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# Beyond Pythagoras  Paper

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 aï¿½+bï¿½=cï¿½ 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

aï¿½ = 5ï¿½ =25

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bï¿½ = 12ï¿½ =144

cï¿½ = 13ï¿½ =169

aï¿½+bï¿½ =25+144 =169

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (5,12,13) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 7ï¿½ =49

bï¿½ = 24ï¿½ =576

cï¿½ = 25ï¿½ =625

aï¿½+bï¿½ =49+576 =625

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (7,24,25) because they satisfy the condition of aï¿½+bï¿½=cï¿½ 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.

By using the differencing method I can see that ‘a’ has a difference of 2 between each pythagorean triplet I predict the next two ‘a’ values to be 9 and 11

From the table I can also see that there is a quadratic sequence for the ‘b’ value so I predict the next two values to be 40 and 60

Also from my table, I can see the ‘c’ value is b+1 so my two next predictions I’m going to prove are:

a

b

c

9

40

41

11

60

61

aï¿½ = 9ï¿½ =81

bï¿½ = 40ï¿½ =1600

cï¿½ = 41ï¿½ =1681

aï¿½+bï¿½ =81+1600 =1681

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (9,40,41) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 11ï¿½ =121

bï¿½ = 60ï¿½ =3600

cï¿½ = 61ï¿½ =3721

aï¿½+bï¿½ =121+3600 =3721

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (11,60,61) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

My table now looks like this:

a

b

c

3

4

5

5

12

13

7

24

25

9

40

41

11

60

61

When looking at the table I found a pattern so I am going to prove if it will work.

3ï¿½ =9 -4-5 (3,4,5)

5ï¿½ =25 -12-13 (5,12,13)

7ï¿½ =49 -24-25 (7,24,25)

9ï¿½ =81 -40-41 (9,40,41)

11ï¿½ =121 -60-61 (11,60,61)

aï¿½ = b+c

aï¿½ = b+b+1

aï¿½ + 2b+1

Now to prove if it will work for Pythagoras

aï¿½ aï¿½+bï¿½ = (b+1) ï¿½

bï¿½ aï¿½+bï¿½ = bï¿½=2b+1

b+1ï¿½ aï¿½ = 2b + 1

aï¿½-1=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) ï¿½ = (2n+1)ï¿½ + b

bï¿½+2b+1 = 4nï¿½+4n+1+bï¿½

bï¿½+2b-bï¿½ = 4nï¿½+4n+1-1

2b = 4nï¿½+4n

b = 2nï¿½+2n

c = 2nï¿½+2n+1

I now have the formulae:

a =2n+1

b = 2nï¿½+2n

c = 2nï¿½+2n+1

I am now going to prove that these formulae work

aï¿½ = (2n+1)ï¿½ )

bï¿½ = (2nï¿½+2n)ï¿½ )

cï¿½ = (2nï¿½+2n+1)ï¿½ =

aï¿½ = (2n+1) (2n+1)

4nï¿½+2n+2n+1

4nï¿½+4n+1

bï¿½ = (2nï¿½+2n) (2nï¿½+2n)

4n +4nï¿½+4nï¿½+4nï¿½

aï¿½+bï¿½ = 4n + 8nï¿½+8nï¿½+2n+1

cï¿½ = (2nï¿½+2n+1) (2nï¿½+2n+1)

4n + 4nï¿½+2nï¿½+ 4nï¿½+4nï¿½+2n+2nï¿½+2n+1

4n + 8nï¿½+8nï¿½+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 = 2nï¿½+2n = 220

c = 2nï¿½+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.

aï¿½ = 6ï¿½ =36

bï¿½ = 8ï¿½ =64

cï¿½ = 10ï¿½ =100

aï¿½+bï¿½ =36+64 =100

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (6,8,10) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 8ï¿½ =64

bï¿½ = 15ï¿½ =225

cï¿½ = 17ï¿½ =289

aï¿½+bï¿½ =64+225 =289

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (8,15,17) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 10ï¿½ =100

bï¿½ = 24ï¿½ =576

cï¿½ = 26ï¿½ =676

aï¿½+bï¿½ =100+576 =676

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (10,24,26) because they satisfy the condition of aï¿½+bï¿½=cï¿½ 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) ï¿½ = (2n+4)ï¿½ + b

bï¿½+4b+1 = 4nï¿½+16n+16+bï¿½

bï¿½+4b-bï¿½ = 4nï¿½+16n+16-4

4b = 4nï¿½+16n+12

b = nï¿½+4n+3

c = nï¿½+4n+5

I now have the formulae:

a =2n+4

b = nï¿½+4n+3

c = nï¿½+4n+5

I am now going to prove that these formulae work

aï¿½ = (2n+4)ï¿½ )

bï¿½ = (nï¿½+4n+3)ï¿½ )

cï¿½ = (nï¿½+4n+5)ï¿½ =

aï¿½ = (2n+4) (2n+4)

4nï¿½+8n+8n+16

4nï¿½+16n+16

bï¿½ = (nï¿½+4n+3) (nï¿½+4n+3)

n +4nï¿½+3nï¿½+4nï¿½+16nï¿½+4n+12n+3nï¿½+12n+9

n +8nï¿½+22nï¿½+24n+9

aï¿½+bï¿½ = n + 8nï¿½+26nï¿½+40n+25

cï¿½ = (nï¿½+4n+5) (nï¿½+4n+5)

n + 4nï¿½+5nï¿½+ 4nï¿½+16nï¿½+20n+5nï¿½+20n+25

n + 8nï¿½+26nï¿½+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 = nï¿½+4n+3= 143

c = nï¿½+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.

aï¿½ = 9ï¿½ =81

bï¿½ = 12ï¿½ =144

cï¿½ = 15ï¿½ =100

aï¿½+bï¿½ =81+144 =225

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (9,12,15) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 15ï¿½ =225

bï¿½ = 36ï¿½ =1296

cï¿½ = 39ï¿½ =1521

aï¿½+bï¿½ =225+1296 =1521

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (15,36,39) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 21ï¿½ =441

bï¿½ = 72ï¿½ =5184

cï¿½ = 35ï¿½ =5625

aï¿½+bï¿½ =441+5184 =5625

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (21,72,35) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 27ï¿½ =729

bï¿½ = 120ï¿½ =14,400

cï¿½ = 123ï¿½ =15,129

aï¿½+bï¿½ =729+14,400 =15,129

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (27,120,123) because they satisfy the condition of aï¿½+bï¿½=cï¿½ in a right angled triangle.

aï¿½ = 33ï¿½ =1089

bï¿½ = 180ï¿½ =32,400

cï¿½ = 183ï¿½ =33,489

aï¿½+bï¿½ =441+5184 =5625

aï¿½+bï¿½=cï¿½ so Pythagoras’s theory holds for (33,180,183) because they satisfy the condition of aï¿½+bï¿½=cï¿½ 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) ï¿½ = (6n+3)ï¿½ + b

bï¿½+6b+1 = 36nï¿½+36n+9+bï¿½

bï¿½+6b-bï¿½ = 36nï¿½+36n+9-9

6b = 36nï¿½+36n

b = 6nï¿½+6n

c = 6nï¿½+6n+3

I now have the formulae:

a = 6n+3

b = 6nï¿½+6n

c = 6nï¿½+6n+3

I am now going to prove that these formulae work:

aï¿½ = (6n+3)ï¿½ )

bï¿½ = (6nï¿½+6n)ï¿½ )

cï¿½ = (6nï¿½+6n+3)ï¿½ =

aï¿½ = (6n+3) (6n+3)

36nï¿½+18n+18n+9

36nï¿½+36n+9

bï¿½ = (6nï¿½+6n) (6nï¿½+6n)

36n +36nï¿½+36nï¿½+36nï¿½

36n +72nï¿½+36nï¿½

aï¿½+bï¿½ = 36n +72nï¿½+72nï¿½+36n+9

cï¿½ = (6nï¿½+6n+3) (6nï¿½+6n+3)

36n + 36nï¿½+18nï¿½+36nï¿½+36nï¿½+18n+18nï¿½+18n+9

36n + 72nï¿½+82nï¿½+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 = 6nï¿½+6n = 3660

c = 6nï¿½+6n+3= 3663

the 10th pythagorean triple in this pythagorean family is (21,220,221)

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