# The function o(n)

#### 1337h4x

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?

#### chiph588@

MHF Hall of Honor
I don't have time to solve this symbolically but I will tell you this:

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

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

#### 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!

#### chiph588@

MHF Hall of Honor
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?

#### Drexel28

MHF Hall of Honor
I don't have time to solve this symbolically but I will tell you this:

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

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

• chiph588@

#### chiph588@

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

#### Drexel28

MHF Hall of Honor
I believe this argument shows there is no $$\displaystyle n$$ such that $$\displaystyle \sigma(n)=2^m+1$$ where $$\displaystyle m>1$$ and $$\displaystyle 2^m+1$$ is prime.
Agreed!

#### chiph588@

MHF Hall of Honor
For the other problem try this:

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

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

So $$\displaystyle n=p^m$$ a prime power means $$\displaystyle 15=p^m+\cdots+1$$. Thus $$\displaystyle 14=p^m+\cdots+p=p\cdot(p^{m-1}+\cdots+1)$$ which forces $$\displaystyle p=2$$ or $$\displaystyle 7$$. Next see that $$\displaystyle p=2$$ 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.

Last edited:

#### 1337h4x

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

#### chiph588@

MHF Hall of Honor
Why is that ridiculous? I'm failing to recognize this...
Well one side is even and the other side is odd.