Download Pythagorean triples from fractions

Pythagorean triples from fractions
Investigation one
1. Choose any pair of consecutive even numbers.
2. Find the sum of their reciprocals, simplifying fractions where appropriate.
3. The numerator and the denominator give the first two numbers in your Pythagorean
triple. Substitute these numbers into Pythagoras’ Theorem to find the third number.
E.g.
1. Use 4 and 6.
2. Sum of reciprocals:
3. Pythagoras’ Theorem:
1 1
5
 
4 6 12
5 2  12 2  169
5 2  12 2  13 2
5, 12, 13 is a Pythagorean triple.
Choose other consecutive even numbers to find five more Pythagorean triples. Investigate
whether this method also works with odd numbers.
Does the method still work if you don’t use the simplest possible fraction?
Investigation two
1.
2.
3.
4.
Choose any pair of fractions that have a product of 2.
Add 2 to each fraction.
Divide one fraction by the other.
The numerator and the denominator give the first two numbers in your Pythagorean
triple. Substitute these numbers into Pythagoras’ Theorem to find the third number.
E.g.
1. Use
8
1
and .
4
1
2. Add 2 to each:
10
9
and
4
1
3. Divide:
9 10
9


4 1
40
4. Pythagoras’ Theorem:
9 2  40 2  1681
9 2  40 2  412
9, 40, 41 is a Pythagorean triple.
How many other fractions can you find which have a product of 2? Which Pythagorean triples do
they produce?
© www.teachitmaths.co.uk 2012
17628
Page 1 of 2
Pythagorean triples from fractions
Teaching notes
A nice idea to extend these two investigations is to ask students to construct the triangles for
the Pythagorean triples they have found. These could be used to create posters.
Extension idea
The method shown in investigation one works for consecutive odd numbers as well as
consecutive even numbers. An interesting extension is to ask students to give general solutions
for consecutive odd and consecutive even numbers, and therefore come up with further
examples of Pythagorean triples.
General solution for odd numbers (easier):
Consecutive odds:
(2n  1),(2n  1)
Sum of reciprocals:
1
1
(2n  1)  (2n  1)


2n  1 2n  1 (2n  1)(2n  1)
4n

4n 2  1
(4 n) 2  (4 n 2  1) 2  16 n 2  16 n 4  8n 2  1
 16 n 4  8n 2  1
Pythagoras’ Theorem:
 (4 n 2  1) 2
So 4 n,(4 n 2  1),(4 n 2  1) is a Pythagorean triple.
General solution for even numbers (harder)
Consecutive evens:
2n,(2n  2)
Sum of reciprocals:
1
1
(2n  2)  2n


2n 2n  2
2n(2n  2)
4n  2

4n 2  4n
2n  2

2 n 2  2n
(2n  1) 2  (2n 2  2n) 2  4 n 2  4 n  1  4 n 4  8n 3  4 n 2
 4 n 4  8n 3  8n 2  4  1
Pythagoras’ Theorem:
 (2n 2  2n  1) 2
So (2n  1),(2n 2  2n),(2n 2  2n  1) is a Pythagorean triple.
© www.teachitmaths.co.uk 2012
17628
Page 2 of 2