+\cdots+1)$ which forces $p=2$ or $7$. Next see that $p=2$ is the only option. Now you can find the exponent fairly easily.

This argument is a bit sloppy but I think I give enough info to piece it together.

9. Originally Posted by Drexel28
What about this? Let $n\in\mathbb{N}$ be arbitrary. Then, $n=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$ and so $\sigma(n)=\prod_{j=1}^{m}\frac{p_m^{\alpha_m+1}-1}{p_m-1}$. So, if $\sigma(n)=17$ it follows that $17=\frac{p_m^{\alpha_m+1}-1}{p-1}$ from it's factorization properties. But, this means that $17=1+\cdots+p_m^{\alpha_m}$ and thus $16=p_m^1+\cdots+p_m^{\alpha_m}$ but this means that $16=p_m(1+\cdots+p_m^{\alpha_m})$ and thus $p_m=2\implies 8=1+\cdots+p_m^{\alpha_m-1}$ which is ridicuoulous why?
Why is that ridiculous? I'm failing to recognize this...

10. Originally Posted by 1337h4x
Why is that ridiculous? I'm failing to recognize this...
Well one side is even and the other side is odd.