Proof in number theory: the sum of all divisors of n

[mathstudent1]

let n be a positive integer show that if n is square then σ(n)( the sum of all divisors of n )is odd.

 

[fresh_42]

Do you have any ideas? What do you know about $\sigma (n)$?

Before you start proving something you must make an inventory. E.g. there is a formula for $\sigma (n)$ if $n$ is given by its prime factor decomposition. If you can use that, then the proof is probably a lot easier as if you do not know this formula.

 

[mathstudent1]

Do you have any ideas? What do you know about $\sigma (n)$?

Before you start proving something you must make an inventory. E.g. there is a formula for $\sigma (n)$ if $n$ is given by its prime factor decomposition. If you can use that, then the proof is probably a lot easier as if you do not know this formula.

you mean this σ(p^a)=p^(a+1) -1/(p-1)? I have another question how can I write the form of square in the form of prime?

 

[fresh_42]

you mean this σ(p^a)=p^(a+1) -1/(p-1)? I have another question how can I write the form of square in the form of prime?

I meant a more complicated formula with more primes.

Let's start with $n$ and what do we know. Say $n=p_1^{r_1}\cdot p_2^{r_2}\cdots p_k^{r_k}$ with pairwise distinct primes $p_1,\ldots,p_k$ and positive integers $r_1,\ldots,r_k.$

What does it mean that $n$ is a square?

 

[mathstudent1]

I meant a more complicated formula with more primes.

Let's start with $n$ and what do we know. Say $n=p_1^{r_1}\cdot p_2^{r_2}\cdots p_k^{r_k}$ with pairwise distinct primes $p_1,\ldots,p_k$ and positive integers $r_1,\ldots,r_k.$

What does it mean that $n$ is a square?

n^2=p1^2r1*p2^2r2*..........*p^2rk?

 

[fresh_42]

Yes, all powers of the prime factors are even. Btw., here is how to write formulas here at PF:
https://www.physicsforums.com/help/latexhelp/

Let's start easy: what is $\sigma(p^{2r})$? Then how do we get to $\sigma (p_1^{2r_1}\cdot p_2^{2r_2})$?

 

[mathstudent1]

Yes, all powers of the prime factors are even. Btw., here is how to write formulas here at PF:
https://www.physicsforums.com/help/latexhelp/

Let's start easy: what is $\sigma(p^{2r})$? Then how do we get to $\sigma (p_1^{2r_1}\cdot p_2^{2r_2})$?

Do you mean we can use this form p^a+1 -1/p-1 ? and since sigma is a multiplicative function, we can distribute the sigma?

 

[fresh_42]

Do you mean we can use this form p^a+1 -1/p-1 ? and since sigma is a multiplicative function, we can distribute the sigma?

Use the formula you gave us:
$$
\sigma (p^a)=\dfrac{p^{a+1}-1}{p-1}
$$
All we know is that $a$ is even. This is not much. Can you perform the long division $(p^{a+1}-1)\, : \,(p-1)$?

Proceed step by step. If we know that it is true with one prime, then we can use multiplicativity, yes.

 

[mathstudent1]

Use the formula you gave us:
$$
\sigma (p^a)=\dfrac{p^{a+1}-1}{p-1}
$$
All we know is that $a$ is even. This is not much. Can you perform the long division $(p^{a+1}-1)\, : \,(p-1)$?

Proceed step by step. If we know that it is true with one prime, then we can use multiplicativity, yes.

so we can use this form? 1+p+P^2+.......+p^a?

 

[fresh_42]

so we can use this form? 1+p+P^2+.......+p^a?

Yes. How many terms are there? You probably have to distinguish between odd and even $p$ if you want to answer whether the sum is even or odd.

What is $1+p+p^2+\ldots+p^a \pmod{2}$?

 

[mathstudent1]

Yes. How many terms are there? You probably have to distinguish between odd and even $p$ if you want to answer whether the sum is even or odd.

a+1 terms and since a must be even we will have even+1= odd and since sigma is multiplicative, we can distribute the sigma so we have this odd*odd*.....= odd that is why sigma is odd right?

 

[fresh_42]

a+1 terms and since a must be even we will have even+1= odd and since sigma is multiplicative, we can distribute the sigma so we have this odd*odd*.....= odd that is why sigma is odd right?

That's the idea but you must be a bit more precise. You should write it like

\begin{align*}
\sigma (p^{2r})=\dfrac{p^{2r+1}-1}{p-1}=\underbrace{1+p^1+p^2+\ldots+p^{2r}}_{2r+1\text{ terms}}
=\begin{cases}
\equiv 1 &\text{ if } p\equiv 0 \pmod{2}\\
\equiv (2r+1)\cdot 1\equiv 1 &\text{ if } p\equiv 1 \pmod{2}
\end{cases}
\end{align*}

because it has two different reasons why the sum is odd depending on whether $p=2$ or not.

Finally, write it step by step.

\begin{align*}
n&=p_1^{r_1}\cdot\ldots\cdot p_k^{r_k}\text{ with pairwise distinct primes }p_i \text{ and } r_i\equiv 0\pmod{2}\\
\sigma (n)&=\sigma (p_1^{r_1}\cdot\ldots\cdot p_k^{r_k})\\
&=\sigma (p_1^{r_1})\cdot\ldots\cdot \sigma (p_k^{r_k})\text{ by multiplicativity of }\sigma \\
&\equiv \underbrace{1\cdot 1\ldots \cdot 1}_{k\text{-times }}\\
&\equiv 1\pmod{2}
\end{align*}

You have had all the ingredients (the formula for $p^a$, the multiplicativity of $\sigma$, and the result of the long division), you only had to put them into the right order. A general guideline is to start with a list of what you have and make conclusions step by step.

Not sure whether this helps (I have written better ones) ...
https://www.physicsforums.com/insights/how-most-proofs-are-structured-and-how-to-write-them/

 

[mathstudent1]

That's the idea but you must be a bit more precise. You should write it like

\begin{align*}
\sigma (p^{2r})=\dfrac{p^{2r+1}-1}{p-1}=\underbrace{1+p^1+p^2+\ldots+p^{2r}}_{2r+1\text{ terms}}
=\begin{cases}
\equiv 1 &\text{ if } p\equiv 0 \pmod{2}\\
\equiv (2r+1)\cdot 1\equiv 1 &\text{ if } p\equiv 1 \pmod{2}
\end{cases}
\end{align*}

because it has two different reasons why the sum is odd depending on whether $p=2$ or not.

Finally, write it step by step.

\begin{align*}
n&=p_1^{r_1}\cdot\ldots\cdot p_k^{r_k}\text{ with pairwise distinct primes }p_i \text{ and } r_i\equiv 0\pmod{2}\\
\sigma (n)&=\sigma (p_1^{r_1}\cdot\ldots\cdot p_k^{r_k})\\
&=\sigma (p_1^{r_1})\cdot\ldots\cdot \sigma (p_k^{r_k})\text{ by multiplicativity of }\sigma \\
&\equiv \underbrace{1\cdot 1\ldots \cdot 1}_{k\text{-times }}\\
&\equiv 1\pmod{2}
\end{align*}

You have had all the ingredients (the formula for $p^a$, the multiplicativity of $\sigma$, and the result of the long division), you only had to put them into the right order. A general guideline is to start with a list of what you have and make conclusions step by step.

Not sure whether this helps (I have written better ones) ...
https://www.physicsforums.com/insights/how-most-proofs-are-structured-and-how-to-write-them/

thank you