Thread: sigma function help

1. sigma function help

I don't know where to start.

$\sigma(n)$ is the sum of the divisors of n, including n itself.

Prove that there are no solutions to $\sigma(n)=17$.

I was thinking of using the formula $\sigma(p^a)=\frac {p^{a+1}-1}{p-1}$ where p is prime and $a\ge1$ to show that it's impossible to get 17, but I'm not sure if this is the right track.

2. Hello,
Originally Posted by Pn0yS0ld13r
I don't know where to start.

$\sigma(n)$ is the sum of the divisors of n, including n itself.

Prove that there are no solutions to $\sigma(n)=17$.

I was thinking of using the formula $\sigma(p^a)=\frac {p^{a+1}-1}{p-1}$ where p is prime and $a\ge1$ to show that it's impossible to get 17, but I'm not sure if this is the right track.
Let $n=p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k}$

Hence we know that $\sigma(n)=\prod_{i=1}^k \sigma \left(p_i^{a_i}\right)=\prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1}$

For $\sigma(n)$ to be equal to 17, 17 must be equal to one of the factors, let's say $\frac{p_1^{a_1+1}-1}{p_1-1}$, because 17=17*1 is the only factorisation for 17.
So it means that the others factors are equal to 1, that is to say $\forall i \ne 1 \quad \frac{p_i^{a_i+1}-1}{p_i-1}=1 \quad \Rightarrow \quad a_i+1=1 \quad \text{that is to say } a_i=0$. Thus $n=p_1^{a_1}$ since all the other powers are 0. (this is an extra part lol)

Let's go back to the problem.
We want $17=\frac{p_1^{a_1+1}-1}{p_1-1}=1+p_1+p_1^2+\dots+p_1^{a_1}$

$\implies p_1+p_1^2+\dots+p_1^{a_1}=16$

This can be factorised : $p_1(1+p_1+\dots+p_1^{a_1-1})=16$

But the only prime number that divides 16 is 2. Therefore $p_1=2$.

Thus $1+p_1+\dots+p_1^{a_1-1}=8$, which is impossible because the left hand side is an odd number and the right hand side is an even number.

We're done