8.3
Application to Pythagorean Triples.
The Simple Identity: (a − b)2+ 4ab = (a + b)2
has been used for many clever applications.
We now show how a clever manipulation of it gives us yet another proof of the circle parameterization.
First we divide it by (a + b)2 to get
(a − b)2
(a + b)2 +
4ab
(a + b)2 = 1.
Replacing a by 1 and b by m2 we see that
1 − m2 1 + m2 2 + 2m 1 + m2 2 = 1.
We now show how it can lead to a new explanation of the parameterization of a circle.
Consider, as above the circle
x2 + y2= r2. Rewrite it after dividing both sides by r2 to get:
x r 2 +y r 2 = 1.
Using our identity, we could simply set: x r = 1 − m2 1 + m2 and y r = 2m 1 + m2
and clearly our circle equation is satisfied! It follows that x = r1 − m 2 1 + m2 , y = r 2m 1 + m2. is a parameterization! 3
We now use the above to discuss: Pythagorean triples.
A Pythagorean triple is a triple of integers x, y, z such that x2+ y2 = z2.
3What is not clear is that every point of the circle corresponds to one and only one value of the
parameter m, provided we make some reasonable assignment for m = ∞.
134 CHAPTER 8. THE CIRCLE
For convenience, one often requires all numbers to be positive. If x, y, z don’t have a common factor, then the triple is said to be primitive.
The same original (simple) identity can also be used to generate lots of examples of Pythagorean triples.
We divide the original identity by 4 and plug in a = s2, b = t2. Thus, we have:
(s2− t2)2
4 + s
2t2 = (s2+ t2)
4 and this simplifies to:
s2− t2 2 2 + (st)2 = s 2+ t2 2 2 .
Then we easily see that
x = s
2− t2
2 , y = st , z =
s2+ t2
2 forms a Pythagorean triple!
What is more exciting is that with a little number theoretic argument, this can be shown to give all possible primitive positive integer solutions (x, y, z). 4
These primitive Pythagorean triples have been rediscovered often, in spite of their knowledge for almost 2000 years. It is apparent that they have been known around the world for a long time.
Try generating some triples by taking suitable values for s, t. (For best results, take odd values s > t without common factors!) For instance, we can take:
s t x y z 3 1 4 3 5 5 1 12 5 13 5 3 8 15 17 7 5 12 35 37 7 3 20 21 29 7 1 24 7 25
What is the moral? A good identity goes a long way!
Optional section: A brief discussion of the General conic.
A conic is a plane curve which is obtained by intersecting a cone with planes in different position. Study of conic sections was an important part of Greek Geometry and several interesting and intriguing properties of conics have been developed over two thousand years.
8.3. APPLICATION TO PYTHAGOREAN TRIPLES. 135
In our algebraic viewpoint, a conic can be described as any degree two curve, i.e. a curve of the form
General conic: ax2+ by2+ 2hxy + 2f x + 2gy + c = 0.
We invite the reader to show that the above idea of parameterizing a circle works just as well for a conic. We simply take a point A(p, q) on the conic and consider a line x = p + t, y = q + mt. The intersection of this line with the conic gives two value of t, namely
t = 0 , t = −2(bqm + f + gm + ap + hpm + hq)
a + 2hm + bm2 .
The reader is invited to carry out this easy but messy calculation as an exercise in algebra.
It is interesting to work out the different types of the conics: circle/ellipse, parabola, hyperbola and a pair of lines. We can work out how the equations can tell us the type of the conic and its properties. But a detailed analysis will take too long. Here we simply illustrate how the idea of parameterization can be worked out for a given equation.
Example of parameterization. Parameterize the parabola y2 = 2x + 2
using the point A(1, 2).
Answer: We shall first change coordinates to bring the point A to the origin. This is not needed, but does simplify our work.
Thus we set x = u + 1, y = v + 2, then the point A(1, 2) gets new coordinates u = 0, v = 0. Our equation becomes:
(v + 2)2= 2(u + 1) + 2 or v2+ 4v = 2u.
Now a line through the origin in (u, v) coordinates is given by u = t, v = mt where m is the slope.
Substitution gives
m2t2+ 4mt = 2t or, by simplifying t(m2t + (4m − 2)) = 0. Thus there are two solutions t = 0, t = 2 − 4m
m2 . We ignore the known solution t = 0
corresponding to our starting point A.
Note that when m = 0, we end up with a single solution t = 0. We use the second solution for the parameterization:
u = t = 2 − 4m
m2 and v = mt =
2m − 4m2
136 CHAPTER 8. THE CIRCLE
When we go back to the original coordinates, we get:
x = 1 +2 − 4m m2 = m2− 4m + 2 m2 and y = 2 + 2m − 4m 2 m2 = 2m − 2m2 m2 .
It is an excellent idea to plug this into the original equation and verify that the equation is satisfied for all values of m. This means that the final equation should be free of m and always true!
Actually, the value m = 0 is a problem, since it appears in the denominator.
However, when plugged into the equation of the curve, the m just cancels! When we take the value m = 0 the other point of intersection of the line simply runs off to infinity. Indeed, this is a situation where a graph is very useful to discover that the line corresponding to m = 0 will be parallel to the axis of the parabola!
m=0 line A(1,2)