xy
= r may or may not have integer solutions x,
x+y
y. For example, if r = 25
, no solutions exist, while for r = 10
we have at least two solutions, since
4
3
4·20
5·10
10
= 4+20 = 5+10 .
3
xy
Show that given a rational number r 6= 0 the equation
= r can have at most a finite
x+y
number of integer solutions.
Given a rational number r, the equation
Solution by Ngoc Ngo Quang Tran
Let s = 1/r, a rational number. Thus we have 1/x + 1/y = s. If |s| > 2 then there are no
solutions at all. Let s = p/q, gcd(p, q) = 1, p 6= 0. Then there is an at-least-equal number of
solutions for the equation 1/x0 + 1/y 0 = 1/q (let x0 = px; y 0 = py). We need to prove that such
equation has a finite number of solution.
Side question to solve this problem: we now try to find every combination of x, y, z that
satisfies the equation 1/x + 1/y = 1/z. Let d = gcd(x, y). Then x = dm, y = dn, gcd(m, n) = 1;
z = dmn/(m + n). Since z is an integer, (m + n)|d, or d = k(m + n). Thus we have the solution:
x = dm(m + n); y = dn(m + n); z = kmn.
Now, for any given integer q (z in our previous side question), there is an unique prime factorization; thus there is a finite number of combinations of (k, m, n) satisfying q = kmn; giving a finite
number of combinations of (x0 , y 0 ) satisfying 1/x0 + 1/y 0 = 1/q.
Solution by David Stone
The nature of the problem gives us a clue to the solution. It asks us to prove that there can be
at most a finite number of solutions for a given r to the equation r = xy/(x + y), with r rational and
x and y integers. This suggests that we’ll likely be dealing with the concept of ”infinite descent.” Is
there some way that we can show that some aspect of our set of solutions corresponds to counting
down through the natural numbers? Since we can’t count down indefinitely, we would know our
number of solutions is finite.
Let’s assume for the moment that x, y, and r are all positive. It makes the argument much
simpler to deal with only one quadrant, and we can then easily generalize to the additional cases
of the other quadrants of the x, y plane.
First, it’s easy to show that for a given x, only one y can satisfy our equation. If xy(1)/(x +
y(1)) = r = xy(2)/(x + y(2)), then cross-multiplying and canceling shows us that y(1) = y(2).
Now, let’s take our set of x, y solutions for a given r, and order that set by the size of x. Let’s
now take an arbitrary x(1) such thatx(1) < x(2).
Then r = x(1)y(1)/(x(1) + y(1)) = x(2)y(2)/(x(2) + y(2)).
Once again, we cross-multiply and find that
x(1)y(1)x(2) + x(1)y(1)y(2) = x(1)x(2)y(2) + x(2)y(2)y(1).
Since x(1) < x(2), we know that the second term on the left is less than the second term on the
right, so the first term on the left must be greater than the first term on the right. That is,
x(1)y(1)x(2) > x(1)x(2)y(2)
so y(1) > y(2).
We then conclude in our solution set that as x goes up, y goes down. Since all x and y are
positive integers, sooner or later we run out of room and the solution set has to be finite. We can
construct similar arguments for the other quadrants, and conclude that the overall solution set is
finite.
Dave Stone ’91
– Dr. David R. Stone Pickett Professor of Military History Director, Institute for Military History and 20th Century Studies 221 Eisenhower Hall Kansas State University Manhattan, KS 66506
(785) 532-6730 http://www-personal.ksu.edu/ stone/ http://amazon.com/author/davidrstone