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Prove that the sum of all divisors of a natural number $n$ is odd if and only if $n = 2^r \cdot k^2$ where $k$ and $r$ are natural numbers.

The first direction: if $k^2$ is an even number, we rewrite $2^r \cdot k^2$ as $2^{(r+m)} \cdot z$ where $z$ is a natural odd number. Now, the sum of all divisors of $2^{(r+m)}$ without $1$ is even, and the sum of of all divisors of $z$ is odd, so we get odd.
This is true? To the other direction, I have no idea...

TMM
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GO VEGAN
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  • You will need to use the "square" part. The sum of the divisors of $15$ is even, as is the sum of the divisors of $27$. – André Nicolas Mar 28 '13 at 18:59
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    Take an odd number $n$ which is not a square, for example $45$. Pair together the divisors $a$ and $n/a$, like $1$ and $45$, $3$ and $15$, $5$ and $9$. The sum of any pair is even, so the sum of the divisors is even. If $n$ is a square, like $225$, then $\sqrt{n}$ remains unpaired. The others divisors pair and add up to an even number. The leftover $15$ makes the sum odd. The same idea works in general. – André Nicolas Mar 28 '13 at 19:26
  • The pairing of a divisor with its cofactor is a prototypical example of a parity information gleaned from pairings via involutions or reflections. It generalizes to various group theoretical versions of Wilson's Theorem. – Math Gems Mar 28 '13 at 20:26

2 Answers2

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A different approach:

Assume that the prime factorization of $n$ is $n=2^{a_0}p_1^{a_1}p_2^{a_2}\cdots p_s^{a_s}$ where $p_i$ are odd primes different with each other. Let's write $p_0=2$.
Any divisor of $n$ is of the form $p_0^{b_0}p_1^{b_1}p_2^{b_2}\cdots p_s^{b_s}$ with $0\leq b_i\leq a_i, \ \forall i$. Therefore the sum of the divisors of $n$ is $\sum p_0^{b_0}p_1^{b_1}p_2^{b_2}\cdots p_s^{b_s}$ where the sum range over all $b_i$ with $0\leq b_i\leq a_i, \ \forall i$.
Equivalently the sum can be written as a product: $$\sum_{0\leq b_i\leq a_i}p_0^{b_0}p_1^{b_1}p_2^{b_2}\cdots p_s^{b_s}=\prod_{i=0}^s(1+p_i+p_i^2+\ldots+p_i^{a_i}).$$

Now the sum of the divisors of $n$ is odd iff $1+p_i+p_i^2+\ldots+p_i^{a_i}$ is odd for all $i=1,2,\ldots, s$. For $i=0, \ 1+p_i+p_i^2+\ldots+p_i^{a_i}$ is odd. For $i>0$ (i.e. $p_i$ odd) when is $1+p_i+p_i^2+\ldots+p_i^{a_i}$ odd?

If $n=2^rk^2$ what does this mean for the $a_i$? Conversely for what $a_i$, $n$ can be written as $2^rk^2?$

P..
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Hint: all divisors of an odd number are odd, so you need to have an odd number of them. Can you see why $k^2$ has an odd number of factors? The sum of divisors function is multiplicative, so you have taken care of $2^{(r+m)}$, you can consider $k$ to be odd.

Ross Millikan
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