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Number theory

σ(n)=dn(d),τ(n)=dn(1) \sigma(n) = \sum_{d \vert n} (d), \tau(n) = \sum_{d \vert n} (1)

If σ(n)=2n \sigma(n) = 2n show that τ(n)0mod(2) \tau(n) \equiv 0 \mod(2)

We're allowed to assume multiplicativity of both functions.
I've found:
τ(n)=iτ(piei)=i(ei+1) \tau(n) = \prod_{i} \tau(p_i^{e_i}) = \prod_{i}(e_i + 1)
And,
σ(n)=iσ(piei)=i(1+pi++piei)=i(1piei+11pi)=2n[br]=ni1pieipi1pi[br]i1pieipi1pi=2[br] \sigma(n) = \prod_i \sigma(p_i^{e_i}) = \prod_i(1 + p_i + \dots + p_i^{e_i}) = \prod_{i}(\dfrac{1-p_i^{e_i+1}}{1-p_i}) = 2n [br]= n \cdot \prod_{i} \dfrac{\dfrac{1}{p_i^{e_i}} - p_i}{1 - p_i} [br]\Rightarrow \prod_{i} \dfrac{\dfrac{1}{p_i^{e_i}} - p_i}{1 - p_i} = 2 [br]

But then for exactly one p_i:

1pkekpk1pk=2 \dfrac{\dfrac{1}{p_k^{e_k}} - p_k}{1 - p_k} = 2

But this has no solutions.


Clearly I'm blundering somewhere... I also get the feeling that this is completely the wrong direction. Any help would be appreciated.

I should say my strategy was to show that n is of the form:

n=2k1(2k1) n = 2^{k-1}(2^k - 1)
(edited 5 years ago)
Original post by Ryanzmw
σ(n)=dn(d),τ(n)=dn(1) \sigma(n) = \sum_{d \vert n} (d), \tau(n) = \sum_{d \vert n} (1)

If σ(n)=2n \sigma(n) = 2n show that τ(n)0mod(2) \tau(n) \equiv 0 \mod(2)

We're allowed to assume multiplicativity of both functions.
I've found:
τ(n)=iτ(piei)=i(ei+1) \tau(n) = \prod_{i} \tau(p_i^{e_i}) = \prod_{i}(e_i + 1)
And,
σ(n)=iσ(piei)=i(1+pi++piei)=i(1piei+11pi)=2n[br]=ni1pieipi1pi[br]i1pieipi1pi=2[br] \sigma(n) = \prod_i \sigma(p_i^{e_i}) = \prod_i(1 + p_i + \dots + p_i^{e_i}) = \prod_{i}(\dfrac{1-p_i^{e_i+1}}{1-p_i}) = 2n [br]= n \cdot \prod_{i} \dfrac{\dfrac{1}{p_i^{e_i}} - p_i}{1 - p_i} [br]\Rightarrow \prod_{i} \dfrac{\dfrac{1}{p_i^{e_i}} - p_i}{1 - p_i} = 2 [br]

But then for exactly one p_i:

1pkekpk1pk=2 \dfrac{\dfrac{1}{p_k^{e_k}} - p_k}{1 - p_k} = 2

But this has no solutions.


Clearly I'm blundering somewhere... I also get the feeling that this is completely the wrong direction. Any help would be appreciated.

I should say my strategy was to show that n is of the form:

n=2k1(2k1) n = 2^{k-1}(2^k - 1)
I don't remember this stuff well, but σ(n)=2n\sigma(n) = 2n is the definition of a perfect number. It is known that if n is a perfect even number it must take the form you describe (and 2^k - 1 must also be prime), but the question of existence of odd perfect numbers is unsolved, so I don't think this is a fruitful direction to persue!

It looks to me it would be more fruitful to note that τ(n)\tau(n) is odd iff n is a perfect square (and then show if n is a perfect square than σ(n)2n\sigma(n) \neq 2n), although this seems to me unlikely to make use of multiplicativity (so probably not what's intended, given the hint).

Edit: A little more thought tells me the last paragraph does work, and does use multiplicativity.

Spoiler

(edited 5 years ago)
Reply 2
Original post by DFranklin
I don't remember this stuff well, but σ(n)=2n\sigma(n) = 2n is the definition of a perfect number. It is known that if n is a perfect even number it must take the form you describe (and 2^k - 1 must also be prime), but the question of existence of odd perfect numbers is unsolved, so I don't think this is a fruitful direction to persue!

It looks to me it would be more fruitful to note that τ(n)\tau(n) is odd iff n is a perfect square (and then show if n is a perfect square than σ(n)2n\sigma(n) \neq 2n), although this seems to me unlikely to make use of multiplicativity (so probably not what's intended, given the hint).

Edit: A little more thought tells me the last paragraph does work, and does use multiplicativity.

Spoiler




That helped a lot, thanks!

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