Multiple Solutions

We have been given two equations:
xyz = 4
and we want to find as many solutions to this system of equations as we can. We could proceed using simple guess and check. We know x, y, and z must all be nonzero, since xyz ≠ 0. So that limits our search (although only slightly). Let's try x = y = 2. We want xyz = 4, so we must have z = 1. Does the point (2,2,1) solve the second equation? Let's see. 3(2) + 2(2)  1 = 9 ≠ 3 So (2,2,1) is not a solution of the system. So let's try x = 1, y = 2. We want xyz = 4, so we must have z = 2. Does the point (1,2,2) solve the second equation? 3(1) + 2(2)  (2) = 3, as desired. So (1,2,2) is a solution of the system. We could continue this process, but it seems pretty slow going. Also, there might be a lot of solutions, and using this process we can never be sure we've found them all. So let's try a new strategy.
Imagine we had been given two twodimensional equations, like
2x  y = 1
One method of finding solutions is to graph both equations and find their points of intersection:
The solutions to the system are the points of intersection, (3,7) and (1,1). We can use a similar strategy to determine some solutions to our threedimensional equations. There are a couple of differences, though. First, since the equations are threedimensional, we will need to graph them in xyz space, rather than in the xy plane. Also, since we have three unknowns but only two equations, we will either get an infinite set of solutions, or none at all. Of course, we have already seen that (1,2,2) is a solution, so we have at least some solutions. Let's graph our two equations:
The two surfaces do indeed intersect, and, as with the twodimensional equations, every point in their intersection is a solution to the system. Let's try to determine a description of the intersection. Note that it is a curve, rather than discrete points. We know 3x + 2y  z = 3, so z = 3x + 2y  3.
So we can substitute this for z into xyz = 4, and we have
Now we can use the Quadratic Formula to solve for y in terms of x:
So our solutions are of the form (x_{0}, y_{0}, z_{0}), where
If we project this onto the xy plane (by setting z = 0), we have the following graph of the x and yvalues of all solutions:
Let's use this information to find some integral solutions to the system.
For our solution to be integral, we need
Using a similar strategy as what we used in the twodimensional example (above), we can graph y = 9x^{4}  18x^{3} + 9x^{2} + 32x and determine its points of intersection with y = 1, y = 4, y = 9, y = 16, etc. I used my TI83 and checked y = n^{2} up to n = 10, and I found that the graph intersects y = 1, y = 9, y = 16, y = 25, y = 36, y = 49, y = 64, and y = 81 at points with irrational xcoordinates. So these will not give us integral solutions to the system.
However, the graph intersects y = 4 at (1,4),
So x = 1 and x = 2 will give us rational (possibly integral) solutions. Let's determine what those solutions are and see if they're truly integral.
x = 1
x = 2
Note that (1,2,2) is the solution we found using guessandcheck (above). But now we've found two more integral solutions:
(1,2), (1,1), (2,2), and (2,0.5) are all points on our projected solution curve:
You can use this Graphing Calculator file to try to locate more points on the projected solution curve. Then you can easily determine the zcoordinate of the solution by finding z = 4 / (xy).
