MAT 128 OCC, © Barry Dayton 2002


Pythagorean Triples Worksheet

Many people think Pythagoras discovered the "Pythagorean Theorem", or at least provided the first proof of it. Neither of these is true, Pythagoras learned of the theorem, or its converse, from the Babylonians (Iraqi's) and had no reason to prove it, at least not in the modern sense of a proof. Pythagoras was interested in whole numbers (positive integers) and the theorem was named after him because he was the first to find many different triples of whole numbers (a, b, c) with a2 + b2 = c2.

Problem 1 Here is Pythagoras' most famous set of triples where a is an odd number. Find the pattern and complete the chart.

abc b + ca2 b2c2
3 4 5 9916 25
5 12 13 2525144  
7 24 25 49 49    
9 40 41 81      
11 60 61        
13 84          

Most right triangles do not have sides of whole number length or even fraction length, in fact in most right triangles the ratio of the sides is irrational. Not knowing about irrational numbers or decimals Pythagoras had a different way of handling this problem: he allowed a2 + b2 to differ slightly from c2, in particular he accepted a triangle as being a right triangle if these two quantities differed by 1. He then came up with another set of integer triples for isosceles right triangles based on the fact that if a isosceles right triangle has equal sides s and hypotenuse d then the triangle with sides S = s + d, and hypotenuse D= 2s + d is also a right triangle (see picture and note that an isosceles right triangle is a 45° 45° 90° triangle).

Problem 2 Find the pattern and complete the chart below. In the last column give a decimal (calculator) value for d / s.

sds2+ s2 d2d / s
1 1 2 1 1.0
2 3 8 9 1.5
5 7 50 49 1.4
12 17 288   1.41666667
29 41   1681  

Given that the square root of 2 is approximately 1.4142135 what do you think is happening to d / s? What would this ratio be in a "real" right triangle?

Problem 3 (Extra Credit) Working backwards Pythagoras may have realized that a "real" isosceles right triangle (one where a2 + b2 = c2 exactly) could not have integer sides (in modern language "the square root of 2 is irrational"). Fill in the details in the following proof:

Suppose there is an isosceles right triangle with integer sides. Among all such triangles there must be one with smallest sides (why?). Suppose this triangle had side S and hypotenuse D. Then let s = D - S and d = 2S - D, the isosceles triangle with equal sides s and third side d is also an isosceles right triangle (why? give an algebraic or geometric proof). But s,d are also integers and are smaller than S, D so our original supposition could not have been true.