I'm trying to solve the problem of how many solutions (x,y) there are to:
for a fixed positive integer n and positive integers x and y.
I've been told that this equates to finding the # of divisors of minus the # of duplicate solutions, which in total equals (where is the divisor function).
My question is, what is the reasoning for being the number of duplicate solutions? Where does this come from?
Thanks a lot!
My understanding is that the solutions are the values of x and y that make 1/x + 1/y = 1/n true, and the "duplicates" are values that appear more than once if you just consider all the divisors of .
The way I got to this was that 1/x + 1/y = 1/n => n(x+y) = xy => (x-n)(y-n) = . So the solutions are all the divisors of , -- is that right? would you approach this in a different way?
I think that's a very nice and promising way to approach it. About "duplicate" I think Chip's idea may be right. For example,
for . We have here
the solutions , and these are, perhaps, called "duplicate". I'd rather call the solutions "different up to order of x,y" ,
but as long as the idea is clear...
In the above example, we've the solutions , 5 different
(up to order) solutions, but I am not able to apply the "duplicate" thingy a priori, i.e. before I know
the solutions...
Tonio
The solutions in your example before removing the "duplicates" - there are 9 of them, which does match the number of divisors of 36. So that makes sense.
If you were asked the question "For n a fixed pos. integer, how many solutions (x,y) are there to the equation 1/x + 1/y = 1/n?", would you think you were meant to consider every solution or just the solutions up to order? Because if it is up to order, I really don't know how derive that number of "duplicate" solutions in the general case. :/
If I've understood your formula correctly, I don't think is the number of duplicate solutions.
From your analysis, you'll notice that to get and , you have to get two factors of . We will call these factor pairs and , which come from the divisors of .
Now, you should note that from the list of divisors, we can pair them up to get the desired product, except for one case, which is when . So, if we add one to the number of divisors of and divide it by two, you'll get the number of solutions of that satisfy your conditions.
In other words, the solution is
You'll notice that this is essentially the same as your formula
The corresponding interpretation of your formula is that you started with the number of divisors, then you removed the corresponding partner from the paired factors