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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

2 + 4n + 4 – 1

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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