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?

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?

\(\displaystyle \sigma(n)>n \) 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's unclear to you?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 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?

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

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

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.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?

Agreed!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.

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.

\(\displaystyle 15=3\cdot5 \) 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.

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Why is that ridiculous? I'm failing to recognize this...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?

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