1. ## The function d(n)

Hello everyone,

I have some questions about the function d(n) which gives the number of positive divisors of n including n itself.

1. How can it be proven that d(n) is odd IFF n is a perfect square? (Kind of like proving that d(n) is odd iff n = k^2 for some integer k.

2. How is the following statement true: The product of all of the positive divisors of n (including n itself) is n^[d(n)/2] . Is there a proof for this ?

Any help is appreciated!

2. Originally Posted by 1337h4x
Hello everyone,

I have some questions about the function d(n) which gives the number of positive divisors of n including n itself.

1. How can it be proven that d(n) IFF n is a perfect square? (Kind of like proving that d(n) is odd iff n = k^2 for some integer k.

2. How is the following statement true: The product of all of the positive divisors of n (including n itself) is n^[d(n)/2] . Is there a proof for this ?

Any help is appreciated!
1. That's not even a sentence! $d(n)$ what if and only if (...)?

2. Look at the power with which a prime dividing $n$ appears in the canonical factorization of the resulting product.

3. Originally Posted by Bruno J.
1. That's not even a sentence! $d(n)$ what if and only if (...)?

2. Look at the power with which a prime dividing $n$ appears in the canonical factorization of the resulting product.
I fixed it, It should say d(n) is odd IFF ...

Anyways, what do you mean by your point of 2? Can you also explain my first question?

4. Can anyone help with this? I'm struggling to understand this.

5. Can anyone please explain the second question at least? I'm really stuck on this concept.

6. For all parts, note that if $d|n$ then $\left(\tfrac{n}{d}\right)|n$, the idea would be the following:

For each divisor $d<\sqrt{n}$ there's a corresponding one $\tfrac{n}{d}>\sqrt{n}$, the only divisor that is not counted here is $d = \tfrac{n}{d}=\sqrt{n}$ in case of this being possible (that is, n is a square).

Thus if $n$ is not a square the divisors come in pairs, hence the number of divisors in that case is even, and in the case of n being a square it is odd (we add one).

For the second part use the above idea and note that $d\cdot \left(\tfrac{n}{d}\right)=n$ -and now try to use it to calculate the product of the divisors-

7. Originally Posted by PaulRS
For all parts, note that if $d|n$ then $\left(\tfrac{n}{d}\right)|n$, the idea would be the following:

For each divisor $d<\sqrt{n}$ there's a corresponding one $\tfrac{n}{d}>\sqrt{n}$, the only divisor that is not counted here is $d = \tfrac{n}{d}=\sqrt{n}$ in case of this being possible (that is, n is a square).

Thus if $n$ is not a square the divisors come in pairs, hence the number of divisors in that case is even, and in the case of n being a square it is odd (we add one).

For the second part use the above idea and note that $d\cdot \left(\tfrac{n}{d}\right)=n$ -and now try to use it to calculate the product of the divisors-

What happens if you calculate the product of the divisors? I can't get this to related to n^(d(n)/2)....

8. Wait, my problem involved d(n), your solution never mentioned d(n) it mentioned d...

9. Originally Posted by 1337h4x
Wait, my problem involved d(n), your solution never mentioned d(n) it mentioned d...
His solution is adequate but look at it like this: I'm going to assume $n$ is not square.

Your product $=\prod_{d|n}d = \prod_{d|n, \; d<\sqrt{n}} d \left(\frac{n}{d}\right) = \prod_{d|n, \; d<\sqrt{n}} n = n^{d(n)/2}$

10. for 1) let $n=\prod_{j=1}^k p_j^{r_j}$ be the prime factorization of $n$. then $d(n)=\prod_{j=1}^k (1+r_j).$ clearly $d(n)$ is odd if and only if $r_j$ is even for all $j,$ i.e. $n$ is a perfect square.

for 2) let $P=\prod_{d \mid n} d.$ then we'll also have $P = \prod_{d \mid n} \frac{n}{d}$ and thus $P^2 = \prod_{d \mid n} d \cdot \frac{n}{d} = \prod_{d \mid n} n = n^{d(n)}.$