I am looking to finalize my solution to the following number theory problem:
Find a rule in terms of n for the number of ways to write integers a and b, with a less than or equal to b, so that $$\mathbf{1/a + 1/b = 1/n}$$ Specifically, I need help finding a rule for even integer values of n.
I began with some simple algebraic manipulation to find that $\mathbf{a = (nb/(b-n))}$ and then graphed the function using x for b and substituting various values of n, then looking for patterns in what values of y (a=y) were integer solutions and less than or equal to b. I first established that positive and negatives of the same integer value of n will always have the same number of ways to write a and b.
--When changing from n to -n the exact values of (b,a) follow a pattern: if (b,a) for n = (x,y), (b,a) for -n = (-y,-x)) For example the solution values of (b,a) for n=4 are:
- (20,5)
- (8,8)
- (3,-12)
- (2-4)
- (12,6)
and the solution values of (b,a) for n=-4 are:
- (-5,-20)
- (-8,-8)
- (12,-3)
- (4,-2)
- (-6,-12)
Knowing this, I chose to strictly test positive integer values of n because I would only have to plug in positive values of x into my table (using the equation $\mathbf{f(x) = nx/(x-n)}$) because any negative value of b that I plugged in would necessarily result in a positive value of a in order for $\mathbf{1/a + 1/b = 1/n}$ to remain true and that would violate the inequality a less than or equal to b.
I found the following rules:
- All $\mathbb {Z}$, excluding $1$ and $-1$, have the following three solutions:
$\mathbf{(b,a) = (n^2+n, n+1)}$
$\mathbf{(b,a) = (2n, 2n)}$
$\mathbf{(b,a) = (n-1, n-n^2)}$
$N$ cannot equal 0
If $\mathbf{abs. value (n)}$ = 1, there will be exactly one solution at $\mathbf{(b,a) = (2n, 2n)}$ The reason that 1 and -1 only have one solution is because neither a nor b can ever equal 0, which eliminates the first and third solutions listed above.
If $\mathbf{GCD ((abs. value n), 2) = 1}$ the three solutions listed above are the only solutions, and so the rule follows that for all odd integer values of $n$ there are 3 ways.
I have not yet found a rule for when $\mathbf{GCD ((abs. value n), 2) = 2}$; that is what I am looking for help with. I collected the following data, I am just failing to see the pattern it presents:
For n = {2,4,6,8,10,12,14,16,18,20}, the number of solutions are {3,5,9,7,9,15,10,10,16,16}, respectively. The set of solutions is therefore also true for {-2,-4,-6,-8,-10,-12,-14,-16,-18,-20}, respectively. I'm hoping someone will see a pattern in this data, or will find a more efficient way to find the number of ways using less trial and error, because my process becomes increasingly more tedious as n grows larger (as I must test all x from 0 to $\mathbf{ n^2+n}$).
Note: for positive $\mathbf{n}, values of $\mathbf{b}, where $\mathbf{a = (nb/(b-n))}$, only exist on the interval $ \mathbf{(0,n^2+n]} $
It may be worth restating that all even integers have the three solutions listed above, as well as some additional ones for which I have not found a rule.
It may also be interesting to note that solutions will occur when the above function is graphed and the function crosses through a lattice point on or above the line $y =x$.