Beyond Pythagoras
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Introduction
Beyond Pythagoras
I have been asked to investigate Pythagorean triplets where the shortest side is an odd number and all the three sides are positive integers. A pythagorean triple is a set of integers (a,b,c) that specifies the lengths of a right angle triangle a²+b²=c² in which ‘a’ is the shortest side ‘b’ is the middle side and ‘c’ is the hypotenuse.
The first set of triples (3,4,5) which has already been proved to satisfy Pythagoras’s theory.
I have also been given two other pythagorean triples (5,12,13) and (7,24,25) I will now prove these to satisfy Pythagoras’s theory
a² = 5² =25
b² = 12² =144
c² = 13² =169
a²+b² =25+144 =169
a²+b²=c² so Pythagoras’s theory holds for (5,12,13) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 7² =49
b² = 24² =576
c² = 25² =625
a²+b² =49+576 =625
a²+b²=c² so Pythagoras’s theory holds for (7,24,25) because they satisfy the condition of a²+b²=c² in a right angled triangle.
I am now going to put my results in to a table so that I can predict more values:
a | b | c |
3 | 4 | 5 |
5 | 12 | 13 |
7 | 24 | 25 |
I will now predict the next two values in the table so I can work out a general formula for this pythagorean family.
Middle

So now with 5 triplets that I have proved I will now find a general formula to find other pythagorean triplets in this family.
By looking at the table it’s obvious that the formulae for ‘a’ is a=2n+1
I’m now going to find the general formulae for the ‘b’ and ‘c’ value.
(b+1) ² = (2n+1)² + b
b²+2b+1 = 4n²+4n+1+b²
b²+2b-b² = 4n²+4n+1-1
2b = 4n²+4n
b = 2n²+2n
c = 2n²+2n+1
I now have the formulae:
a =2n+1
b = 2n²+2n
c = 2n²+2n+1
I am now going to prove that these formulae work
a² = (2n+1)² )
b² = (2n²+2n)² )
c² = (2n²+2n+1)² =
a² = (2n+1) (2n+1)
4n²+2n+2n+1
4n²+4n+1
b² = (2n²+2n) (2n²+2n)
4n +4n³+4n³+4n²
a²+b² = 4n + 8n³+8n²+2n+1
c² = (2n²+2n+1) (2n²+2n+1)
4n + 4n³+2n²+ 4n³+4n²+2n+2n²+2n+1
4n + 8n³+8n²+2n+1
I will give an example of how to use the formulae for the 10th as the nth term:
a =2n+1 =21
b = 2n²+2n = 220
c = 2n²+2n+1 = 221
the 10th pythagorean triple in this pythagorean family is (21,220,221)
I am now going to investigate the 2nd pythagorean family. I will get this by doubling the values of the first family.
(3,4,5) = (6,8,10)
(5,12,13) = (10,24,26)
(7,24,25) = (14,48,50)
(9,40,41) = (18,80,82)
(11,60,61) = (22,120,122)
Conclusion
c² = 123² =15,129
a²+b² =729+14,400 =15,129
a²+b²=c² so Pythagoras’s theory holds for (27,120,123) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 33² =1089
b² = 180² =32,400
c² = 183² =33,489
a²+b² =441+5184 =5625
a²+b²=c² so Pythagoras’s theory holds for (33,180,183) because they satisfy the condition of a²+b²=c² in a right angled triangle.
I am now going to find a general formula for the 3rd pythagorean family that I will name the b+3 family.
From the values I have I have used the differencing method to find out that a = 6n+3 I will now work out the ‘b’ and ‘c’ values.
(b+3) ² = (6n+3)² + b
b²+6b+1 = 36n²+36n+9+b²
b²+6b-b² = 36n²+36n+9-9
6b = 36n²+36n
b = 6n²+6n
c = 6n²+6n+3
I now have the formulae:
a = 6n+3
b = 6n²+6n
c = 6n²+6n+3
I am now going to prove that these formulae work:
a² = (6n+3)² )
b² = (6n²+6n)² )
c² = (6n²+6n+3)² =
a² = (6n+3) (6n+3)
36n²+18n+18n+9
36n²+36n+9
b² = (6n²+6n) (6n²+6n)
36n +36n³+36n³+36n²
36n +72n³+36n²
a²+b² = 36n +72n³+72n²+36n+9
c² = (6n²+6n+3) (6n²+6n+3)
36n + 36n³+18n²+36n³+36n²+18n+18n²+18n+9
36n + 72n³+82n²+36n+9
I will give an example of how to use the formulae for the 10th as the nth term:
a = 6n+3 =63
b = 6n²+6n = 3660
c = 6n²+6n+3= 3663
the 10th pythagorean triple in this pythagorean family is (21,220,221)
This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.
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