The function d(n)

Apr 2009
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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!
 
Last edited:

Bruno J.

MHF Hall of Honor
Jun 2009
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Canada
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! \(\displaystyle d(n)\) what if and only if (...)?

2. Look at the power with which a prime dividing \(\displaystyle n\) appears in the canonical factorization of the resulting product.
 
Apr 2009
96
0
1. That's not even a sentence! \(\displaystyle d(n)\) what if and only if (...)?

2. Look at the power with which a prime dividing \(\displaystyle n\) 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?
 
Apr 2009
96
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Can anyone help with this? I'm struggling to understand this.
 
Apr 2009
96
0
Can anyone please explain the second question at least? I'm really stuck on this concept.
 

PaulRS

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Oct 2007
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For all parts, note that if \(\displaystyle d|n\) then \(\displaystyle \left(\tfrac{n}{d}\right)|n\), the idea would be the following:

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

Thus if \(\displaystyle n\) 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 \(\displaystyle d\cdot \left(\tfrac{n}{d}\right)=n\) -and now try to use it to calculate the product of the divisors-
 
Apr 2009
96
0
For all parts, note that if \(\displaystyle d|n\) then \(\displaystyle \left(\tfrac{n}{d}\right)|n\), the idea would be the following:

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

Thus if \(\displaystyle n\) 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 \(\displaystyle d\cdot \left(\tfrac{n}{d}\right)=n\) -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)....
 
Apr 2009
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0
Wait, my problem involved d(n), your solution never mentioned d(n) it mentioned d...
 

chiph588@

MHF Hall of Honor
Sep 2008
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Champaign, Illinois
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 \(\displaystyle n \) is not square.

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

NonCommAlg

MHF Hall of Honor
May 2008
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for 1) let \(\displaystyle n=\prod_{j=1}^k p_j^{r_j}\) be the prime factorization of \(\displaystyle n\). then \(\displaystyle d(n)=\prod_{j=1}^k (1+r_j).\) clearly \(\displaystyle d(n)\) is odd if and only if \(\displaystyle r_j\) is even for all \(\displaystyle j,\) i.e. \(\displaystyle n\) is a perfect square.

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