LOGIN/REGISTER
Please Wait...
It is currently Sep 02, 2010, 11:23 am
Post new topic Reply to topic  [ 3 posts ]  Share: Facebook
Message
Post Posted: May 24, 2007, 5:24 pm • # 1 


Show that for all prime numbers p, Q(p)=\prod^{p-1}_{k=1}k^{2k-p-1} is an integer.
 
 
Post Posted: Jul 21, 2007, 7:25 pm • # 2 


nicetry007 wrote:
For p = 2, it is trivially true. Let p \;\geq\; 3.


Q(p) = \prod_{i = 1}^{p - 1}k^{2k - p - 1} = \left( \frac {\prod_{i = 1}^{p - 1}k^{k}}{\left( (p - 1)! \right)^{(p + 1)/2}}\r...

It is enough to show that the term inside the parenthesis is an integer.

Let q be a prime less than p. The exponent of q in the denominator is given by

\frac {(p + 1)}{2}\left (\lfloor\frac {p - 1}{q}\rfloor + \lfloor\frac {p - 1}{q^{2}}\rfloor + \cdots + \lfloor\frac {p - 1}{...

In the numerator, the terms that are multiples of q are as follows:

q^{q}, (2q)^{2q}, \cdots, \left( \lfloor\frac {p - 1}{q}\rfloor q \right)^{\lfloor\frac {p - 1}{q}\rfloor q}

Factoring out a q from all these terms gives us

q^{(q + 2q + \cdots + \lfloor\frac {p - 1}{q}\rfloor q )} = q^{\frac {q}{2}\lfloor\frac {(p - 1)}{q}\rfloor \left( \lfloor\fr...

In all, there are \lfloor\frac {p - 1}{q}\rfloor multiples of q in the numerator. Of these \lfloor\frac {p - 1}{q^{2}}\rfloor are multiples of q^{2}.

Factoring out a q from these terms gives us

q^{(q^{2} + 2q^{2} + \cdots + \lfloor\frac {p - 1}{q^{2}}\rfloor q^{2})} = q^{\frac {q^{2}}{2}\lfloor\frac {p - 1}{q^{2}}\rfl....

Continuing in this fashion, we can compute the exponent of q in the numerator as follows:

\frac {q}{2}\lfloor\frac {p - 1}{q}\rfloor \left( \lfloor\frac {p - 1}{q}\rfloor + 1 \right) + \frac {q^{2}}{2}\lfloor\frac {....

Comparing the exponent of q in the numerator and the denominator , it suffices to show

\frac {(p + 1)}{2}\left(\lfloor\frac {p - 1}{q^{i}}\rfloor \right) \leq \frac {q^{i}}{2}\lfloor\frac {p - 1}{q^{i}}\rfloor \l....

It is enough to show

(p + 1) \leq q^{i}\left( \lfloor\frac {p - 1}{q^{i}}\rfloor + 1 \right) \;\text{\;(i.e.)\;}\; \frac {(p + 1)}{q^{i}}\leq \lfl....

Suppose q^{i} divides p + 1. Then \lfloor\frac {p - 1}{q^{i}}\rfloor = \frac {(p + 1)}{q^{i}} - 1 (i.e.) \frac {(p + 1)}{q^{i}} = \lfloor\frac {p - 1}{q^{i}}\r....

Suppose q^{i} does not divide p + 1. Then \lfloor\frac {(p + 1)}{q^{i}}\rfloor = \lfloor \frac {p - 1}{q^{i}}\rfloor (i.e.) \frac {(p + 1)}{q^{i}} = \lfloor\frac {p - 1}{q^{i}}\rfloor + f where f is a fraction less than 1.
Hence, \frac {(p + 1)}{q^{i}}\leq \lfloor\frac {p - 1}{q^{i}}\rfloor + 1.

Therefore, Q(p) is an integer. In fact, a perfect square.

_________________
Boo!
 
 
Post Posted: Jan 24, 2010, 9:00 pm • # 3 


Nice problem .

AMM problem always takes my interest . :lol:

Solution :

Fristly , we will recall an well-known identity : \prod_{k = 1}^{n - 1}k! \ = \ \prod_{k = 1}^{n - 1} k^{n - k} \ \ (*) \ \ (n \ge 2)

Indeed , \prod_{k = 1}^{n - 1}k! \ = \ (1 ) . (1.2)....(1.2.....(n - 1))

we can see that in this product , the numbers 1 apprears n - 1 times , the numbers 2 appears n - 2 times ,....,

the numbers n - 1 appears only one time . So , we can deduce that :

\prod_{k = 1}^{n - 1}k! \ = \prod_{k = 1}^{n - 1}(n - k)! \ = \ 1^{n - 1} . 2^{n - 2} \cdots (n - 1)^1 \ = \ \prod_{k = 1}^{n... , Done

Now return to our problem :

We can rewrite \mathbb{Q} (p) as follows :

\mathbb{Q} (p) = \prod_{k = 1}^{p - 1} k^{2k - p - 1} = \frac {\prod_{k = 1}^{p - 1} k^{k - 1} }{ \prod_{k = 1}^{p - 1} k^{p ...

= \frac {\prod_{k = 1}^{p - 1} k^{k - 1} \cdot \prod_{k = 1}^{p - 1} k^{p - k} }{ \left(\prod_{k = 1}^{p - 1} k^{p - k} \righ... ( Using (**) )

= \frac { ((p - 1)!)^{p - 1}}{\prod_{k - 1}^{p - 1} k! (p - k)! } \ = \ \frac {1}{p^{p - 1}} \cdot \frac { (p!)^{p - 1}}{\pro...

So , we can have the beautiful identity : \mathbb{Q} (p) = \prod_{k = 1}^{p - 1} \frac {\binom{p}{k} }{p}

We have known that : if p is a prime , then : p | \binom{p}{k} \ \ \forall \ \ k = \overline{ 1 ; p - 1}

\rightarrow \frac {\binom{p}{k} }{p} \in \mathbb{N} \ \ \forall \ \ k = \overline{ 1 ; p - 1} \rightarrow \mathbb{Q} (p) \in ... , QED

_________________
Love the simplicity
 
 
Display posts from previous:  Sort by  

All times are UTC - 8 hours [ DST ]

Share: Facebook
Post new topic Reply to topic  [ 3 posts ] 

Login

Username:   Password:   Log me on automatically each visit  

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum