Sunday Function

Happy birthday to the United States! It's one of the younger countries on the planet and yet has still managed to have one of the oldest continuous systems of government. Not too shabby. Here's hoping our current wobbles get straightened out and our next few Independence Days occur under more pleasant conditions.

I'm not sure what function might fit very well in the context of this holiday, but I'll make sort of a stretch and do a function about the Pythagorean theorem. The Pythagorean theorem is a very ancient discovery with numerous different methods of proof. One of them was discovered by James Garfield, who would later become a president of the United States. I don't think many of our 44 presidents have made independent nontrivial mathematical discoveries, so good for him. Here's a figure from his proof:


The Pythagorean theorem states that for the sides of a right triangle A, B, and C, the identity A^2 + B^2 = C^2 will be satisfied. You may be familiar with the Scarecrow's statement of the theorem in the Wizard of Oz, though he actually butchers it somewhat and what he says isn't actually true.

For certain side lengths, you'll find that the lengths can be specified with pure integers. For example, the triple {3, 4, 5} is a so-called Pythagorean triple because 3^2 + 4+2 = 5^2. Go ahead, try it out. There's an infinite number of these triples, and they can be generated with our Sunday Function:


Given any two integers m and n (with m > n), and this formula will generate three numbers a, b, and c which are Pythagorean triples. For instance, with n = 1776 and m = 2010, you'll get the triple {885924, 7139520, 7194276}.



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First counterexample I found is
12^2 + 9^2 = 15^2



Viva La Revolution!

@7, you need a better calculator:

= 6083057661870775743244762534472426132278375097

= 45405004576037486373919670125222681810267716171

Add the above and get:
= 51488062237908262117164432659695107942546091268

= 51488062237908262117145710842129274774286712832

= 18721817565833168259378436

By Andrew G. (not verified) on 05 Jul 2010 #permalink

common divisors. For such a solution exactly one of a or
Then the formulae give a bijective correspondence between
such pairs and all primitive solutions with b even.
The other primitive solutions are obtained by interchanging
Here only solutions in positive integers are considered.
Obviously if negative integers or zero are allowed there
are other solutions.
The solution mentioned by Paul Murray is non-primitive.

By Annonymous (not verified) on 05 Jul 2010 #permalink

I should have said "common factor" instead of "common multiple"

By Annonymous (not verified) on 05 Jul 2010 #permalink


By Annonymous (not verified) on 05 Jul 2010 #permalink

Hey Matt! I just checked SB's traffic stats for the past 6 months and found that you and me have pretty much the same kind of figures. We're bang in the middle of the distribution. We define normal around here. (-;