a² = 9² =81
b² = 40² =1600
c² = 41² =1681
a²+b² =81+1600 =1681
a²+b²=c² so Pythagoras’s theory holds for (9,40,41) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 11² =121
b² = 60² =3600
c² = 61² =3721
a²+b² =121+3600 =3721
a²+b²=c² so Pythagoras’s theory holds for (11,60,61) because they satisfy the condition of a²+b²=c² in a right angled triangle.
My table now looks like this:
When looking at the table I found a pattern so I am going to prove if it will work.
3² =9 -4-5 (3,4,5)
5² =25 -12-13 (5,12,13)
7² =49 -24-25 (7,24,25)
9² =81 -40-41 (9,40,41)
11² =121 -60-61 (11,60,61)
a² = b+c
a² = b+b+1
a² + 2b+1
Now to prove if it will work for Pythagoras
a² a²+b² = (b+1) ²
b² a²+b² = b²=2b+1
b+1² a² = 2b + 1
a²-1=b
2
Pythagoras theory is satisfied
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)
Using the internet for research I found out that there are in-between values so these are the pythagorean triples I have for this family now:
(6,8,10)
(8,15,17)
(10,24,26)
(12,35,37)
(14,48,50)
(16,63,65)
(18,80,82)
(20,99,101)
(22,120,122)
I will now prove that Pythagoras theorem holds for some of these pythagorean triplets.
a² = 6² =36
b² = 8² =64
c² = 10² =100
a²+b² =36+64 =100
a²+b²=c² so Pythagoras’s theory holds for (6,8,10) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 8² =64
b² = 15² =225
c² = 17² =289
a²+b² =64+225 =289
a²+b²=c² so Pythagoras’s theory holds for (8,15,17) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 10² =100
b² = 24² =576
c² = 26² =676
a²+b² =100+576 =676
a²+b²=c² so Pythagoras’s theory holds for (10,24,26) 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 2nd pythagorean family that I will name the b+2 family.
From the values I have I know a = 2n+4 by using the differencing method
I will now work out the ‘b’ and ‘c’ values.
(b+2) ² = (2n+4)² + b
b²+4b+1 = 4n²+16n+16+b²
b²+4b-b² = 4n²+16n+16-4
4b = 4n²+16n+12
b = n²+4n+3
c = n²+4n+5
I now have the formulae:
a =2n+4
b = n²+4n+3
c = n²+4n+5
I am now going to prove that these formulae work
a² = (2n+4)² )
b² = (n²+4n+3)² )
c² = (n²+4n+5)² =
a² = (2n+4) (2n+4)
4n²+8n+8n+16
4n²+16n+16
b² = (n²+4n+3) (n²+4n+3)
n +4n³+3n²+4n³+16n²+4n+12n+3n²+12n+9
n +8n³+22n²+24n+9
a²+b² = n + 8n³+26n²+40n+25
c² = (n²+4n+5) (n²+4n+5)
n + 4n³+5n²+ 4n³+16n²+20n+5n²+20n+25
n + 8n³+26n²+40n+25
I will give an example of how to use the formulae for the 10th as the nth term:
a =2n+4 =24
b = n²+4n+3= 143
c = n²+4n+5= 145
the 10th pythagorean triple in this pythagorean family is (21,220,221)
I am now going to investigate the 3rd pythagorean family. I will get this by tripling the values of the first family.
(3,4,5) = (9,12,15)
(5,12,13) = (15,36,39)
(7,24,25) = (21,72,75)
(9,40,41) = (27,120,123)
(11,60,61) = (33,180,183)
I will now prove that Pythagoras theorem holds for these pythagorean triplets.
a² = 9² =81
b² = 12² =144
c² = 15² =100
a²+b² =81+144 =225
a²+b²=c² so Pythagoras’s theory holds for (9,12,15) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 15² =225
b² = 36² =1296
c² = 39² =1521
a²+b² =225+1296 =1521
a²+b²=c² so Pythagoras’s theory holds for (15,36,39) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 21² =441
b² = 72² =5184
c² = 35² =5625
a²+b² =441+5184 =5625
a²+b²=c² so Pythagoras’s theory holds for (21,72,35) because they satisfy the condition of a²+b²=c² in a right angled triangle.
a² = 27² =729
b² = 120² =14,400
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)
