Conference rusure::math

Equivalently, we're asked for

	(p1+1)/p1 * (p2+1)/p2 ...

to be an integer (p's are increasing primes).


For this to be an integer, I would expect some sort of canceling on top
and bottom (i.e. divide numerator and denominator by same value) to be
possible.

So, what cancelations are possible, or are definitely not possible ?

Well, let's leave out 2 for a moment so we can say all the p's are odd.
Hence all the p'n + 1 are even and hence none will cancel with any of the p
on the bottom.

What is possible, is that p'j+1 could equal p'k*n.  For example, if 5
is one of the primes and 19 is another, we could have

	(5+1)/5 * ... * (19+1)/19

and now the 5 on the bottom cancels the 4 which divides 19+1.

This leads me to wonder in general what sort of primes divide numbers
of the form p+1.  For example, 19+1 is divisible by 2,5.  37+1 is
divisible by 2,19.  43+1 is divisible by 2,11.  11+1 is divisible by
2,3.  Certainly p+1 is divisible by 2.  What else ?  Let's ignore the 2,
and look at the others.  59+1 is divisible by 3,5.

Is this helping ?

/Eric

>   (a) Find all sequences p1 p2 < ... < pn of distinct prime numbers such
>   that (1+1/p1) (1+1/p2) ... (1+1/pn) is an integer.
        
        As Eric said in .-1, take p1 < p2 < ... < pn and rewrite this
        as
        
        	p1 + 1   p2 + 1         pn + 1
        	------ * ------ * ... * ------
        	  p1       p2             pn
        
        The largest prime in the denominator, pn, can only be
        cancelled by one of the p1 + 1, p2 + 1, etc. terms, and then
        only if pn = pk + 1 for some k.  This only happens when pn = 3
        and pk = 2.  The only solution is is {2,3} with (1 + 1/2)(1 + 1/3)
        = (3/2)(4/3) = 2.
        
        Dan

But other types of cancelation are possible.  I already gave the example of

	5+1         19+1
	--- + ... + ---- + ...
	 5           19


Note that the 5 can be canceled, changing the 19+1 to 4.


So I don't see yet how we can conclude that {2,3} is the only solution.

/Eric

What can cancel the largest denominator, which is prime?

I give up.  What ?

    Re .5:
    
    The point is that nothing cancels the largest prime.  Sure, you can
    cancel smaller primes pi with a larger factor pj+1, but what cancels
    the largest prime?  Nothing.  So you can't do it.
    
    
    				-- edp
    
    
Public key fingerprint:  8e ad 63 61 ba 0c 26 86  32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.

oh well, sorry

>    (b) Can (1+1/a1^2) (1+1/a2^2) ... (1+1/an^2) be an integer where a1,
>    a2, ..., an are distinct integers greater than 1?

	If p is a prime dividing ai, for some i, then p = 2 or p == 1 mod 4.

	Proof: p must divide 1 + aj^2 for some j. ie: -1 is a square mod 4.
This can happen for odd p exactly when p == 1 mod 4.

	So if ai = bi*2^di, where bi is odd, then bi == 1 mod 4. Does this
help us to construct an upper bound on the value of (1+1/a1^2)*...*(1+1/an^2)?
Perhaps we can show that it's less than 2?

Andrew

The answer is no, because the product must be between 1 and 2.

The product is clearly greater than 1, and is less than

  ___       2    ___        2                2
  | | (1+1/n ) < | | exp(1/n ) = exp( sum(1/n ) - 1)
  n>1            n>1                  n>=1

         2
 = exp(pi /6 - 1) = approx. 1.90586.

(b) can be generalized so that ai=1 is allowed, i.e., "distinct integers greater
than 1" can be replaced by "distinct positive integers."

We have just shown that if the product is an integer, it must include some ai=1.
In this case, the product is greater than 2 and less than 3.9, so it must be
exactly 3.  But 3 cannot divide any of the numerators, so this is impossible.

Sorry, I meant except for the obvious case where n=1 and a1=1 is the only ai.
Then, the product is 2, famous for being an integer.  :-)

    > Then, the product is 2, famous for being an integer.  :-)
    
    Famous because it is a rare even prime. That makes it interesting, even.
    
      John

        One might almost say, it makes it odd. :-)
        
        Dan

>    (b) Can (1+1/a1^2) (1+1/a2^2) ... (1+1/an^2) be an integer where a1,
>    a2, ..., an are distinct integers greater than 1?

	Suppose we remove the restriction that the ai be distinct. What
then?

Andrew.

How about just asking for what collections of integer n, greater than 1 and not
necessarily distinct, is the following an integer:

	(n1+1)/n1 * (n2+1)/n2 ...

The generalization being introduced here is we're no longer forcing the n's to be squares.

/Eric