Thoh's half baked beam
Quote:
confession: i don't know the answer
Listen you
.. if I ever get hold of you I'm gonna rip your head off and s
no wait;
"i was waiting for mark"
Well, in that case pull up a seat and have a beer. I am waiting for Pythagoras:
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
(n + im)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 (except of course for the trivial ones - (0, 1) and (1, 0).) 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 and
t2(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 is
t2(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)
Formulas (1)-(3) show that t is rational iff both x and y are rational.
What more needs to be said read it and weep.
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and laughing 
