# Investigating families of Pythagorean triples.

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Introduction

I was set the task of investigating families of Pythagorean triples. I started off by investigating families of Pythagorean triples where the shortest side is an odd number and c is always equal to b+1. I then extended my research into other families of triples. Finally, I came up with a formula that can calculate all Pythagorean triples.

Pythagoras’s formula is as follows:

a2 + b2 = c2

This is referring to such a triangle:

The triangle always has one right angle and c is always the longest side, also known as the hypotenuse.

An example of this is with the triple 3,4 and 5:

Here, a is 3, b is 4 and c is 5.

52 = 32 + 42

25 = 9 + 16

I will now investigate the Pythagorean triples where c is always equal to b+1. The following table shows the first ten cases, along with the areas and perimeters:

N | a | b | c | P | A | a2 | b2 | c2 |

1 | 3 | 4 | 5 | 12 | 6 | 9 | 16 | 25 |

2 | 5 | 12 | 13 | 30 | 30 | 25 | 144 | 169 |

3 | 7 | 24 | 25 | 56 | 84 | 49 | 576 | 625 |

4 | 9 | 40 | 41 | 90 | 180 | 81 | 1600 | 1681 |

5 | 11 | 60 | 61 | 132 | 330 | 121 | 3600 | 3721 |

6 | 13 | 84 | 85 | 182 | 546 | 169 | 7056 | 7225 |

7 | 15 | 112 | 113 | 240 | 840 | 225 | 12544 | 12769 |

8 | 17 | 144 | 145 | 306 | 1224 | 289 | 20736 | 21025 |

9 | 19 | 180 | 181 | 380 | 1710 | 361 | 32400 | 32761 |

10 | 21 | 220 | 221 | 462 | 2310 | 441 | 48400 | 48841 |

By looking at the numbers for a, I noticed that they are all odd numbers. The basic formula for odd numbers is:

2n + 1

When I applied this formula to the numbers in the table, they turned out correct. As a result,

a = 2n + 1

The difference between the numbers in column b is 8 at first, then 12, then 16, then 20, and so on, adding 4 to the difference each time. Therefore, the second difference is 4. If we divide 4 by 2, we get 2.

Middle

1936

233289

235225

In the values for a, the difference is 4, meaning that the formula starts with:

4n

By looking carefully at the formula and numbers, I noticed that by adding 4 to this formula, I would get the right numbers.

a = 4n + 4

Factorising this, gave me:

4 ( n + 1 )

I found b by halving the difference between the numbers in row b and adding 8n + 3. This gave me:

4n2 + 8n + 3

Or

( 2n + 1 ) ( 2n + 3 )

As c was 2 numbers bigger than b, all I had to do was to add 2 to b’s formula:

4n2 + 8n + 5

When factorised, it became:

( 2n + 1 ) ( 2n + 3 ) + 2

The formulas for the b+2 family are shown below:

a = 4n + 4

b = 4n2 + 8n + 3 or ( 2n + 1 ) ( 2n + 3 )

c = 4n2 + 8n + 5 or ( 2n + 1 ) ( 2n + 3 ) + 2

Once again, I used my previous techniques to show that the left hand side was equal to the right hand side:

a2 + b2 = c2

( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2 = ( 4n2 + 8n + 5 ) 2

Left Hand Side:

( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2

Right Hand Side:

( 4n2 + 8n + 5 ) 2

( a + b ) 2 = ( a + b ) x ( a + b )

= a ( a + b ) + b ( a + b )

= a2 + ab + ab + b2

= a2 + b2 + 2ab

( a + b+ c )2 = ( a + b+ c ) x ( a + b+ c )

= a ( a + b+ c ) + b ( a + b+ c ) + c ( a + b+ c )

= a2 + b2 + c2 + 2ab + 2ac + 2bc

Left Hand Side:

( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2

= 16n2 + 32n + 16 + 16n4 + 64n3 + 88n2 + 48n + 9

= 16n4 + 64n3 + 104n2 + 80n + 25

Right Hand Side:

( 4n2 + 8n + 5 ) 2

= 16n4 + 64n3 + 104n2 + 80n + 25

After I had finished exploring the b+2 triples, I decided to experiment with b+3 triples. I didn’t manage to find any that weren’t multiples. The ones I did find are shown in the table below. All of the numbers can be divided by three to produce the b+1 family:

n | a | b | c | a2 | b2 | c2 |

1 | 9 | 12 | 15 | 81 | 144 | 225 |

2 | 15 | 36 | 39 | 225 | 1296 | 1521 |

3 | 21 | 72 | 75 | 441 | 5184 | 5625 |

4 | 27 | 120 | 123 | 729 | 14400 | 15129 |

5 | 33 | 180 | 183 | 1089 | 32400 | 33489 |

6 | 39 | 252 | 255 | 1521 | 63504 | 65025 |

7 | 45 | 336 | 339 | 2025 | 112896 | 114921 |

8 | 51 | 432 | 435 | 2601 | 186624 | 189225 |

9 | 57 | 540 | 543 | 3249 | 291600 | 294849 |

10 | 63 | 660 | 663 | 3969 | 435600 | 439569 |

I then moved straight on to the b+4 family, but yet again I only found multiples. The same applied for the b+5, b+6 and b+7 families. The table for b+4 is shown below. The numbers can all be divided by 4 to give the b+1 family:

n | a | b | c | a2 | b2 | c2 |

1 | 12 | 16 | 20 | 144 | 256 | 400 |

2 | 20 | 48 | 52 | 400 | 2304 | 2704 |

3 | 28 | 96 | 100 | 784 | 9216 | 10000 |

4 | 36 | 160 | 164 | 1296 | 25600 | 26896 |

5 | 44 | 240 | 244 | 1936 | 57600 | 59536 |

6 | 52 | 336 | 340 | 2704 | 112896 | 115600 |

7 | 60 | 448 | 452 | 3600 | 200704 | 204304 |

8 | 68 | 576 | 580 | 4624 | 331776 | 336400 |

9 | 76 | 720 | 724 | 5776 | 518400 | 524176 |

10 | 84 | 880 | 884 | 7056 | 774400 | 781456 |

Conclusion

a = ( n + 2 )2 – 1

This was the same as the formula for a in the b+1 family, the only difference was that it was (n+2) instead of (n+1) and 1 instead of n. I then went on to b:

b = 2n + 4

b = 2 ( n + 2 )

b = 2 x 1 x ( n + 2 )

Once again, the formula was the same as the formula for b in the b+1 family, the only difference was that it was (n+2) instead of (n+1) and 1 instead of n. I finally tried c:

c = n2 + 4n + 5

c = n2 + 4n + 4 + 1

c = ( n + 2 )2 + 1

I then came up with the following formula for the b+2 family:

y = 1

z = ( n + 2 )

I then generalised the formulae for the two families above and I got:

a = z2 – y2

b = 2zy

c = z2 + y2

This method only works for numbers where z is bigger than y and both z and y are positive integers. Z must be bigger than y so that we avoid getting any negative numbers. To prove this method works, I then tried to solve the equation algebraically:

a2 + b2 = c2

( z2 – y2 )2 + ( 2zy )2 = ( z2 + y2 )2

Left Hand Side:

( z2 – y2 )2 + ( 2zy )2

Right Hand Side:

( z2 + y2 )2

Left Hand Side:

( z2 – y2 )2 + ( 2zy )2

= z4 + y4 – 2z2y2 + 4z2n2

=

Right Hand Side:

( z2 + y2 )2

= 2z2y2 + z4 + y4

As a result, the general formula for finding any Pythagorean triple, as long as z is bigger than y, and they are both positive integers is:

a = z2 – y2

b = 2zy

c = z2 + y2

This student written piece of work is one of many that can be found in our GCSE Pythagorean Triples section.

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