Pythagorean Triples
Pythagorean Triples Three integers a, b, and c that satisfy a2 + b2 = c2 are
called Pythagorean Triples. There are infinitely many such numbers and there
also exists a way to generate all the triples. Let n and m be integers, n*m.
Then define(*) a = n2 - m2, b = 2nm, c = n2 + m2. The three number a, b, and c
always form a Pythagorean triple. The proof is simple: (n2 - m2)2 + (2mn)2 = n4
- 2n2m2 + m4 + 4n2m2 = n4 + 2n2m2 + m4 = (n2 + m2)2. The formulas were known to
Euclid and used by Diophantus to obtain Pythagorean triples with special
properties. However, he never raised the question whether in this way one can
obtain all possible triples.The fact is that for m and n coprime of different
parities, (*) yields coprime numbers a, b, and c. Conversely, all coprime
triples can indeed be obtained in this manner. All others are multiples of
coprime triples: ka, kb, kc.As an aside, those who mastered the arithmetic of
complex numbers might have noticed that (m + in)2 = (n2 - m2) + i2mn. Which
probably indicates that (*) has a source in trigonometry. But the proof below
only uses simple geometry and algebra.First of all, note that if a2 + b2 = c2,
then (a/c)2 + (b/c)2 = 1. With x = a/c and y = b/c we get x2 + y2 = 1.
This is
the well known equation of the unit circle with center at the origin. Finding
Pythagorean triples is therefore equivalent to locating rational points (i.e.,
points (x,y) for which both x and y are rational) on the unit circle. For if
(p/q)2 + (r/s)2 = 1, multiplication by a common denominator leads to an identity
between integers.Rational numbers approximate irrational to any degree of
accuracy. Therefore, the set of rational pairs is dense in the whole plane. So,
perhaps, one might expect that any curve should contain a lot of rational pairs
or meander wildly to avoid them. But this is not the case. The recent proof of
Fermat's Last Theorem lets us claim that the curves xN + yN = 1 with N*2 contain
no rational points. But there are simpler examples. From Lindemann's theorem, we
conclude that the graph of a perfectly smooth function y = ex contains a single
rational point, (0,1). Moreover, pulling the unit circle even a little aside may
change the picture drastically. Let (xk, yk) = ( 2/k, 3/k), and consider a unit
circle with center at (xk, yk). As k grows, the point approaches the origin, but
for no k, such a circle contains a rational point.Let t be defined by(1) t =
y/(x+1). Then t(x+1) = y andt2(x + 1)2 = y2 = 1 - x2 = (1 + x)(1 - x). We are
not interested in negative x. So let's cancel (1+x) on both sides. The result
ist2(x + 1) = (1 - x). Solving for x we get(2) x = (1 - t2)/(1 + t2) From y =
t(1+x) we also obtain(3) y = 2t/(1 + t2) Formula (1)-(3) show that t is rational
iff both x and y are rational.There is another way to look at the just described
configuration.
