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Pythagorean TriplesAlmost everyone knows of the "3-4-5 triangle," one of the right triangles found in every draftsman's toolkit (along with the 45-45-90). This triangle is different from most right triangles because it has three integer edges. Pythagoras' theorem tells us that the squares of the sides of a right triangle sum to give to the square of the hypotenuse: 32 + 42 = 52 I am often asked whether this relationship is unique, or if there are other right triangles with three integer edges as well. When we randomly select
two integers and add their squares, we usually acquire a non-integer square
as a result; thus 32 + 52 = 34, or 42
+ 72 = 65, and so on. Neither 34 nor 65 are integer squares.
This type of result seems to be the general go of things; so the question
posed is not without merit. a2 + b2 = c2 The terms a and b are the sides of the right triangle so that a < c and b < c. Thus, we can subtract either a2 or b2 from both sides of the equation. Let's choose b2: a2 = c2 - b2 Next, from basic algebra we see that we can factor the right-hand side: a2 = (c + b)(c - b) Now let's assume that a, b, and c are integers. Then a2, (c + b), and (c - b) must also be integers. Furthermore, since the product (c + b)(c - b) is equal to an integer square, both (c + b), and (c - b) must be integer squares. Let u and v be integers and set c +
b = 2u2 where u2
> v2. (Why do we place a 2 in front of u2 and
v2? Keep going-the reason should become clear shortly.) Then
a2 = 4u2v2 = (2uv)2, and a
= ± 2uv. Since a is a length, we can, without loss of generality,
choose the + sign and set a = + 2uv. 2b
= 2(u2 - v2) and 2c = 2(u2 + v2) (Now do you see the reason for the 2's? Try the same calculation again without them and evaluate the results as we do below!) With these two relations, and a = 2uv, we can set out to discover as many new right triangles with integer edges as we please. The set of numbers, {a, b, c}, is called a Pythagorean triple. Here are a few examples to start you off. You can find as many more as you please.
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