Conference rusure::math

    Eric:
    
    I'm missing something.  If
    
    		a = c + 1
    		b = c * (c + 1)
    
    then
    
    	1/a + 1/b = 1/(c+1) + 1/(c*(c+1))
    		  = c/(c*(c+1)) + 1/(c*(c+1))
    		  = (c+1)/(c*(c+1))
    		  = 1/c
    
    John

    That's pretty clever.
    
    Actually, I've got an idea for a puzzle, but I'm not sure whether it's
    a "good" puzzle or not.
    
    Here's the idea:
    
    	I've taken the reciprocal of five positive integers, and added
    	them together, and got a sum.  Like this:
    
    		1/a + 1/b + 1/c + 1/d + 1/e = 296573/1590680
    
    	What are a,b,c,d,e ?
    
    What sorts of things would make it a "good" puzzle?  Well, one nice
    thing would be if we could easily pick enough integers to make the
    problem unfeasible for the trivial exhaustive search by computer.
    
    And of course another property would be if it's not too easy to figure
    out the answer without a computer.
    
    And another nice property might be if there aren't too many answers,
    although this might not be important, since if there are many answers
    but finding even one of them is hard, it might be a nice puzzle.
                                                                    
    /Eric

     re .2     1/a + 1/b + 1/c + 1/d + 1/e = 296573/1590680
     
     The least common denominator of a,b,c,d,e is a most likely
     small multiple of the denominator on the right hand side. 
     So factor 1590680 = 23 * 19 * 13 * 7 * 5 * 2 * 2 * 2.
     Trial and error within LISP shows that taking a=13 and b=35
     leaves 1/c + 1/d + 1/e = 283/3496 = 283/(23 * 19 * 2 * 2 * 2).
     Eventually I got c = 19, d = 23 * 2 = 46, e = 19 * 8 = 152,
     1/13 + 1/35 + 1/19 + 1/46 + 1/152 = 296573/1590680.
     
     re .1     1/c = 1/(c+1) + 1/(c(c+1))
     
     That's good.  I searched for all (a,b,c) such that
     1/a + 1/b = 1/c with a,b,c integers and 1 <= b < a <= 500.
     Many of the c, 1 <= c <= 500 that didn't show up were
     primes, starting with 23.  Since 19 * 20 < 500 the smaller
     primes were represented, but 23 * 24 > 500 so many [all?] of
     the larger ones weren't.  This is similar to what Eric
     stated in .0, so I would conclude that he did something
     similar.
     
     Dan

    Fascinating.  My mileage was different.  I started with:
    
    	1/13 + 1/24 + 1/35 + 1/46 + 1/57 = 296573/1590680
    
    You got:
    
    	1/13 + 1/35 + 1/19 + 1/46 + 1/152 = 296573/1590680
    
    Hence we now know
    
    	1/24 + 1/57 = 1/19 + 1/152
    
    This brings up some questions:
    
    o	Given a fraction, how many ways can it be formed as a sum
    	two reciprocals of integers ?  What sorts of fractions
    	can only be formed in one way as the sum of two reciprocals.
    
    	(Again, ignore the trivial case of 1/(2a) + 1/(2a) = 1/a)
    
    o	Can we find fractions that are the sum of five reciprocals in
    	only one way, and hence have a "hard" puzzle ?
    
    /Eric

The example
    	1/24 + 1/57 = 1/19 + 1/152
is the special case {p=3, q=8, r=19} of the following:

If {p,q} are integers >0 and r=q(p-1)+p, then dividing by pqr 
gives
	1/pq + 1/pr = 1/r + 1/rq

Taking this as a basis, you should be able to develop some other general 
formulas.

Lynn Yarbrough 

    I've been trying to work on this fraction stuff without much luck.
    
    For example, I want to know how to find the complete set of
    sums of two reciprocals of whole numbers that add to a given fraction.
    
    For example, consider
    
    	1/a  +  1/b  =  5/12
    
    By inspection, we see
    
    	1/3 + 1/12 = 5/12
    
    and
    
    	1/6 + 1/4 = 5/12
    
    But how do we get the full set ?
    
    I tried
    
    	1/a + 1/b = 5/12
    
    which I can write as
    
    	(a+b)/(ab) = 5/12
    
    I can cross-multiply and say
    
    	12(a+b) = 5ab
    
    How do we get full set of solutions from this ?
    
    I'm still interested in the more general question of whether
    we can find fractions that have only ONE way of being decomposed
    into a sum of two reciprocals.
    
    /Eric

     re .6
     
     Consider your example of 1/a + 1/b = 5/12 = p/q
     
     Take a and b such that 0 < a <= b.  Write b in the form b = ax,
     where obviously x >= 1.
     
     Then from (a + b)/ab = p/q or q(a + b) = pab you get
     
          q(a + ax) = pa(ax)
     
                        2
          qa(1 + x) = pa x
     
     Now a is not zero so we can divide through by a to get
     
          q(1 + x) = pax
     
          q + qx = pax
     
          q = (pa - q)x
     
          x = q / (pa - q)
     
     Now we know that x >= 1, so we need only consider a such
     that
     
          0 < pa - q <= q
     
          q < pa <= 2q
     
          q/p < a <= 2q/p
     
     So for integral a in the given range, compute x and see if b = ax
     is an integer.  This should generate all of the possible
     solutions.
     
     Dan

     All that .-1 really says is that if 1/a + 1/b = p/q and all
     of the variables are positive integers, then the larger
     fraction on the left must be at least half the one on the
     right, but still be less than the one on the right.
     
     Dan