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}.

Ok, enough math for the day. Now go cook some burgers, celebrate your freedoms, and try not to lose any fingers to fireworks!

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Minor nitpick, but does that function actually generate the Pythagorean triples? That is, is the mapping (m, n) |-> (a, b, c) a bijection?

I would have sworn that it was Abe Lincoln. Too bad there isn't some sort of global electronic repository of public knowledge that I could tap to figure that out.

HJ

First counterexample I found is

12^2 + 9^2 = 15^2

we have to treat b as 12 (because be must be even), so that gives us m and n of (1,6) or (2,3). These give us an a^2 of 35 and 5 - neither of which is 9.

Paul:

m=6 and n=1 gives the triple {35,12,37}

http://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple

Viva La Revolution!

(3,472,073)^7 + (4,627,011)^7 = (4,710,868)^7

@7, you need a better calculator:

(3,472,073)^7

= 6083057661870775743244762534472426132278375097

(4,627,011)^7

= 45405004576037486373919670125222681810267716171

Add the above and get:

= 51488062237908262117164432659695107942546091268

(4,710,868)^7

= 51488062237908262117145710842129274774286712832

Difference:

= 18721817565833168259378436

A solution a,b,c is called primitive if a,b,c have no

common divisors. For such a solution exactly one of a or

b is even. Let S be the set of all pairs of positive integers m and n such that m and n are relatively prime,

m is greater than n, and m and n have opposite parities.

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

a and b. Non-primitive solutions are obtained by multiplying a,b,c by a common multiple.

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.

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

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

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. (-;