How to make the Tau function odd

[napoleon67]

Homework Statement

What are the necessary conditions on n for T(n) {the Tau function} to be odd.

Homework Equations

Tau function counts the number of divisors of n
T(1) = 1, T(p) = 2, and T(p^k) = k + 1

The Attempt at a Solution

I have that n cannot be prime, also if n = p^k then k must be even or of the from k = 2c for some c > 0.

I cannot figure out how to describe evens with were are multiple primes such as 1x3x7x8 = 231. That number has an even number of divisors but I'm and lost at how to prevent that.

 

[matt grime]

I don't understand why k is even OR k=2c. Those are identical conditions...What is T(xy) for x,y coprime?

 

[napoleon67]

I meant k is even i.e k =2c... I was just adding more definition.

T(xy) for x,y is coprime is an even number of divisors. Thanks.

 

[AKG]

I meant k is even i.e k =2c... I was just adding more definition.

T(xy) for x,y is coprime is an even number of divisors. Thanks.

No, that's not right.

 

[napoleon67]

T(xy) for x,y is coprime means T(x)T(y)

 

[napoleon67]

Is is sufficient to say that T(n) is only odd when n is a perfect square?

 

[AKG]

T(n) is odd if and only if n is a perfect square. Can you prove it?

 

[napoleon67]

I was just supposed to make a conjecture, but I'll try to work on a proof

 

[napoleon67]

Assume that n is an odd tau number. Let n = p^a1*p^a2...* p^ak. By the definition of tau number (a1 + 1)(a2 + 1)(a3 + 1)... (ak + 1) | n. Therefore for any 0 < i < k + 1, ai + 1 is odd, and hence ai is even. Since every prime in the factorization of n is raised to an even power, n is a perfect square.

 

[AKG]

Assume that n is an odd tau number. Let n = p^a1*p^a2...* p^ak. By the definition of tau number (a1 + 1)(a2 + 1)(a3 + 1)... (ak + 1) | n.

What you meant to say was

$$(a_1 + 1)(a_2 + 1)(a_3 + 1)\dots (a_k + 1) = \tau (n)$$

not

$$(a_1 + 1)(a_2 + 1)(a_3 + 1)\dots (a_k + 1) | n$$

Therefore for any 0 < i < k + 1, ai + 1 is odd, and hence ai is even. Since every prime in the factorization of n is raised to an even power, n is a perfect square.

The question only asked you to find a necessary condition, and you've done this. But the stuff you've said above goes both ways, so it's actually a sufficient condition as well.